Initializing floating value of UNION in C
0
I have below union,
typedef union
{
struct
{
float x;
float y;
float z;
float Backup;
} pt;
float Max[4];
} Var3D;
When I try to initialize above union like Var3D= { 0.0, 0.0, 0.0, 0.0 };
It shows build error like
suggest braces around initialization of subobject
[-Werror,-Wmissing-braces]
Var3D= {0.0, 0.0, 0.0, 0.0};
How can I fix this?
c
asked Nov 24 '18 at 6:46
PerryPerry
11
|
show 3 more comments
0
I have below union,
typedef union
{
struct
{
float x;
float y;
float z;
float Backup;
} pt;
float Max[4];
} Var3D;
When I try to initialize above union like Var3D= { 0.0, 0.0, 0.0, 0.0 };
It shows build error like
suggest braces around initialization of subobject
[-Werror,-Wmissing-braces]
Var3D= {0.0, 0.0, 0.0, 0.0};
How can I fix this?
c
asked Nov 24 '18 at 6:46
PerryPerry
11
What happens when your follow the advice "suggest braces around initialization of subobject" and put{}around{0.0, 0.0, 0.0, 0.0}?
– chux
Nov 24 '18 at 6:50
1
It would help if you gave an actual code sample.3DVaris not a valid identifier, so yourtypedefwould not compile. In any event, assuming your ACTUAL code (as distinct from the code you have posted) has a validtypedefname other than3DVar, try doing as the error message suggests:YourType some_variable = {{0.0f, 0.0f, 0.0f, 0.0f}};. Note there are two sets of{}. In future, read up on how to ask a question in a way that is more likely to get useful responses, in particular provision of a Minimal, Complete, and Verifiable example.
– Peter
Nov 24 '18 at 6:52
Adding braces like below resolves, {{0.0, 0.0, 0.0, 0.0}} but i don't understand the {{0.0, 0.0, 0.0, 0.0}}; --> Resolve this. But {0.0, 0.0, 0.0, 0.0} is already initialization. I don't understand this clearly. Braces are already present and this is of union type. So when we initialize larger size member.
– Perry
Nov 24 '18 at 7:07
2
Note that you can't use the type name3Dvar— identifiers cannot start with digits. You also don't show a valid variable definition. Maybe you need to use a type name such asVar3Dand then the variable definition would beVar3D var3d = { .pt = { 0.0, 0.0, 0.0, 0.0 } };using a designated initializer, or you can drop the.pt =and leave the doubled braces to initialize correctly. Technically, that undesignated notation initializes the first element of the union, which ispt. You could designate the alternative with.Max =in place of.pt =. The net result is the same here.
– Jonathan Leffler
Nov 24 '18 at 7:15
2
@n.m.: In C, unlike C++, it is not undefined to access a member other than the last one stored. Per the C standard, the bytes are reinterpreted as the new type. There may be implementation-defined effects.
– Eric Postpischil
Nov 24 '18 at 12:06
|
show 3 more comments
0
0
0
I have below union,
typedef union
{
struct
{
float x;
float y;
float z;
float Backup;
} pt;
float Max[4];
} Var3D;
When I try to initialize above union like Var3D= { 0.0, 0.0, 0.0, 0.0 };
It shows build error like
suggest braces around initialization of subobject
[-Werror,-Wmissing-braces]
Var3D= {0.0, 0.0, 0.0, 0.0};
How can I fix this?
c
asked Nov 24 '18 at 6:46
PerryPerry
11
I have below union,
typedef union
{
struct
{
float x;
float y;
float z;
float Backup;
} pt;
float Max[4];
} Var3D;
When I try to initialize above union like Var3D= { 0.0, 0.0, 0.0, 0.0 };
It shows build error like
suggest braces around initialization of subobject
[-Werror,-Wmissing-braces]
Var3D= {0.0, 0.0, 0.0, 0.0};
How can I fix this?
c
c
asked Nov 24 '18 at 6:46
PerryPerry
11
asked Nov 24 '18 at 6:46
PerryPerry
11
edited Nov 24 '18 at 7:24
Perry
asked Nov 24 '18 at 6:46
PerryPerry
11
asked Nov 24 '18 at 6:46
PerryPerry
11
asked Nov 24 '18 at 6:46
PerryPerry
11
11
What happens when your follow the advice "suggest braces around initialization of subobject" and put{}around{0.0, 0.0, 0.0, 0.0}?
– chux
Nov 24 '18 at 6:50
1
It would help if you gave an actual code sample.3DVaris not a valid identifier, so yourtypedefwould not compile. In any event, assuming your ACTUAL code (as distinct from the code you have posted) has a validtypedefname other than3DVar, try doing as the error message suggests:YourType some_variable = {{0.0f, 0.0f, 0.0f, 0.0f}};. Note there are two sets of{}. In future, read up on how to ask a question in a way that is more likely to get useful responses, in particular provision of a Minimal, Complete, and Verifiable example.
