Is it possible to get a pointer to one subobject via a pointer to a different, unreleated subobject?Where in the C++ Standard is the memory layout of objects documented?What are the differences between a pointer variable and a reference variable in C++?Is the Empty Base Class Optimization now a mandatory optimization (at least for standard-layout classes)?Pointer arithmetic across subobject boundariesWhy is f(i = -1, i = -1) undefined behavior?What makes it possible for glibc malloc to compare pointers from different “objects”?Can an equality comparison of unrelated pointers evaluate to true?Compatibility between structs with differently const-qualified pointersGuaranteed memory layout for standard layout struct with a single array member of primitive typeUsing std::launder to get a pointer to an active union member from a pointer to an inactive union member?Accessing the members of a struct referenced by a void pointer

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Is it possible to get a pointer to one subobject via a pointer to a different, unreleated subobject?

Where in the C++ Standard is the memory layout of objects documented?What are the differences between a pointer variable and a reference variable in C++?Is the Empty Base Class Optimization now a mandatory optimization (at least for standard-layout classes)?Pointer arithmetic across subobject boundariesWhy is f(i = -1, i = -1) undefined behavior?What makes it possible for glibc malloc to compare pointers from different “objects”?Can an equality comparison of unrelated pointers evaluate to true?Compatibility between structs with differently const-qualified pointersGuaranteed memory layout for standard layout struct with a single array member of primitive typeUsing std::launder to get a pointer to an active union member from a pointer to an inactive union member?Accessing the members of a struct referenced by a void pointer

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

Look at this simple code:

struct Point 
 int x;
 int y;
;

void something(int *);

int main() 
 Point p1, 2;

 something(&p.x);

 return p.y;

I expect, that main's return value can be optimized to return 2;, as something doesn't have access to p.y, it only gets a pointer to p.x.

But, none of the major compilers optimize the return value of main to 2. Godbolt.

Is there something in the standard, which allows something to modify p.y, if we only give access to p.x? If yes, does this depend on whether Point has standard layout?

What if I use something(&p.y);, and return p.x; instead?

edited 8 hours ago

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

1

I don't understand the downvote and the flag. This is perfectly reasonable question to me.

– Zereges
8 hours ago

stackoverflow.com/questions/50803202/… "Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other"

– alter igel
8 hours ago

@alterigel: this shouldn't matter, as we can use offsetof. But I'm not sure that we can actually get a pointer to p.y from p.x, with the new pointer semantic rules of C++17. std::launder cannot be used, as a pointer to p.x can only reach p.x. So even if I could move the pointer to the correct location, maybe that pointer won't point to the object of p.y.

– geza
8 hours ago

If you change it to void something(int * x) *(x+1) = 6; it optimizes the return value to a fixed 6, so the compiler at least sees that as a valid option to do.

– t.niese
8 hours ago

1

@zereges, didn't downvote, but almost did. The question is very hard to parse. I believe the OP is asking: "why can't the compiler realize the something() function will not modify p.y ?" If the OP asked that directly, there'd be less downvotes.

– Jeffrey
8 hours ago

add a comment
|

Look at this simple code:

struct Point 
 int x;
 int y;
;

void something(int *);

int main() 
 Point p1, 2;

 something(&p.x);

 return p.y;

I expect, that main's return value can be optimized to return 2;, as something doesn't have access to p.y, it only gets a pointer to p.x.

But, none of the major compilers optimize the return value of main to 2. Godbolt.

Is there something in the standard, which allows something to modify p.y, if we only give access to p.x? If yes, does this depend on whether Point has standard layout?

What if I use something(&p.y);, and return p.x; instead?

edited 8 hours ago

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

1

I don't understand the downvote and the flag. This is perfectly reasonable question to me.

– Zereges
8 hours ago

stackoverflow.com/questions/50803202/… "Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other"

– alter igel
8 hours ago

@alterigel: this shouldn't matter, as we can use offsetof. But I'm not sure that we can actually get a pointer to p.y from p.x, with the new pointer semantic rules of C++17. std::launder cannot be used, as a pointer to p.x can only reach p.x. So even if I could move the pointer to the correct location, maybe that pointer won't point to the object of p.y.

– geza
8 hours ago

If you change it to void something(int * x) *(x+1) = 6; it optimizes the return value to a fixed 6, so the compiler at least sees that as a valid option to do.

