c programming, memory lost after reading file

This is part of my c code, the program is to read a file of keywords separated by lines and store them into an array of string.....

int keyno = 4;

char *keywords[keyno];

char var[100];

i=0;

while(fgets(var, sizeof(var), k)!=NULL){   //k is the file

    printf("var: %s", var);

    if(i>0)

        keywords[i-1]=var;

    printf("keyword: %s", keywords[i-1]);

    i++;

}



for (i=0; i<keyno;i++)

    printf("keyword: %s", keywords[i]);

result:

var:

keyword: (null)

var: abc

keyword: abc

var: def

keyword: def

var: ghi

keyword:ghi

var: jkl

keyword: jkl

var:

keyword: @-g(?)



keyword: @-g(?) keyword: @-g(?) keyword: @-g(?) keyword: @-g(?)

why the keywords are gone in the forloop...? what line should i add?

asked Nov 24 '18 at 6:14

ycsjose

The pointers of keywords char pointer array pointing to same variable var. Hence, you are getting the output as last value of var. Make keywords 2D array and copy the var to keywords array.

– H.S.
Nov 24 '18 at 6:21

After you now have been given some hints, you might show what you tried (Minimal, Complete, and Verifiable example) and update the code in your question.

– Swordfish
Nov 24 '18 at 6:44

add a comment |

This is part of my c code, the program is to read a file of keywords separated by lines and store them into an array of string.....

int keyno = 4;

char *keywords[keyno];

char var[100];

i=0;

while(fgets(var, sizeof(var), k)!=NULL){   //k is the file

    printf("var: %s", var);

    if(i>0)

        keywords[i-1]=var;

    printf("keyword: %s", keywords[i-1]);

    i++;

}



for (i=0; i<keyno;i++)

    printf("keyword: %s", keywords[i]);

result:

var:

keyword: (null)

var: abc

keyword: abc

var: def

keyword: def

var: ghi

keyword:ghi

var: jkl

keyword: jkl

var:

keyword: @-g(?)



keyword: @-g(?) keyword: @-g(?) keyword: @-g(?) keyword: @-g(?)

why the keywords are gone in the forloop...? what line should i add?

asked Nov 24 '18 at 6:14

ycsjose

The pointers of keywords char pointer array pointing to same variable var. Hence, you are getting the output as last value of var. Make keywords 2D array and copy the var to keywords array.

– H.S.
Nov 24 '18 at 6:21

After you now have been given some hints, you might show what you tried (Minimal, Complete, and Verifiable example) and update the code in your question.

– Swordfish
Nov 24 '18 at 6:44

add a comment |

This is part of my c code, the program is to read a file of keywords separated by lines and store them into an array of string.....

int keyno = 4;

char *keywords[keyno];

char var[100];

i=0;

while(fgets(var, sizeof(var), k)!=NULL){   //k is the file

    printf("var: %s", var);

    if(i>0)

        keywords[i-1]=var;

    printf("keyword: %s", keywords[i-1]);

    i++;

}



for (i=0; i<keyno;i++)

    printf("keyword: %s", keywords[i]);

result:

var:

keyword: (null)

var: abc

keyword: abc

var: def

keyword: def

var: ghi

keyword:ghi

var: jkl

keyword: jkl

var:

keyword: @-g(?)



keyword: @-g(?) keyword: @-g(?) keyword: @-g(?) keyword: @-g(?)

why the keywords are gone in the forloop...? what line should i add?

asked Nov 24 '18 at 6:14

ycsjose

This is part of my c code, the program is to read a file of keywords separated by lines and store them into an array of string.....

int keyno = 4;

char *keywords[keyno];

char var[100];

i=0;

while(fgets(var, sizeof(var), k)!=NULL){   //k is the file

    printf("var: %s", var);

    if(i>0)

        keywords[i-1]=var;

    printf("keyword: %s", keywords[i-1]);

    i++;

}



for (i=0; i<keyno;i++)

    printf("keyword: %s", keywords[i]);

result:

var:

keyword: (null)

var: abc

keyword: abc

var: def

keyword: def

var: ghi

keyword:ghi

var: jkl

keyword: jkl

var:

keyword: @-g(?)



keyword: @-g(?) keyword: @-g(?) keyword: @-g(?) keyword: @-g(?)

why the keywords are gone in the forloop...? what line should i add?

asked Nov 24 '18 at 6:14

ycsjose

asked Nov 24 '18 at 6:14

ycsjose

asked Nov 24 '18 at 6:14

ycsjose

asked Nov 24 '18 at 6:14

ycsjose

asked Nov 24 '18 at 6:14

ycsjose

The pointers of keywords char pointer array pointing to same variable var. Hence, you are getting the output as last value of var. Make keywords 2D array and copy the var to keywords array.

