算术的铸造指针 - C

Question

If I have a void * x, but I want to cast it to a char *, so that the ++ operator will make it point to the next byte and not the next 4 byte block.

However, when I do:

(char *) x -= byte_length;

The compiler complains:

Error, lvalue required as left value of assignment.

Where am I going wrong please? Thanks.

Answer 1

I would do it like this:

 x = (char*)x - byte_length;

Cast x to char* , then apply the offset, then assign back to x . Since void* is assignment compatible with all pointer types no further cast is needed.

Answer 2

(char *)x evaluates to a temporary with the same value as x but a different type. The compiler won't allow -= on a temporary. Do

x = (char *)x - byte_length;

The situation is analogous to the following:

short x = 0;
(long)x += 1;     // invalid; (long)x is a temporary
x = (long)x + 1;  // works

Answer 3

Do something like this:

void* x;
char* cx = (char*)x;
cx -= byte_length;
x = cx;

Answer 4

(char*) x -= byte_length;

In this statement, (char*) x creates a temporary value with char* as type. You can only assign to variables, or more precisely, references (or an lvalue according to your compiler).

One way to get a reference to it is casting the address of x to char* and dereferencing that:

*(char*) &x -= byte_length;

Answer 5

(char*)x is a temporary value that has not address in memory. You can't do a -- to it, for the same reason that if you have int i , you can't do (i+5)++ .

You should do x=(char*)x-byte_length; instead.

See also http://c-faq.com/ptrs/castincr.html