script in C is not adding Floats correctly












1















This is my first time posting here so this post might be a little bit of a mess but I'll try my best to explain everything.



I have to make a program that acts kinda like Self-checkout in a store.
However i run into two issues when people are inserting money into the machine.
People insert money like this:
10 20 50 0.10 .... and the paying ends with 0 or by using ctrl+d.
(people can only use coins of this value: 100, 50, 20, 10, 5, 2, 1, 0.50, 0.20, 0.10, 0.05, 0.02 a 0.01 as you can see in the code)



Well when I end the payment with 0 program exits normally.(like this: 50 20 0)



However when I end it with ctrl+d it causes infinite loop and i don't understand why.



And the second issue is that for some reason it won't add numbers like 0.10, 0.20 and so on. Any ideas how to fix this or what might be causing the error?



And dont mind the printf in the middle that was just me checking the value.



float bill,x,payment=0,k=0;



printf("coins ");
while(k==0)
{
scanf("%f", &x);
if(x==0 )
{
goto END;
}

if(x ==100 || x ==50 || x ==20 || x ==10 || x ==5 || x ==2 || x ==1 || x ==0.50 || x==0.10 || x ==0.20 || x ==0.05 || x ==0.02|| x ==0.01 )
{
payment += x;
printf("==============");
printf("%.2f n",payment);
}

else{
printf("%.2f is invalid",x);
k = 1;
goto END2;
}
}
END:
printf("%.2f n", payment);

END2:
return 0;









share|improve this question




















  • 3





    You should check the return value of scanf, expecially if you want to "end it with ctrl+d".

    – Bob__
    Nov 24 '18 at 22:17











  • I wanted to point out the same as @Bob__ did cplusplus.com/reference/cstdio/scanf the scanf function can fail and you do not consider it.

    – petrch
    Nov 24 '18 at 22:18








  • 2





    You would be better off keeping the money in pennies rather than using floating point for fractions.

    – Retired Ninja
    Nov 24 '18 at 22:38











  • Yeah, It could be said, that never use floating point numbers for money (unless you intentionally want rounding errors to add up to visible amounts...).

    – hyde
    Nov 24 '18 at 22:40











  • @hyde: First, there are decimal floating-point formats in both hardware (including Intel and IBM) and software (including Microsoft) that are perfectly fine for doing the kind of currency calculations shown in this question. So the advice never to use floating-point for money is wrong on that count. Second, floating-point formats, whether binary or decimal, are appropriate for working on financial calculations such as stock market option evaluation, properly designed mortgage calculations, or even calculating sales taxes when done correctly.

    – Eric Postpischil
    Nov 25 '18 at 0:30


















1















This is my first time posting here so this post might be a little bit of a mess but I'll try my best to explain everything.



I have to make a program that acts kinda like Self-checkout in a store.
However i run into two issues when people are inserting money into the machine.
People insert money like this:
10 20 50 0.10 .... and the paying ends with 0 or by using ctrl+d.
(people can only use coins of this value: 100, 50, 20, 10, 5, 2, 1, 0.50, 0.20, 0.10, 0.05, 0.02 a 0.01 as you can see in the code)



Well when I end the payment with 0 program exits normally.(like this: 50 20 0)



However when I end it with ctrl+d it causes infinite loop and i don't understand why.



And the second issue is that for some reason it won't add numbers like 0.10, 0.20 and so on. Any ideas how to fix this or what might be causing the error?



And dont mind the printf in the middle that was just me checking the value.



float bill,x,payment=0,k=0;



printf("coins ");
while(k==0)
{
scanf("%f", &x);
if(x==0 )
{
goto END;
}

if(x ==100 || x ==50 || x ==20 || x ==10 || x ==5 || x ==2 || x ==1 || x ==0.50 || x==0.10 || x ==0.20 || x ==0.05 || x ==0.02|| x ==0.01 )
{
payment += x;
printf("==============");
printf("%.2f n",payment);
}

else{
printf("%.2f is invalid",x);
k = 1;
goto END2;
}
}
END:
printf("%.2f n", payment);

END2:
return 0;









share|improve this question




















  • 3





    You should check the return value of scanf, expecially if you want to "end it with ctrl+d".