– Peter
Nov 24 '18 at 6:52
Adding braces like below resolves, {{0.0, 0.0, 0.0, 0.0}} but i don't understand the {{0.0, 0.0, 0.0, 0.0}}; --> Resolve this. But {0.0, 0.0, 0.0, 0.0} is already initialization. I don't understand this clearly. Braces are already present and this is of union type. So when we initialize larger size member.
– Perry
Nov 24 '18 at 7:07
2
Note that you can't use the type name3Dvar— identifiers cannot start with digits. You also don't show a valid variable definition. Maybe you need to use a type name such asVar3Dand then the variable definition would beVar3D var3d = { .pt = { 0.0, 0.0, 0.0, 0.0 } };using a designated initializer, or you can drop the.pt =and leave the doubled braces to initialize correctly. Technically, that undesignated notation initializes the first element of the union, which ispt. You could designate the alternative with.Max =in place of.pt =. The net result is the same here.
– Jonathan Leffler
Nov 24 '18 at 7:15
2
@n.m.: In C, unlike C++, it is not undefined to access a member other than the last one stored. Per the C standard, the bytes are reinterpreted as the new type. There may be implementation-defined effects.
– Eric Postpischil
Nov 24 '18 at 12:06
|
show 3 more comments
What happens when your follow the advice "suggest braces around initialization of subobject" and put{}around{0.0, 0.0, 0.0, 0.0}?
– chux
Nov 24 '18 at 6:50
1
It would help if you gave an actual code sample.3DVaris not a valid identifier, so yourtypedefwould not compile. In any event, assuming your ACTUAL code (as distinct from the code you have posted) has a validtypedefname other than3DVar, try doing as the error message suggests:YourType some_variable = {{0.0f, 0.0f, 0.0f, 0.0f}};. Note there are two sets of{}. In future, read up on how to ask a question in a way that is more likely to get useful responses, in particular provision of a Minimal, Complete, and Verifiable example.
– Peter
Nov 24 '18 at 6:52
Adding braces like below resolves, {{0.0, 0.0, 0.0, 0.0}} but i don't understand the {{0.0, 0.0, 0.0, 0.0}}; --> Resolve this. But {0.0, 0.0, 0.0, 0.0} is already initialization. I don't understand this clearly. Braces are already present and this is of union type. So when we initialize larger size member.
– Perry
Nov 24 '18 at 7:07
2
Note that you can't use the type name3Dvar— identifiers cannot start with digits. You also don't show a valid variable definition. Maybe you need to use a type name such asVar3Dand then the variable definition would beVar3D var3d = { .pt = { 0.0, 0.0, 0.0, 0.0 } };using a designated initializer, or you can drop the.pt =and leave the doubled braces to initialize correctly. Technically, that undesignated notation initializes the first element of the union, which ispt. You could designate the alternative with.Max =in place of.pt =. The net result is the same here.
– Jonathan Leffler
Nov 24 '18 at 7:15
2
@n.m.: In C, unlike C++, it is not undefined to access a member other than the last one stored. Per the C standard, the bytes are reinterpreted as the new type. There may be implementation-defined effects.
– Eric Postpischil
Nov 24 '18 at 12:06
What happens when your follow the advice "suggest braces around initialization of subobject" and put
{} around {0.0, 0.0, 0.0, 0.0}?– chux
Nov 24 '18 at 6:50
What happens when your follow the advice "suggest braces around initialization of subobject" and put
{} around {0.0, 0.0, 0.0, 0.0}?– chux
Nov 24 '18 at 6:50
1
1
It would help if you gave an actual code sample.
3DVar is not a valid identifier, so your typedef would not compile. In any event, assuming your ACTUAL code (as distinct from the code you have posted) has a valid typedef name other than 3DVar, try doing as the error message suggests: YourType some_variable = {{0.0f, 0.0f, 0.0f, 0.0f}};. Note there are two sets of {}. In future, read up on how to ask a question in a way that is more likely to get useful responses, in particular provision of a Minimal, Complete, and Verifiable example.– Peter
Nov 24 '18 at 6:52
It would help if you gave an actual code sample.
3DVar is not a valid identifier, so your typedef would not compile. In any event, assuming your ACTUAL code (as distinct from the code you have posted) has a valid typedef name other than 3DVar, try doing as the error message suggests: YourType some_variable = {{0.0f, 0.0f, 0.0f, 0.0f}};. Note there are two sets of {}. In future, read up on how to ask a question in a way that is more likely to get useful responses, in particular provision of a Minimal, Complete, and Verifiable example.– Peter
Nov 24 '18 at 6:52
Adding braces like below resolves, {{0.0, 0.0, 0.0, 0.0}} but i don't understand the {{0.0, 0.0, 0.0, 0.0}}; --> Resolve this. But {0.0, 0.0, 0.0, 0.0} is already initialization. I don't understand this clearly. Braces are already present and this is of union type. So when we initialize larger size member.
