What is the size of void?

c c++


What would this statement yield?

void *p = malloc(sizeof(void));

Edit: An extension to the question.

If sizeof(void) yields 1 in GCC compiler, then 1 byte of memory is allocated and the pointer p points to that byte and would p++ be incremented to 0x2346? Suppose p was 0x2345. I am talking about p and not *p.

2/9/2019 8:44:48 PM

Accepted Answer

The type void has no size; that would be a compilation error. For the same reason you can't do something like:

void n;

EDIT. To my surprise, doing sizeof(void) actually does compile in GNU C:

$ echo 'int main() { printf("%d", sizeof(void)); }' | gcc -xc -w - && ./a.out 

However, in C++ it does not:

$ echo 'int main() { printf("%d", sizeof(void)); }' | gcc -xc++ -w - && ./a.out 
<stdin>: In function 'int main()':
<stdin>:1: error: invalid application of 'sizeof' to a void type
<stdin>:1: error: 'printf' was not declared in this scope
2/1/2015 3:10:56 AM

If you are using GCC and you are not using compilation flags that remove compiler specific extensions, then sizeof(void) is 1. GCC has a nonstandard extension that does that.

In general, void is a incomplete type, and you cannot use sizeof for incomplete types.

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