    – Bob__
    Nov 24 '18 at 22:17











  • I wanted to point out the same as @Bob__ did cplusplus.com/reference/cstdio/scanf the scanf function can fail and you do not consider it.

    – petrch
    Nov 24 '18 at 22:18








  • 2





    You would be better off keeping the money in pennies rather than using floating point for fractions.

    – Retired Ninja
    Nov 24 '18 at 22:38











  • Yeah, It could be said, that never use floating point numbers for money (unless you intentionally want rounding errors to add up to visible amounts...).

    – hyde
    Nov 24 '18 at 22:40











  • @hyde: First, there are decimal floating-point formats in both hardware (including Intel and IBM) and software (including Microsoft) that are perfectly fine for doing the kind of currency calculations shown in this question. So the advice never to use floating-point for money is wrong on that count. Second, floating-point formats, whether binary or decimal, are appropriate for working on financial calculations such as stock market option evaluation, properly designed mortgage calculations, or even calculating sales taxes when done correctly.

    – Eric Postpischil
    Nov 25 '18 at 0:30
















1












1








1








This is my first time posting here so this post might be a little bit of a mess but I'll try my best to explain everything.



I have to make a program that acts kinda like Self-checkout in a store.
However i run into two issues when people are inserting money into the machine.
People insert money like this:
10 20 50 0.10 .... and the paying ends with 0 or by using ctrl+d.
(people can only use coins of this value: 100, 50, 20, 10, 5, 2, 1, 0.50, 0.20, 0.10, 0.05, 0.02 a 0.01 as you can see in the code)



Well when I end the payment with 0 program exits normally.(like this: 50 20 0)



However when I end it with ctrl+d it causes infinite loop and i don't understand why.



And the second issue is that for some reason it won't add numbers like 0.10, 0.20 and so on. Any ideas how to fix this or what might be causing the error?



And dont mind the printf in the middle that was just me checking the value.



float bill,x,payment=0,k=0;



printf("coins ");
while(k==0)
{
scanf("%f", &x);
if(x==0 )
{
goto END;
}

if(x ==100 || x ==50 || x ==20 || x ==10 || x ==5 || x ==2 || x ==1 || x ==0.50 || x==0.10 || x ==0.20 || x ==0.05 || x ==0.02|| x ==0.01 )
{
payment += x;
printf("==============");
printf("%.2f n",payment);
}

else{
printf("%.2f is invalid",x);
k = 1;
goto END2;
}
}
END:
printf("%.2f n", payment);

END2:
return 0;









share|improve this question
















This is my first time posting here so this post might be a little bit of a mess but I'll try my best to explain everything.



I have to make a program that acts kinda like Self-checkout in a store.
However i run into two issues when people are inserting money into the machine.
People insert money like this:
10 20 50 0.10 .... and the paying ends with 0 or by using ctrl+d.
(people can only use coins of this value: 100, 50, 20, 10, 5, 2, 1, 0.50, 0.20, 0.10, 0.05, 0.02 a 0.01 as you can see in the code)



Well when I end the payment with 0 program exits normally.(like this: 50 20 0)



However when I end it with ctrl+d it causes infinite loop and i don't understand why.



And the second issue is that for some reason it won't add numbers like 0.10, 0.20 and so on. Any ideas how to fix this or what might be causing the error?



And dont mind the printf in the middle that was just me checking the value.



float bill,x,payment=0,k=0;



printf("coins ");
while(k==0)
{
scanf("%f", &x);
if(x==0 )
{
goto END;
}

if(x ==100 || x ==50 || x ==20 || x ==10 || x ==5 || x ==2 || x ==1 || x ==0.50 || x==0.10 || x ==0.20 || x ==0.05 || x ==0.02|| x ==0.01 )
{
payment += x;
printf("==============");
printf("%.2f n",payment);
}

else{
printf("%.2f is invalid",x);
k = 1;
goto END2;
}
}
END:
printf("%.2f n", payment);

END2:
return 0;






c while-loop floating-point






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 1 '18 at 0:37







Thomas

















asked Nov 24 '18 at 22:08









ThomasThomas

83




83








  • 3





    You should check the return value of scanf, expecially if you want to "end it with ctrl+d".