– t.niese
8 hours ago

1

@zereges, didn't downvote, but almost did. The question is very hard to parse. I believe the OP is asking: "why can't the compiler realize the something() function will not modify p.y ?" If the OP asked that directly, there'd be less downvotes.

– Jeffrey
8 hours ago

add a comment
|

Look at this simple code:

struct Point 
 int x;
 int y;
;

void something(int *);

int main() 
 Point p1, 2;

 something(&p.x);

 return p.y;

I expect, that main's return value can be optimized to return 2;, as something doesn't have access to p.y, it only gets a pointer to p.x.

But, none of the major compilers optimize the return value of main to 2. Godbolt.

Is there something in the standard, which allows something to modify p.y, if we only give access to p.x? If yes, does this depend on whether Point has standard layout?

What if I use something(&p.y);, and return p.x; instead?

edited 8 hours ago

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

Look at this simple code:

struct Point 
 int x;
 int y;
;

void something(int *);

int main() 
 Point p1, 2;

 something(&p.x);

 return p.y;

I expect, that main's return value can be optimized to return 2;, as something doesn't have access to p.y, it only gets a pointer to p.x.

But, none of the major compilers optimize the return value of main to 2. Godbolt.

Is there something in the standard, which allows something to modify p.y, if we only give access to p.x? If yes, does this depend on whether Point has standard layout?

What if I use something(&p.y);, and return p.x; instead?

c++ language-lawyer c++17

edited 8 hours ago

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

edited 8 hours ago

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

edited 8 hours ago

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

asked 8 hours ago

geza

16.6k3 gold badges40 silver badges97 bronze badges

1

I don't understand the downvote and the flag. This is perfectly reasonable question to me.

– Zereges
8 hours ago

stackoverflow.com/questions/50803202/… "Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other"

– alter igel
8 hours ago

@alterigel: this shouldn't matter, as we can use offsetof. But I'm not sure that we can actually get a pointer to p.y from p.x, with the new pointer semantic rules of C++17. std::launder cannot be used, as a pointer to p.x can only reach p.x. So even if I could move the pointer to the correct location, maybe that pointer won't point to the object of p.y.

– geza
8 hours ago

If you change it to void something(int * x) *(x+1) = 6; it optimizes the return value to a fixed 6, so the compiler at least sees that as a valid option to do.

– t.niese
8 hours ago

1

@zereges, didn't downvote, but almost did. The question is very hard to parse. I believe the OP is asking: "why can't the compiler realize the something() function will not modify p.y ?" If the OP asked that directly, there'd be less downvotes.

– Jeffrey
8 hours ago

add a comment
|

1

I don't understand the downvote and the flag. This is perfectly reasonable question to me.

– Zereges
8 hours ago

stackoverflow.com/questions/50803202/… "Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other"

– alter igel
8 hours ago

@alterigel: this shouldn't matter, as we can use offsetof. But I'm not sure that we can actually get a pointer to p.y from p.x, with the new pointer semantic rules of C++17. std::launder cannot be used, as a pointer to p.x can only reach p.x. So even if I could move the pointer to the correct location, maybe that pointer won't point to the object of p.y.

– geza
8 hours ago

If you change it to void something(int * x) *(x+1) = 6; it optimizes the return value to a fixed 6, so the compiler at least sees that as a valid option to do.

– t.niese
8 hours ago

1

@zereges, didn't downvote, but almost did. The question is very hard to parse. I believe the OP is asking: "why can't the compiler realize the something() function will not modify p.y ?" If the OP asked that directly, there'd be less downvotes.

– Jeffrey
8 hours ago

I don't understand the downvote and the flag. This is perfectly reasonable question to me.

– Zereges
8 hours ago

stackoverflow.com/questions/50803202/… "Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other"

– alter igel
8 hours ago

@alterigel: this shouldn't matter, as we can use offsetof. But I'm not sure that we can actually get a pointer to p.y from p.x, with the new pointer semantic rules of C++17. std::launder cannot be used, as a pointer to p.x can only reach p.x. So even if I could move the pointer to the correct location, maybe that pointer won't point to the object of p.y.

– geza
8 hours ago

If you change it to void something(int * x) *(x+1) = 6; it optimizes the return value to a fixed 6, so the compiler at least sees that as a valid option to do.