– H.S.
Nov 24 '18 at 6:21

After you now have been given some hints, you might show what you tried (Minimal, Complete, and Verifiable example) and update the code in your question.

– Swordfish
Nov 24 '18 at 6:44

add a comment |

The pointers of keywords char pointer array pointing to same variable var. Hence, you are getting the output as last value of var. Make keywords 2D array and copy the var to keywords array.

– H.S.
Nov 24 '18 at 6:21

After you now have been given some hints, you might show what you tried (Minimal, Complete, and Verifiable example) and update the code in your question.

– Swordfish
Nov 24 '18 at 6:44

The pointers of keywords char pointer array pointing to same variable var. Hence, you are getting the output as last value of var. Make keywords 2D array and copy the var to keywords array.

– H.S.
Nov 24 '18 at 6:21

After you now have been given some hints, you might show what you tried (Minimal, Complete, and Verifiable example) and update the code in your question.

– Swordfish
Nov 24 '18 at 6:44

add a comment |

3 Answers
3

active

oldest

votes

char *keywords[keyno];

is an array of pointers.

In the loop at this line

keywords[i-1]=var;

you make all pointers in keywords point to var. So when you print keywords you actually just print var multiple times.

Instead you could do

char keywords[keyno][100];

and copy like:

strcpy(keywords[i-1], var);

edited Nov 24 '18 at 7:04

answered Nov 24 '18 at 6:23

4386427

21.1k31845

The key point is that each of the keyword pointers ends up pointing at the same space for the string, which can only hold the last value stored in it. You're right that a 2D array of characters is one to deal with. Using POSIX function strdup() is also a way to handle it — though the memory would need to be freed too. You can easily implement strdup() if it isn't available: char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }.

– Jonathan Leffler
Nov 24 '18 at 6:33

@ycsjose: I recommend reading up on how to create an MCVE (Minimal, Complete, and Verifiable example). There are all sorts of ways you could be running into compilation errors or warnings, or perhaps runtime errors if you ignore the compile time warnings. We can't begin to guess which clever and non-obvious tricks you've used to fool the compiler — novice programmers are far too clever to be easily predicted by us.

– Jonathan Leffler
Nov 24 '18 at 6:39

add a comment |

When you declare an array like this

char* keywords[keyno];

You are declaring an array of pointers to strings. However the pointers need to point somewhere
and each pointer needs to point to a different location.

keywors[i-1]=var;

makes each pointer point to the same address, the address of the array var

to fix this, allocate memory when you get something from the file.

e.g.

keywords[i-1] = malloc(strlen(var)+1); /* allocate memory */

strcpy(keywords[i-1], var);  /* copy to the newly allocated memory */

later you need to free that memory

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

as a precaution make sure all pointers are set to NULL at the beginning in your program since free on an uninitialized pointer is undefined behavior but freeing a NULL pointer is fine.

char* keywords[key];

for (int i = 0; i < keyno; ++i)

  keywords[i] = NULL;

...

while(fgets(var, sizeof(var), k)!=NULL){  ... }

...

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

answered Nov 24 '18 at 7:19

Anders

30.2k64872

add a comment |

-2

Simply put, var is just a pointer (it's actually an array, but the difference is not relevant here). It means that:

keywords[i-1]=var assigns all array entries to this pointer (so they store the same value).

After var becomes something strange (which happens when fgets returns null), all entries also become strange.

A simple solution is to declare keywords as char keyword[num][len], IIRC, and read directly inside keyword[i].

edited Nov 24 '18 at 6:50

answered Nov 24 '18 at 6:24

dyukha

582511

2

No; var is defined as char var[100]; so it is not a pointer. It easily becomes a pointer, but it is an array, not a pointer.

– Jonathan Leffler
Nov 24 '18 at 6:30

Well, this is why I said "simply put". The difference is irrelevant here.

– dyukha
Nov 24 '18 at 6:32

@dyukha Why not just do it right(tm)?

