Void function as pointer

Question

What's the difference between this:

void copydata(void *, void *, size_t);

and this:

void *copydata(void *, void *, size_t);

I use pointer functions when they must return a pointer, for example:

char *myName(int i)
{
  switch (i) { case 0: return "Jack"; break; default: return "Adam"; }
}

but in the "void *copydata" example, the function doesn't have to return anything, and it compiles correctly in both ways, and the resulting asm code appears to be the same in both cases.

The difference is that if you don't return anything from a function returning non-`void` that's undefined behavior. — The Paramagnetic Croissant, Apr 15 '14 at 06:52
As you said it yourself, if you declare a function with pointer return type, it *must* return a pointer. End of story. So, what is your question then? — AnT stands with Russia, Apr 15 '14 at 08:07

score 3 · Answer 1 · edited Jun 20 '20 at 09:12

What's the difference between this:

void copydata(void *, void *, size_t); // (1)

and this:

void *copydata(void *, void *, size_t); // (2)

(1) is a void function. That means it does not return anything. (2) is a function returning a pointer-to-void, i.e. a void*. It has to return something.

but in the "void *copydata" esample, the function doesn't have to return anything,

That is incorrect. It has to return something that is a void*, or is convertible to one.

and it compiles correctly in both ways, and the resulting asm code appears to be the same in both cases.

Not returning from a non-void function is undefined behaviour. The compiler doesn't have to produce an error, but the resulting code cannot be trusted.

See this related question.

score 2 · Answer 2 · edited May 23 '17 at 12:27

The first one is a void function: it returns no value.

The second one returns a pointer to void: it returns a pointer.

In the code posted you're returning the addresses of char literals. If you don't return anything while your code is expecting a pointer you're getting undefined behavior since you're not even touching the pointer at the "return end" of the call and dirty memory will be in place (or in the best case zero-initialized if it's a global).

More info here: https://stackoverflow.com/a/1610454/1938163

score 0 · Answer 3 · answered Apr 15 '14 at 06:52

0

The first function would return nothing while the second one would return a void pointer.

answered Apr 15 '14 at 06:52

Codor

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score 0 · Answer 4 · answered Apr 15 '14 at 06:53

0

void copydata(void *, void *, size_t);

This means the function doesnt return anything. where as,

void *copydata(void *, void *, size_t);

means that the function returns a pointer. In 2nd case the function should return a pointer to any thing.

answered Apr 15 '14 at 06:53

LearningC

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score 0 · Answer 5 · answered Apr 15 '14 at 08:00

0

If return type is void * you must return a pointer with no type, where functions having void return type does not need to return anything. However that can use return; within the function (not necessary at the end) to return the control conditionally.

answered Apr 15 '14 at 08:00

Dr. Debasish Jana

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score 0 · Answer 6 · answered Apr 15 '14 at 11:47

In terms of memory allocations: 'void *copydata()' will behave exactly the same as 'int *copydata()' and 'double copydata()' etc. That means there will some memory allocated on the stack to store the return value of the function (subject to the compiler implementation and optimization). If you don't do 'return result;' that memory won't be correctly initialized. I think the result will be the same as trying to use uninitialized variable. Although it won't break your program if you don't use the result.

Void function as pointer

6 Answers6