– t.niese
8 hours ago

@zereges, didn't downvote, but almost did. The question is very hard to parse. I believe the OP is asking: "why can't the compiler realize the something() function will not modify p.y ?" If the OP asked that directly, there'd be less downvotes.

– Jeffrey
8 hours ago

add a comment
|

2 Answers
2

active

oldest

votes

This is perfectly well-defined:

void something(int *x) 
 reinterpret_cast<Point*>(x)->y = 42;

The Point object (p) and its x member are pointer-interconvertible, from [basic.compound]:

Two objects a and b are pointer-interconvertible if:

[...]

one is a standard-layout class object and the other is the first non-static data member of that object, or, if the object has no non-static data members, any base class subobject of that object ([class.mem]), or:

[...]

If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

That reinterpret_cast<Point*>(x) is valid and does end up with a pointer that points to p. Hence, modifying it directly is fine. As you can see, the standard-layout part and the first non-static data member part are significant.

Although it's not like the compilers in question optimize out the extra load if you pass a pointer to p.y in and return p.x instead.

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

1

Thanks, I knew that I left a hole in my question. So, if I used something(&p.y); return p.x;, then it is clearly a missed optimization?

– geza
8 hours ago

@geza For some definition of "clearly". There almost certainly exists a lot of code that does things like this that such an optimization would break.

– Barry
7 hours ago

1

@geza Technically, yes, it would be a missed optimization then, but I don't think major compilers, at this time, are prepared to break all the existing code that uses offsetof.

– Brian
7 hours ago

Doesn't that violate the strict-aliasing rule? Or is that a C-only rule?

– sturcotte06
7 hours ago

1

@sturcotte06 it does not because of If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

– NathanOliver
7 hours ago

add a comment
|

-4

Given a POD structure (trivial with a standard layout, such as Point), you could change y as follows, even if x was not the first field in the object:

void something(int *data)

 data[(offsetof(Point, y)- offsetof(Point, x))/sizeof(int)] = 3;

edited 6 hours ago

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

You might want to reread the question. The OP is not asking about this.

– NathanOliver
7 hours ago

The OP says Is there something in the standard, which allows something to modify p.y, if we only give access to p.x?

– Gonen I
7 hours ago

Correct. You are talking about an array, not whether or not something could modify y given a pointer to x.

– NathanOliver
7 hours ago

The result will be the same. It is not possible for the compiler to optimize, as the data is freely changeable by something given its prototype.

– Gonen I
7 hours ago

Sure, but arrays are different than structs and you cannot treat a pointer to a member as a pointer to an array. You also don't back up you claim with any text from the standard.

– NathanOliver
7 hours ago

|
show 2 more comments

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

This is perfectly well-defined:

void something(int *x) 
 reinterpret_cast<Point*>(x)->y = 42;

The Point object (p) and its x member are pointer-interconvertible, from [basic.compound]:

Two objects a and b are pointer-interconvertible if:

[...]

one is a standard-layout class object and the other is the first non-static data member of that object, or, if the object has no non-static data members, any base class subobject of that object ([class.mem]), or:

[...]

If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

Although it's not like the compilers in question optimize out the extra load if you pass a pointer to p.y in and return p.x instead.

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

1

Thanks, I knew that I left a hole in my question. So, if I used something(&p.y); return p.x;, then it is clearly a missed optimization?

– geza
8 hours ago

@geza For some definition of "clearly". There almost certainly exists a lot of code that does things like this that such an optimization would break.

– Barry
7 hours ago

1

@geza Technically, yes, it would be a missed optimization then, but I don't think major compilers, at this time, are prepared to break all the existing code that uses offsetof.

– Brian
7 hours ago

Doesn't that violate the strict-aliasing rule? Or is that a C-only rule?

– sturcotte06
7 hours ago

1

@sturcotte06 it does not because of If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

– NathanOliver
7 hours ago

add a comment
|

This is perfectly well-defined:

void something(int *x) 
 reinterpret_cast<Point*>(x)->y = 42;

The Point object (p) and its x member are pointer-interconvertible, from [basic.compound]:

Two objects a and b are pointer-interconvertible if:

[...]

one is a standard-layout class object and the other is the first non-static data member of that object, or, if the object has no non-static data members, any base class subobject of that object ([class.mem]), or:

[...]