– Swordfish
Nov 24 '18 at 6:33

1

Misleading people who are struggling with pointers and arrays is not, IMO, helpful. All else apart, if var were just a pointer, you'd have to worry about the storage it points out. At least with an array, you don't have to deal with that detail. That is a crucial and major difference between a pointer and an array.

– Jonathan Leffler
Nov 24 '18 at 6:34

1

@dyukha Because it's irrelevant – No, spreading misinformation (knowingly!!) is hardly irrelevant but a crime against every poor soul that happens to read it.

– Swordfish
Nov 24 '18 at 6:38

|
show 8 more comments

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3 Answers
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active

oldest

votes

3 Answers
3

active

oldest

votes

char *keywords[keyno];

is an array of pointers.

In the loop at this line

keywords[i-1]=var;

you make all pointers in keywords point to var. So when you print keywords you actually just print var multiple times.

Instead you could do

char keywords[keyno][100];

and copy like:

strcpy(keywords[i-1], var);

edited Nov 24 '18 at 7:04

answered Nov 24 '18 at 6:23

4386427

21.1k31845

The key point is that each of the keyword pointers ends up pointing at the same space for the string, which can only hold the last value stored in it. You're right that a 2D array of characters is one to deal with. Using POSIX function strdup() is also a way to handle it — though the memory would need to be freed too. You can easily implement strdup() if it isn't available: char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }.

– Jonathan Leffler
Nov 24 '18 at 6:33

@ycsjose: I recommend reading up on how to create an MCVE (Minimal, Complete, and Verifiable example). There are all sorts of ways you could be running into compilation errors or warnings, or perhaps runtime errors if you ignore the compile time warnings. We can't begin to guess which clever and non-obvious tricks you've used to fool the compiler — novice programmers are far too clever to be easily predicted by us.

– Jonathan Leffler
Nov 24 '18 at 6:39

add a comment |

char *keywords[keyno];

is an array of pointers.

In the loop at this line

keywords[i-1]=var;

you make all pointers in keywords point to var. So when you print keywords you actually just print var multiple times.

Instead you could do

char keywords[keyno][100];

and copy like:

strcpy(keywords[i-1], var);

edited Nov 24 '18 at 7:04

answered Nov 24 '18 at 6:23

4386427

21.1k31845

The key point is that each of the keyword pointers ends up pointing at the same space for the string, which can only hold the last value stored in it. You're right that a 2D array of characters is one to deal with. Using POSIX function strdup() is also a way to handle it — though the memory would need to be freed too. You can easily implement strdup() if it isn't available: char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }.

– Jonathan Leffler
Nov 24 '18 at 6:33

@ycsjose: I recommend reading up on how to create an MCVE (Minimal, Complete, and Verifiable example). There are all sorts of ways you could be running into compilation errors or warnings, or perhaps runtime errors if you ignore the compile time warnings. We can't begin to guess which clever and non-obvious tricks you've used to fool the compiler — novice programmers are far too clever to be easily predicted by us.

– Jonathan Leffler
Nov 24 '18 at 6:39

add a comment |

char *keywords[keyno];

is an array of pointers.

In the loop at this line

keywords[i-1]=var;

you make all pointers in keywords point to var. So when you print keywords you actually just print var multiple times.

Instead you could do

char keywords[keyno][100];

and copy like:

strcpy(keywords[i-1], var);

edited Nov 24 '18 at 7:04

answered Nov 24 '18 at 6:23

4386427

21.1k31845

char *keywords[keyno];

is an array of pointers.

In the loop at this line

keywords[i-1]=var;

you make all pointers in keywords point to var. So when you print keywords you actually just print var multiple times.

Instead you could do

char keywords[keyno][100];

and copy like:

strcpy(keywords[i-1], var);

edited Nov 24 '18 at 7:04

answered Nov 24 '18 at 6:23

4386427

21.1k31845

edited Nov 24 '18 at 7:04

answered Nov 24 '18 at 6:23

4386427

21.1k31845

answered Nov 24 '18 at 6:23

4386427

21.1k31845

answered Nov 24 '18 at 6:23

4386427

21.1k31845

The key point is that each of the keyword pointers ends up pointing at the same space for the string, which can only hold the last value stored in it. You're right that a 2D array of characters is one to deal with. Using POSIX function strdup() is also a way to handle it — though the memory would need to be freed too. You can easily implement strdup() if it isn't available: char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }.