    – Bob__
    Nov 24 '18 at 22:17











  • I wanted to point out the same as @Bob__ did cplusplus.com/reference/cstdio/scanf the scanf function can fail and you do not consider it.

    – petrch
    Nov 24 '18 at 22:18








  • 2





    You would be better off keeping the money in pennies rather than using floating point for fractions.

    – Retired Ninja
    Nov 24 '18 at 22:38











  • Yeah, It could be said, that never use floating point numbers for money (unless you intentionally want rounding errors to add up to visible amounts...).

    – hyde
    Nov 24 '18 at 22:40











  • @hyde: First, there are decimal floating-point formats in both hardware (including Intel and IBM) and software (including Microsoft) that are perfectly fine for doing the kind of currency calculations shown in this question. So the advice never to use floating-point for money is wrong on that count. Second, floating-point formats, whether binary or decimal, are appropriate for working on financial calculations such as stock market option evaluation, properly designed mortgage calculations, or even calculating sales taxes when done correctly.

    – Eric Postpischil
    Nov 25 '18 at 0:30
















  • 3





    You should check the return value of scanf, expecially if you want to "end it with ctrl+d".

    – Bob__
    Nov 24 '18 at 22:17











  • I wanted to point out the same as @Bob__ did cplusplus.com/reference/cstdio/scanf the scanf function can fail and you do not consider it.

    – petrch
    Nov 24 '18 at 22:18








  • 2





    You would be better off keeping the money in pennies rather than using floating point for fractions.

    – Retired Ninja
    Nov 24 '18 at 22:38











  • Yeah, It could be said, that never use floating point numbers for money (unless you intentionally want rounding errors to add up to visible amounts...).

    – hyde
    Nov 24 '18 at 22:40











  • @hyde: First, there are decimal floating-point formats in both hardware (including Intel and IBM) and software (including Microsoft) that are perfectly fine for doing the kind of currency calculations shown in this question. So the advice never to use floating-point for money is wrong on that count. Second, floating-point formats, whether binary or decimal, are appropriate for working on financial calculations such as stock market option evaluation, properly designed mortgage calculations, or even calculating sales taxes when done correctly.

    – Eric Postpischil
    Nov 25 '18 at 0:30










3




3





You should check the return value of scanf, expecially if you want to "end it with ctrl+d".

– Bob__
Nov 24 '18 at 22:17





You should check the return value of scanf, expecially if you want to "end it with ctrl+d".

– Bob__
Nov 24 '18 at 22:17













I wanted to point out the same as @Bob__ did cplusplus.com/reference/cstdio/scanf the scanf function can fail and you do not consider it.

– petrch
Nov 24 '18 at 22:18







I wanted to point out the same as @Bob__ did cplusplus.com/reference/cstdio/scanf the scanf function can fail and you do not consider it.

– petrch
Nov 24 '18 at 22:18






2




2





You would be better off keeping the money in pennies rather than using floating point for fractions.

– Retired Ninja
Nov 24 '18 at 22:38





You would be better off keeping the money in pennies rather than using floating point for fractions.

– Retired Ninja
Nov 24 '18 at 22:38













Yeah, It could be said, that never use floating point numbers for money (unless you intentionally want rounding errors to add up to visible amounts...).

– hyde
Nov 24 '18 at 22:40





Yeah, It could be said, that never use floating point numbers for money (unless you intentionally want rounding errors to add up to visible amounts...).

– hyde
Nov 24 '18 at 22:40













@hyde: First, there are decimal floating-point formats in both hardware (including Intel and IBM) and software (including Microsoft) that are perfectly fine for doing the kind of currency calculations shown in this question. So the advice never to use floating-point for money is wrong on that count. Second, floating-point formats, whether binary or decimal, are appropriate for working on financial calculations such as stock market option evaluation, properly designed mortgage calculations, or even calculating sales taxes when done correctly.

