指针算法C和强制转换

Question

I've seen a very strange code snippet and i am not quite sure if i understood it right:

#include <stdio.h>

int main(char *argc, char **argv)
{
   char a[50];
   *(char *) (a + 2) = 'b'; // <== THE LINE WHICH CONFUSES ME

   printf("value: %c\n", a[2]);
   return 1;
}

Is it right that we go 2 buckets further cast the 'b' into a pointer to the b and then dereference it?

Answer 1

That's exactly equivalent of

*(a + 2) = 'b';

The cast is unnecessary.

All it does is add two to the array-which-decays-to-a-pointer a , dereference the resulting pointer, and assign the character 'b' to that memory location.

When a is a pointer, the code a[x] is exactly equivalent of *(a + x) . So in your case, *(a + 2) = 'b' is exactly the same as a[2] = 'b' .

Answer 2

*(char *) (a + 2)

is equivalent to

a[2]

By definition, a[2] is *(a + 2) . The value of a + 2 is already of type char * so a cast of a + 2 to char * , as is (char *) (a + 2) , is a no operation.

Answer 3

  *(char *) (a + 2) = 'b'; // <== THE LINE WHICH CONFUSES ME 

This line literally means the very same as

a[2] = 'b'

The first cast (char*) is redundant, since a is already of type char . And indexing in fact translates to addition and dereferenciation, ie

a[n] === *(a + n)

A little known fact about C: You could write as well

n[a]

and get the same result.

Answer 4

You aren't casting the 'b' .

You cast (a+2) to char* (Which does nothing, since it's already char* ), deference it, and put there 'b' .

And yes, it is right that we go 2 buckets further.

Answer 5

No, you treat a as a pointer, increment it by two, then cast it to (char*) (useless cast, it already is char* ), dereference it and then store 'b' into that.

It is exactly the same as this:

a[2] = 'b';