– Jonathan Leffler
Nov 24 '18 at 6:33

@ycsjose: I recommend reading up on how to create an MCVE (Minimal, Complete, and Verifiable example). There are all sorts of ways you could be running into compilation errors or warnings, or perhaps runtime errors if you ignore the compile time warnings. We can't begin to guess which clever and non-obvious tricks you've used to fool the compiler — novice programmers are far too clever to be easily predicted by us.

– Jonathan Leffler
Nov 24 '18 at 6:39

add a comment |

The key point is that each of the keyword pointers ends up pointing at the same space for the string, which can only hold the last value stored in it. You're right that a 2D array of characters is one to deal with. Using POSIX function strdup() is also a way to handle it — though the memory would need to be freed too. You can easily implement strdup() if it isn't available: char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }.

– Jonathan Leffler
Nov 24 '18 at 6:33

@ycsjose: I recommend reading up on how to create an MCVE (Minimal, Complete, and Verifiable example). There are all sorts of ways you could be running into compilation errors or warnings, or perhaps runtime errors if you ignore the compile time warnings. We can't begin to guess which clever and non-obvious tricks you've used to fool the compiler — novice programmers are far too clever to be easily predicted by us.

– Jonathan Leffler
Nov 24 '18 at 6:39

The key point is that each of the keyword pointers ends up pointing at the same space for the string, which can only hold the last value stored in it. You're right that a 2D array of characters is one to deal with. Using POSIX function strdup() is also a way to handle it — though the memory would need to be freed too. You can easily implement strdup() if it isn't available:

char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }

.

– Jonathan Leffler
Nov 24 '18 at 6:33

char *strdup(const char *str) { size_t len = strlen(str) + 1; char *dup = malloc(len); if (dup != 0) memmove(dup, str, len); return dup; }

.

– Jonathan Leffler
Nov 24 '18 at 6:33

@ycsjose: I recommend reading up on how to create an MCVE (Minimal, Complete, and Verifiable example). There are all sorts of ways you could be running into compilation errors or warnings, or perhaps runtime errors if you ignore the compile time warnings. We can't begin to guess which clever and non-obvious tricks you've used to fool the compiler — novice programmers are far too clever to be easily predicted by us.

– Jonathan Leffler
Nov 24 '18 at 6:39

add a comment |

When you declare an array like this

char* keywords[keyno];

You are declaring an array of pointers to strings. However the pointers need to point somewhere
and each pointer needs to point to a different location.

keywors[i-1]=var;

makes each pointer point to the same address, the address of the array var

to fix this, allocate memory when you get something from the file.

e.g.

keywords[i-1] = malloc(strlen(var)+1); /* allocate memory */

strcpy(keywords[i-1], var);  /* copy to the newly allocated memory */

later you need to free that memory

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

as a precaution make sure all pointers are set to NULL at the beginning in your program since free on an uninitialized pointer is undefined behavior but freeing a NULL pointer is fine.

char* keywords[key];

for (int i = 0; i < keyno; ++i)

  keywords[i] = NULL;

...

while(fgets(var, sizeof(var), k)!=NULL){  ... }

...

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

answered Nov 24 '18 at 7:19

Anders

30.2k64872

add a comment |

When you declare an array like this

char* keywords[keyno];

You are declaring an array of pointers to strings. However the pointers need to point somewhere
and each pointer needs to point to a different location.

keywors[i-1]=var;

makes each pointer point to the same address, the address of the array var

to fix this, allocate memory when you get something from the file.

e.g.

keywords[i-1] = malloc(strlen(var)+1); /* allocate memory */

strcpy(keywords[i-1], var);  /* copy to the newly allocated memory */

later you need to free that memory

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

as a precaution make sure all pointers are set to NULL at the beginning in your program since free on an uninitialized pointer is undefined behavior but freeing a NULL pointer is fine.

char* keywords[key];

for (int i = 0; i < keyno; ++i)

  keywords[i] = NULL;

...

while(fgets(var, sizeof(var), k)!=NULL){  ... }

...