If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

Although it's not like the compilers in question optimize out the extra load if you pass a pointer to p.y in and return p.x instead.

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

1

Thanks, I knew that I left a hole in my question. So, if I used something(&p.y); return p.x;, then it is clearly a missed optimization?

– geza
8 hours ago

@geza For some definition of "clearly". There almost certainly exists a lot of code that does things like this that such an optimization would break.

– Barry
7 hours ago

1

@geza Technically, yes, it would be a missed optimization then, but I don't think major compilers, at this time, are prepared to break all the existing code that uses offsetof.

– Brian
7 hours ago

Doesn't that violate the strict-aliasing rule? Or is that a C-only rule?

– sturcotte06
7 hours ago

1

@sturcotte06 it does not because of If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

– NathanOliver
7 hours ago

add a comment
|

This is perfectly well-defined:

void something(int *x) 
 reinterpret_cast<Point*>(x)->y = 42;

The Point object (p) and its x member are pointer-interconvertible, from [basic.compound]:

Two objects a and b are pointer-interconvertible if:

[...]

one is a standard-layout class object and the other is the first non-static data member of that object, or, if the object has no non-static data members, any base class subobject of that object ([class.mem]), or:

[...]

If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

Although it's not like the compilers in question optimize out the extra load if you pass a pointer to p.y in and return p.x instead.

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

This is perfectly well-defined:

void something(int *x) 
 reinterpret_cast<Point*>(x)->y = 42;

The Point object (p) and its x member are pointer-interconvertible, from [basic.compound]:

Two objects a and b are pointer-interconvertible if:

[...]

one is a standard-layout class object and the other is the first non-static data member of that object, or, if the object has no non-static data members, any base class subobject of that object ([class.mem]), or:

[...]

If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

Although it's not like the compilers in question optimize out the extra load if you pass a pointer to p.y in and return p.x instead.

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

edited 7 hours ago

T.C.

111k14 gold badges229 silver badges338 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

answered 8 hours ago

Barry

200k22 gold badges373 silver badges665 bronze badges

1

Thanks, I knew that I left a hole in my question. So, if I used something(&p.y); return p.x;, then it is clearly a missed optimization?

– geza
8 hours ago

@geza For some definition of "clearly". There almost certainly exists a lot of code that does things like this that such an optimization would break.

– Barry
7 hours ago

1

@geza Technically, yes, it would be a missed optimization then, but I don't think major compilers, at this time, are prepared to break all the existing code that uses offsetof.

– Brian
7 hours ago

Doesn't that violate the strict-aliasing rule? Or is that a C-only rule?

– sturcotte06
7 hours ago

1

@sturcotte06 it does not because of If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

– NathanOliver
7 hours ago

add a comment
|

1

Thanks, I knew that I left a hole in my question. So, if I used something(&p.y); return p.x;, then it is clearly a missed optimization?

– geza
8 hours ago

@geza For some definition of "clearly". There almost certainly exists a lot of code that does things like this that such an optimization would break.

– Barry
7 hours ago

1

@geza Technically, yes, it would be a missed optimization then, but I don't think major compilers, at this time, are prepared to break all the existing code that uses offsetof.

– Brian
7 hours ago

Doesn't that violate the strict-aliasing rule? Or is that a C-only rule?

– sturcotte06
7 hours ago

1

@sturcotte06 it does not because of If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

– NathanOliver
7 hours ago

Thanks, I knew that I left a hole in my question. So, if I used something(&p.y); return p.x;, then it is clearly a missed optimization?

– geza
8 hours ago

@geza For some definition of "clearly". There almost certainly exists a lot of code that does things like this that such an optimization would break.

– Barry
7 hours ago

@geza Technically, yes, it would be a missed optimization then, but I don't think major compilers, at this time, are prepared to break all the existing code that uses offsetof.

– Brian
7 hours ago

Doesn't that violate the strict-aliasing rule? Or is that a C-only rule?

– sturcotte06
7 hours ago

@sturcotte06 it does not because of If two objects are pointer-interconvertible, then they have the same address, and it is possible to obtain a pointer to one from a pointer to the other via a reinterpret_cast.

– NathanOliver
7 hours ago

add a comment
|

-4

Given a POD structure (trivial with a standard layout, such as Point), you could change y as follows, even if x was not the first field in the object:

void something(int *data)

 data[(offsetof(Point, y)- offsetof(Point, x))/sizeof(int)] = 3;

edited 6 hours ago

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

You might want to reread the question. The OP is not asking about this.

