为什么在这种情况下将void指针强制转换为char指针不起作用

Question

I've got void pointer which I want to do pointer arithmetic on it. to do so, I want to cast it to a char pointer.

( char * ) buf += current_size;

However when I do so, I get the following error:

error: lvalue required as left operand of assignment

I tried also adding parenthesis on the the whole left side, with no success. Why do I get this ?

Answer 1

Why do I get this

Simply because the language rules state that a cast does not produce an lvalue even if the operand is one .

ISO/IEC 9899:201x - n1570

In a footnote, page 91

A cast does not yield an lvalue.

Answer 2

You can avoid the warning by writing the assignment operator out:

buf = (char *)buf + current_size;

Part of the reason you get the warning is that sizeof(void) is undefined and you cannot do arithmetic on void * . Beware: GCC has an extension whereby it treats the code as if sizeof(void) == 1 , but that is not in the C standard.

The cast on the LHS of the assignment has no effect on the assignment.

Answer 3

Because it's invalid, you can't do this. If you want to achieve this, either

declare buf as a char pointer; or use a temporary variable:

char *tmp = buf; // note you don't need the casting, void * works with everything
tmp += current_size;
buf = tmp;

Edit: as others also suggested, you can even remove that ugly tmp:

buf = (char *)buf + current_size;