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

answered Nov 24 '18 at 7:19

Anders

30.2k64872

add a comment |

When you declare an array like this

char* keywords[keyno];

You are declaring an array of pointers to strings. However the pointers need to point somewhere
and each pointer needs to point to a different location.

keywors[i-1]=var;

makes each pointer point to the same address, the address of the array var

to fix this, allocate memory when you get something from the file.

e.g.

keywords[i-1] = malloc(strlen(var)+1); /* allocate memory */

strcpy(keywords[i-1], var);  /* copy to the newly allocated memory */

later you need to free that memory

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

as a precaution make sure all pointers are set to NULL at the beginning in your program since free on an uninitialized pointer is undefined behavior but freeing a NULL pointer is fine.

char* keywords[key];

for (int i = 0; i < keyno; ++i)

  keywords[i] = NULL;

...

while(fgets(var, sizeof(var), k)!=NULL){  ... }

...

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

answered Nov 24 '18 at 7:19

Anders

30.2k64872

When you declare an array like this

char* keywords[keyno];

You are declaring an array of pointers to strings. However the pointers need to point somewhere
and each pointer needs to point to a different location.

keywors[i-1]=var;

makes each pointer point to the same address, the address of the array var

to fix this, allocate memory when you get something from the file.

e.g.

keywords[i-1] = malloc(strlen(var)+1); /* allocate memory */

strcpy(keywords[i-1], var);  /* copy to the newly allocated memory */

later you need to free that memory

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

as a precaution make sure all pointers are set to NULL at the beginning in your program since free on an uninitialized pointer is undefined behavior but freeing a NULL pointer is fine.

char* keywords[key];

for (int i = 0; i < keyno; ++i)

  keywords[i] = NULL;

...

while(fgets(var, sizeof(var), k)!=NULL){  ... }

...

for (int i = 0; i < keyno; ++i)    

  free(keywords[i[);

answered Nov 24 '18 at 7:19

Anders

30.2k64872

answered Nov 24 '18 at 7:19

Anders

30.2k64872

answered Nov 24 '18 at 7:19

Anders

30.2k64872

answered Nov 24 '18 at 7:19

Anders

30.2k64872

add a comment |

-2

Simply put, var is just a pointer (it's actually an array, but the difference is not relevant here). It means that:

keywords[i-1]=var assigns all array entries to this pointer (so they store the same value).

After var becomes something strange (which happens when fgets returns null), all entries also become strange.

A simple solution is to declare keywords as char keyword[num][len], IIRC, and read directly inside keyword[i].

edited Nov 24 '18 at 6:50

answered Nov 24 '18 at 6:24

dyukha

582511

2

No; var is defined as char var[100]; so it is not a pointer. It easily becomes a pointer, but it is an array, not a pointer.

– Jonathan Leffler
Nov 24 '18 at 6:30

Well, this is why I said "simply put". The difference is irrelevant here.

– dyukha
Nov 24 '18 at 6:32

@dyukha Why not just do it right(tm)?

– Swordfish
Nov 24 '18 at 6:33

1

Misleading people who are struggling with pointers and arrays is not, IMO, helpful. All else apart, if var were just a pointer, you'd have to worry about the storage it points out. At least with an array, you don't have to deal with that detail. That is a crucial and major difference between a pointer and an array.

– Jonathan Leffler
Nov 24 '18 at 6:34

1

@dyukha Because it's irrelevant – No, spreading misinformation (knowingly!!) is hardly irrelevant but a crime against every poor soul that happens to read it.

– Swordfish
Nov 24 '18 at 6:38

|
show 8 more comments

-2

Simply put, var is just a pointer (it's actually an array, but the difference is not relevant here). It means that:

keywords[i-1]=var assigns all array entries to this pointer (so they store the same value).

After var becomes something strange (which happens when fgets returns null), all entries also become strange.

A simple solution is to declare keywords as char keyword[num][len], IIRC, and read directly inside keyword[i].

edited Nov 24 '18 at 6:50

answered Nov 24 '18 at 6:24

dyukha

582511

2

No; var is defined as char var[100]; so it is not a pointer. It easily becomes a pointer, but it is an array, not a pointer.

– Jonathan Leffler
Nov 24 '18 at 6:30

Well, this is why I said "simply put". The difference is irrelevant here.

– dyukha
Nov 24 '18 at 6:32

@dyukha Why not just do it right(tm)?

– Swordfish
Nov 24 '18 at 6:33

1

Misleading people who are struggling with pointers and arrays is not, IMO, helpful. All else apart, if var were just a pointer, you'd have to worry about the storage it points out. At least with an array, you don't have to deal with that detail. That is a crucial and major difference between a pointer and an array.