– Eric Postpischil
Nov 25 '18 at 0:30







@hyde: First, there are decimal floating-point formats in both hardware (including Intel and IBM) and software (including Microsoft) that are perfectly fine for doing the kind of currency calculations shown in this question. So the advice never to use floating-point for money is wrong on that count. Second, floating-point formats, whether binary or decimal, are appropriate for working on financial calculations such as stock market option evaluation, properly designed mortgage calculations, or even calculating sales taxes when done correctly.

– Eric Postpischil
Nov 25 '18 at 0:30














3 Answers
3






active

oldest

votes


















2














When you send Ctrl+D in your terminal, you register an EOF character to stdin, therefore scanf won't read anything anymore, and x will never be set to 0, so you won't escape your loop.



You would need to check the result of scanf, reading man scanf you can see that you are expecting a return value > 0 in case of success.



if (scanf("%f", &x) < 1 || x = 0) // Check return value of scanf, then check x
GOTO: END;


Edit:
Note that if scanf was reading input items, you would need to check that the return value of scanf is not less that the number of input items.






share|improve this answer


























  • This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

    – Neil Edelman
    Nov 24 '18 at 22:32











  • Thanks for the answer. It works perfectly.

    – Thomas
    Nov 25 '18 at 0:25



















1














Fixing this answer based on hyde's and alamit's comments



The reason your program "refuses" to add small numbers like 0.10 is because your variable x and the decimal literals like 0.10 you are comparing it to have different precision.



So, to fix the problem, add f to the values x == 0.10f, or switch to using double and reading and printing with %lf



If you want to know the math behind the problem, read this.






share|improve this answer





















  • 1





    It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

    – hyde
    Nov 24 '18 at 22:39













  • @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

    – Lev M.
    Nov 24 '18 at 22:46






  • 1





    The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

    – alamit
    Nov 24 '18 at 22:49











  • @alamit thanks, I forgot C defaults to double for decimal literals.

    – Lev M.
    Nov 24 '18 at 22:54













  • If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

    – hyde
    Nov 24 '18 at 22:56





















1














When making comparisons with floating point numbers, one must be careful. Floating point representation of 0.1 is different between 32 bits and 64 bits(float vs double). You have defined x to be float which is compared with 0.1(which is double). So the comparison will not work correctly. Either you must define x as double and change nothing else in your code or, make x == 0.1 to x == 0.1f



Further information about this topic can be found this answer and this topic






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    3 Answers
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    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    When you send Ctrl+D in your terminal, you register an EOF character to stdin, therefore scanf won't read anything anymore, and x will never be set to 0, so you won't escape your loop.



    You would need to check the result of scanf, reading man scanf you can see that you are expecting a return value > 0 in case of success.



    if (scanf("%f", &x) < 1 || x = 0) // Check return value of scanf, then check x
    GOTO: END;


    Edit:
    Note that if scanf was reading input items, you would need to check that the return value of scanf is not less that the number of input items.






    share|improve this answer


























    • This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

      – Neil Edelman
      Nov 24 '18 at 22:32











    • Thanks for the answer. It works perfectly.

      – Thomas
      Nov 25 '18 at 0:25
















    2














    When you send Ctrl+D in your terminal, you register an EOF character to stdin, therefore scanf won't read anything anymore, and x will never be set to 0, so you won't escape your loop.



    You would need to check the result of scanf, reading man scanf you can see that you are expecting a return value > 0 in case of success.



    if (scanf("%f", &x) < 1 || x = 0) // Check return value of scanf, then check x
    GOTO: END;


    Edit:
    Note that if scanf was reading input items, you would need to check that the return value of scanf is not less that the number of input items.






    share|improve this answer


























    • This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

      – Neil Edelman
      Nov 24 '18 at 22:32











    • Thanks for the answer. It works perfectly.

      – Thomas
      Nov 25 '18 at 0:25














    2












    2








    2







    When you send Ctrl+D in your terminal, you register an EOF character to stdin, therefore scanf won't read anything anymore, and x will never be set to 0, so you won't escape your loop.