– NathanOliver
7 hours ago

The OP says Is there something in the standard, which allows something to modify p.y, if we only give access to p.x?

– Gonen I
7 hours ago

Correct. You are talking about an array, not whether or not something could modify y given a pointer to x.

– NathanOliver
7 hours ago

The result will be the same. It is not possible for the compiler to optimize, as the data is freely changeable by something given its prototype.

– Gonen I
7 hours ago

Sure, but arrays are different than structs and you cannot treat a pointer to a member as a pointer to an array. You also don't back up you claim with any text from the standard.

– NathanOliver
7 hours ago

|
show 2 more comments

-4

Given a POD structure (trivial with a standard layout, such as Point), you could change y as follows, even if x was not the first field in the object:

void something(int *data)

 data[(offsetof(Point, y)- offsetof(Point, x))/sizeof(int)] = 3;

edited 6 hours ago

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

You might want to reread the question. The OP is not asking about this.

– NathanOliver
7 hours ago

The OP says Is there something in the standard, which allows something to modify p.y, if we only give access to p.x?

– Gonen I
7 hours ago

Correct. You are talking about an array, not whether or not something could modify y given a pointer to x.

– NathanOliver
7 hours ago

The result will be the same. It is not possible for the compiler to optimize, as the data is freely changeable by something given its prototype.

– Gonen I
7 hours ago

Sure, but arrays are different than structs and you cannot treat a pointer to a member as a pointer to an array. You also don't back up you claim with any text from the standard.

– NathanOliver
7 hours ago

|
show 2 more comments

-4

Given a POD structure (trivial with a standard layout, such as Point), you could change y as follows, even if x was not the first field in the object:

void something(int *data)

 data[(offsetof(Point, y)- offsetof(Point, x))/sizeof(int)] = 3;

edited 6 hours ago

answered 7 hours ago

Gonen I

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Given a POD structure (trivial with a standard layout, such as Point), you could change y as follows, even if x was not the first field in the object:

void something(int *data)

 data[(offsetof(Point, y)- offsetof(Point, x))/sizeof(int)] = 3;

edited 6 hours ago

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

edited 6 hours ago

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

answered 7 hours ago

Gonen I

2,07714 silver badges31 bronze badges

You might want to reread the question. The OP is not asking about this.

– NathanOliver
7 hours ago

The OP says Is there something in the standard, which allows something to modify p.y, if we only give access to p.x?

– Gonen I
7 hours ago

Correct. You are talking about an array, not whether or not something could modify y given a pointer to x.

– NathanOliver
7 hours ago

The result will be the same. It is not possible for the compiler to optimize, as the data is freely changeable by something given its prototype.

– Gonen I
7 hours ago

Sure, but arrays are different than structs and you cannot treat a pointer to a member as a pointer to an array. You also don't back up you claim with any text from the standard.

– NathanOliver
7 hours ago

|
show 2 more comments

You might want to reread the question. The OP is not asking about this.

– NathanOliver
7 hours ago

The OP says Is there something in the standard, which allows something to modify p.y, if we only give access to p.x?

– Gonen I
7 hours ago

Correct. You are talking about an array, not whether or not something could modify y given a pointer to x.

– NathanOliver
7 hours ago

The result will be the same. It is not possible for the compiler to optimize, as the data is freely changeable by something given its prototype.

– Gonen I
7 hours ago

Sure, but arrays are different than structs and you cannot treat a pointer to a member as a pointer to an array. You also don't back up you claim with any text from the standard.

– NathanOliver
7 hours ago

You might want to reread the question. The OP is not asking about this.

– NathanOliver
7 hours ago

The OP says Is there something in the standard, which allows something to modify p.y, if we only give access to p.x?

– Gonen I
7 hours ago

Correct. You are talking about an array, not whether or not something could modify y given a pointer to x.

– NathanOliver
7 hours ago

The result will be the same. It is not possible for the compiler to optimize, as the data is freely changeable by something given its prototype.

– Gonen I
7 hours ago

Sure, but arrays are different than structs and you cannot treat a pointer to a member as a pointer to an array. You also don't back up you claim with any text from the standard.

– NathanOliver
7 hours ago

|
show 2 more comments

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