– Jonathan Leffler
Nov 24 '18 at 6:34

1

@dyukha Because it's irrelevant – No, spreading misinformation (knowingly!!) is hardly irrelevant but a crime against every poor soul that happens to read it.

– Swordfish
Nov 24 '18 at 6:38

|
show 8 more comments

-2

Simply put, var is just a pointer (it's actually an array, but the difference is not relevant here). It means that:

keywords[i-1]=var assigns all array entries to this pointer (so they store the same value).

After var becomes something strange (which happens when fgets returns null), all entries also become strange.

A simple solution is to declare keywords as char keyword[num][len], IIRC, and read directly inside keyword[i].

edited Nov 24 '18 at 6:50

answered Nov 24 '18 at 6:24

dyukha

582511

Simply put, var is just a pointer (it's actually an array, but the difference is not relevant here). It means that:

keywords[i-1]=var assigns all array entries to this pointer (so they store the same value).

After var becomes something strange (which happens when fgets returns null), all entries also become strange.

A simple solution is to declare keywords as char keyword[num][len], IIRC, and read directly inside keyword[i].

edited Nov 24 '18 at 6:50

answered Nov 24 '18 at 6:24

dyukha

582511

edited Nov 24 '18 at 6:50

answered Nov 24 '18 at 6:24

dyukha

582511

answered Nov 24 '18 at 6:24

dyukha

582511

answered Nov 24 '18 at 6:24

dyukha

582511

2

No; var is defined as char var[100]; so it is not a pointer. It easily becomes a pointer, but it is an array, not a pointer.

– Jonathan Leffler
Nov 24 '18 at 6:30

Well, this is why I said "simply put". The difference is irrelevant here.

– dyukha
Nov 24 '18 at 6:32

@dyukha Why not just do it right(tm)?

– Swordfish
Nov 24 '18 at 6:33

1

Misleading people who are struggling with pointers and arrays is not, IMO, helpful. All else apart, if var were just a pointer, you'd have to worry about the storage it points out. At least with an array, you don't have to deal with that detail. That is a crucial and major difference between a pointer and an array.

– Jonathan Leffler
Nov 24 '18 at 6:34

1

@dyukha Because it's irrelevant – No, spreading misinformation (knowingly!!) is hardly irrelevant but a crime against every poor soul that happens to read it.

– Swordfish
Nov 24 '18 at 6:38

|
show 8 more comments

2

No; var is defined as char var[100]; so it is not a pointer. It easily becomes a pointer, but it is an array, not a pointer.

– Jonathan Leffler
Nov 24 '18 at 6:30

Well, this is why I said "simply put". The difference is irrelevant here.

– dyukha
Nov 24 '18 at 6:32

@dyukha Why not just do it right(tm)?

– Swordfish
Nov 24 '18 at 6:33

1

Misleading people who are struggling with pointers and arrays is not, IMO, helpful. All else apart, if var were just a pointer, you'd have to worry about the storage it points out. At least with an array, you don't have to deal with that detail. That is a crucial and major difference between a pointer and an array.

– Jonathan Leffler
Nov 24 '18 at 6:34

1

@dyukha Because it's irrelevant – No, spreading misinformation (knowingly!!) is hardly irrelevant but a crime against every poor soul that happens to read it.

– Swordfish
Nov 24 '18 at 6:38

No; var is defined as char var[100]; so it is not a pointer. It easily becomes a pointer, but it is an array, not a pointer.

– Jonathan Leffler
Nov 24 '18 at 6:30

Well, this is why I said "simply put". The difference is irrelevant here.

– dyukha
Nov 24 '18 at 6:32

@dyukha Why not just do it right(tm)?

– Swordfish
Nov 24 '18 at 6:33

Misleading people who are struggling with pointers and arrays is not, IMO, helpful. All else apart, if var were just a pointer, you'd have to worry about the storage it points out. At least with an array, you don't have to deal with that detail. That is a crucial and major difference between a pointer and an array.

– Jonathan Leffler
Nov 24 '18 at 6:34

@dyukha Because it's irrelevant – No, spreading misinformation (knowingly!!) is hardly irrelevant but a crime against every poor soul that happens to read it.

– Swordfish
Nov 24 '18 at 6:38

|
show 8 more comments

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