    You would need to check the result of scanf, reading man scanf you can see that you are expecting a return value > 0 in case of success.



    if (scanf("%f", &x) < 1 || x = 0) // Check return value of scanf, then check x
    GOTO: END;


    Edit:
    Note that if scanf was reading input items, you would need to check that the return value of scanf is not less that the number of input items.






    share|improve this answer















    When you send Ctrl+D in your terminal, you register an EOF character to stdin, therefore scanf won't read anything anymore, and x will never be set to 0, so you won't escape your loop.



    You would need to check the result of scanf, reading man scanf you can see that you are expecting a return value > 0 in case of success.



    if (scanf("%f", &x) < 1 || x = 0) // Check return value of scanf, then check x
    GOTO: END;


    Edit:
    Note that if scanf was reading input items, you would need to check that the return value of scanf is not less that the number of input items.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 24 '18 at 22:38

























    answered Nov 24 '18 at 22:27









    alamitalamit

    36719




    36719













    • This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

      – Neil Edelman
      Nov 24 '18 at 22:32











    • Thanks for the answer. It works perfectly.

      – Thomas
      Nov 25 '18 at 0:25



















    • This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

      – Neil Edelman
      Nov 24 '18 at 22:32











    • Thanks for the answer. It works perfectly.

      – Thomas
      Nov 25 '18 at 0:25

















    This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

    – Neil Edelman
    Nov 24 '18 at 22:32





    This will also work if you have a read error; you could really get fancy by checking if errno is set on END and perror.

    – Neil Edelman
    Nov 24 '18 at 22:32













    Thanks for the answer. It works perfectly.

    – Thomas
    Nov 25 '18 at 0:25





    Thanks for the answer. It works perfectly.

    – Thomas
    Nov 25 '18 at 0:25













    1














    Fixing this answer based on hyde's and alamit's comments



    The reason your program "refuses" to add small numbers like 0.10 is because your variable x and the decimal literals like 0.10 you are comparing it to have different precision.



    So, to fix the problem, add f to the values x == 0.10f, or switch to using double and reading and printing with %lf



    If you want to know the math behind the problem, read this.






    share|improve this answer





















    • 1





      It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

      – hyde
      Nov 24 '18 at 22:39













    • @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

      – Lev M.
      Nov 24 '18 at 22:46






    • 1





      The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

      – alamit
      Nov 24 '18 at 22:49











    • @alamit thanks, I forgot C defaults to double for decimal literals.

      – Lev M.
      Nov 24 '18 at 22:54













    • If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

      – hyde
      Nov 24 '18 at 22:56


















    1














    Fixing this answer based on hyde's and alamit's comments



    The reason your program "refuses" to add small numbers like 0.10 is because your variable x and the decimal literals like 0.10 you are comparing it to have different precision.



    So, to fix the problem, add f to the values x == 0.10f, or switch to using double and reading and printing with %lf



    If you want to know the math behind the problem, read this.






    share|improve this answer





















    • 1





      It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

      – hyde
      Nov 24 '18 at 22:39













    • @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

      – Lev M.
      Nov 24 '18 at 22:46






    • 1





      The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

      – alamit
      Nov 24 '18 at 22:49











    • @alamit thanks, I forgot C defaults to double for decimal literals.

      – Lev M.
      Nov 24 '18 at 22:54













    • If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

      – hyde
      Nov 24 '18 at 22:56
















    1












    1








    1







    Fixing this answer based on hyde's and alamit's comments



    The reason your program "refuses" to add small numbers like 0.10 is because your variable x and the decimal literals like 0.10 you are comparing it to have different precision.



    So, to fix the problem, add f to the values x == 0.10f, or switch to using double and reading and printing with %lf



    If you want to know the math behind the problem, read this.






    share|improve this answer















    Fixing this answer based on hyde's and alamit's comments



    The reason your program "refuses" to add small numbers like 0.10 is because your variable x and the decimal literals like 0.10 you are comparing it to have different precision.



    So, to fix the problem, add f to the values x == 0.10f, or switch to using double and reading and printing with %lf



    If you want to know the math behind the problem, read this.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 24 '18 at 23:04

























    answered Nov 24 '18 at 22:36









    Lev M.Lev M.