– Perry
Nov 24 '18 at 7:07
Adding braces like below resolves, {{0.0, 0.0, 0.0, 0.0}} but i don't understand the {{0.0, 0.0, 0.0, 0.0}}; --> Resolve this. But {0.0, 0.0, 0.0, 0.0} is already initialization. I don't understand this clearly. Braces are already present and this is of union type. So when we initialize larger size member.
– Perry
Nov 24 '18 at 7:07
2
2
Note that you can't use the type name
3Dvar — identifiers cannot start with digits. You also don't show a valid variable definition. Maybe you need to use a type name such as Var3D and then the variable definition would be Var3D var3d = { .pt = { 0.0, 0.0, 0.0, 0.0 } }; using a designated initializer, or you can drop the .pt = and leave the doubled braces to initialize correctly. Technically, that undesignated notation initializes the first element of the union, which is pt. You could designate the alternative with .Max = in place of .pt =. The net result is the same here.– Jonathan Leffler
Nov 24 '18 at 7:15
Note that you can't use the type name
3Dvar — identifiers cannot start with digits. You also don't show a valid variable definition. Maybe you need to use a type name such as Var3D and then the variable definition would be Var3D var3d = { .pt = { 0.0, 0.0, 0.0, 0.0 } }; using a designated initializer, or you can drop the .pt = and leave the doubled braces to initialize correctly. Technically, that undesignated notation initializes the first element of the union, which is pt. You could designate the alternative with .Max = in place of .pt =. The net result is the same here.– Jonathan Leffler
Nov 24 '18 at 7:15
2
2
@n.m.: In C, unlike C++, it is not undefined to access a member other than the last one stored. Per the C standard, the bytes are reinterpreted as the new type. There may be implementation-defined effects.
– Eric Postpischil
Nov 24 '18 at 12:06
@n.m.: In C, unlike C++, it is not undefined to access a member other than the last one stored. Per the C standard, the bytes are reinterpreted as the new type. There may be implementation-defined effects.
– Eric Postpischil
Nov 24 '18 at 12:06
|
show 3 more comments
1 Answer
1
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oldest
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0
So make an answer:
As Peter mentioned you need two pair of braces: one for the union and one for the array included in the union.
So this initialization should work:
Var3D= {{ 0.0, 0.0, 0.0, 0.0 }};
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
add a comment |
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So make an answer:
As Peter mentioned you need two pair of braces: one for the union and one for the array included in the union.
So this initialization should work:
Var3D= {{ 0.0, 0.0, 0.0, 0.0 }};
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
add a comment |
0
So make an answer:
As Peter mentioned you need two pair of braces: one for the union and one for the array included in the union.
So this initialization should work:
Var3D= {{ 0.0, 0.0, 0.0, 0.0 }};
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
add a comment |
0
0
0
So make an answer:
As Peter mentioned you need two pair of braces: one for the union and one for the array included in the union.
So this initialization should work:
Var3D= {{ 0.0, 0.0, 0.0, 0.0 }};
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
So make an answer:
As Peter mentioned you need two pair of braces: one for the union and one for the array included in the union.
So this initialization should work:
Var3D= {{ 0.0, 0.0, 0.0, 0.0 }};
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
answered Dec 4 '18 at 13:42
MikeMike
2,0031722
2,0031722
add a comment |
add a comment |
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What happens when your follow the advice "suggest braces around initialization of subobject" and put
{}around{0.0, 0.0, 0.0, 0.0}?– chux
Nov 24 '18 at 6:50
1
It would help if you gave an actual code sample.
3DVaris not a valid identifier, so yourtypedefwould not compile. In any event, assuming your ACTUAL code (as distinct from the code you have posted) has a validtypedefname other than3DVar, try doing as the error message suggests:YourType some_variable = {{0.0f, 0.0f, 0.0f, 0.0f}};. Note there are two sets of{}. In future, read up on how to ask a question in a way that is more likely to get useful responses, in particular provision of a Minimal, Complete, and Verifiable example.– Peter
Nov 24 '18 at 6:52
Adding braces like below resolves, {{0.0, 0.0, 0.0, 0.0}} but i don't understand the {{0.0, 0.0, 0.0, 0.0}}; --> Resolve this. But {0.0, 0.0, 0.0, 0.0} is already initialization. I don't understand this clearly. Braces are already present and this is of union type. So when we initialize larger size member.
– Perry
Nov 24 '18 at 7:07
2
Note that you can't use the type name
3Dvar— identifiers cannot start with digits. You also don't show a valid variable definition. Maybe you need to use a type name such asVar3Dand then the variable definition would beVar3D var3d = { .pt = { 0.0, 0.0, 0.0, 0.0 } };using a designated initializer, or you can drop the.pt =and leave the doubled braces to initialize correctly. Technically, that undesignated notation initializes the first element of the union, which ispt. You could designate the alternative with.Max =in place of.pt =. The net result is the same here.– Jonathan Leffler
Nov 24 '18 at 7:15
2
@n.m.: In C, unlike C++, it is not undefined to access a member other than the last one stored. Per the C standard, the bytes are reinterpreted as the new type. There may be implementation-defined effects.
– Eric Postpischil
Nov 24 '18 at 12:06