CODEWITHSUNDEEP
cpointerspointer-arithmetic
Alvin
Alvin

Reputation: 159

I don't understand C pointers

I don't really understand the reason why I obtain a different address from what I'm expecting.

I've tried to build this small C code with -m32 flag option.

#include <stdio.h>
#include <stdlib.h>

char *Buffer[10];

int main (void){
 printf("%p\n", Buffer);
 char *Buffer2 = Buffer + 6;
 printf("%p\n", Buffer2);
}

Expected output:

Buffer = 0x56559040
Buffer2 = 0x56559046

Obtained output:

Buffer = 0x56559040
Buffer2 = 0x56559058

Why the obtained output is different from the expected one (0x56559040 + 6 = 0x56559046)?

Upvotes: 1

Views: 134

Answers (1)

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

The difference between these two values

Buffer = 0x56559040
Buffer2 = 0x56559058

is 0x18 or in decimal 24.

In this declaration

char *Buffer2 = Buffer + 6;

the array designator Buffer is converted to pointer to its first element. As the element type of the array Buffer is char * then the expression has the type char **.

There is no implicit conversion between types char * (the type of the variable Buffer2) and char ** (the type of the initializer)

So the compiler should issue at least a warning.

Nevertheless using the pointer arithmetic this expression

Buffer + 6

is evaluated like 

the value of the address pointed to by Buffer + 6 * sizeof( char * )

as the size of a pointer of the type char * (the size of element of the array) in your system is equal to 4 then you get the value 0x56559058 that is 

0x56559040 + 6 * sizeof( char * )
                 ^^^^^^^^^^^^^^^^
                       4

That is the expression

Buffer + 6

points to the sixth element of the array Buffer.

Upvotes: 1

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