    628112




    628112








    • 1





      It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

      – hyde
      Nov 24 '18 at 22:39













    • @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

      – Lev M.
      Nov 24 '18 at 22:46






    • 1





      The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

      – alamit
      Nov 24 '18 at 22:49











    • @alamit thanks, I forgot C defaults to double for decimal literals.

      – Lev M.
      Nov 24 '18 at 22:54













    • If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

      – hyde
      Nov 24 '18 at 22:56
















    • 1





      It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

      – hyde
      Nov 24 '18 at 22:39













    • @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

      – Lev M.
      Nov 24 '18 at 22:46






    • 1





      The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

      – alamit
      Nov 24 '18 at 22:49











    • @alamit thanks, I forgot C defaults to double for decimal literals.

      – Lev M.
      Nov 24 '18 at 22:54













    • If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

      – hyde
      Nov 24 '18 at 22:56










    1




    1





    It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

    – hyde
    Nov 24 '18 at 22:39







    It's not about precision of float. 0.1 can't be accurately represented with binary "decimal", same way 1/3 can not be represented with base 10 decimals (it becomes 0.33...). Using double will only seem to help, but doesn't for real.

    – hyde
    Nov 24 '18 at 22:39















    @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

    – Lev M.
    Nov 24 '18 at 22:46





    @hyde So why does it work with double? I know ranged comparison < > is recommended for floating point, but it seems like an overkill for this exercise. I compiled and ran the code with double and it passes the if condition. Do you know why?

    – Lev M.
    Nov 24 '18 at 22:46




    1




    1





    The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

    – alamit
    Nov 24 '18 at 22:49





    The issue comes from comparing float with double, which doesn't have the same precision, and therefore won't store the same result for a repeating binary number such as 0.1. To compare with a float literal OP should use 0.1f for example.

    – alamit
    Nov 24 '18 at 22:49













    @alamit thanks, I forgot C defaults to double for decimal literals.

    – Lev M.
    Nov 24 '18 at 22:54







    @alamit thanks, I forgot C defaults to double for decimal literals.

    – Lev M.
    Nov 24 '18 at 22:54















    If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

    – hyde
    Nov 24 '18 at 22:56







    If you have matching types (either both float or both double), then 0.1 parsed as that will equal 0.1 parsed as that. But it won't be equal to 0.1 decimal, ever (because neither float or double has a bit pattern which would be 0.1 decimal). You can change the equality comparison to use float by comparing to 1.0f, if you don't want to change the variable type to double, and it would "work" again (for suitably defined values of "work").

    – hyde
    Nov 24 '18 at 22:56













    1














    When making comparisons with floating point numbers, one must be careful. Floating point representation of 0.1 is different between 32 bits and 64 bits(float vs double). You have defined x to be float which is compared with 0.1(which is double). So the comparison will not work correctly. Either you must define x as double and change nothing else in your code or, make x == 0.1 to x == 0.1f



    Further information about this topic can be found this answer and this topic






    share|improve this answer




























      1














      When making comparisons with floating point numbers, one must be careful. Floating point representation of 0.1 is different between 32 bits and 64 bits(float vs double). You have defined x to be float which is compared with 0.1(which is double). So the comparison will not work correctly. Either you must define x as double and change nothing else in your code or, make x == 0.1 to x == 0.1f



      Further information about this topic can be found this answer and this topic






      share|improve this answer


























        1












        1








        1







        When making comparisons with floating point numbers, one must be careful. Floating point representation of 0.1 is different between 32 bits and 64 bits(float vs double). You have defined x to be float which is compared with 0.1(which is double). So the comparison will not work correctly. Either you must define x as double and change nothing else in your code or, make x == 0.1 to x == 0.1f



        Further information about this topic can be found this answer and this topic






        share|improve this answer













        When making comparisons with floating point numbers, one must be careful. Floating point representation of 0.1 is different between 32 bits and 64 bits(float vs double). You have defined x to be float which is compared with 0.1(which is double). So the comparison will not work correctly. Either you must define x as double and change nothing else in your code or, make x == 0.1 to x == 0.1f



        Further information about this topic can be found this answer and this topic







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 24 '18 at 23:07









        Talip Tolga SarıTalip Tolga Sarı

        836




        836






























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