*a = *b (what does it mean, how it works)-CodePudding

What would be the values after performing this operation?

#include <stdio.h>
int main() {
    int *a = 0;
    int *b = 3;
    *a   = *b  ;
    printf("%d", a);
    printf("%d", b);
    return 0;
}

The code above gives me a segmentation fault.

CodePudding user response：

*a = *b (what does it mean, how it works)

*a   = *b  ;

means

*(a  ) = *(b  );

x increments x and returns the original value. So the following is equivalent:

*a = *b;     // Copy the `int` to which `b` points into the `int` to which `a` points.
a = a   1;   // Make `a` point to the following `int`.
b = b   1;   // Make `b` point to the following `int`.

Before:                                     After:

a                                           a
 ----------          ----------              ----------          ---------- 
|        ---------->| x        |            |        ------     | p        |
 ----------          ----------              ----------    |     ---------- 
                    | y        |                            --->| y        |
                     ----------                                  ---------- 
                    |          |                                |          |


b                                           b
 ----------          ----------              ----------          ---------- 
|        ---------->| p        |            |        ------     | p        |
 ----------          ----------              ----------    |     ---------- 
                    | q        |                            --->| q        |
                     ----------                                  ---------- 
                    |          |                                |          |

The code above gives me a segmentation fault.

You assigned garbage to a and b. 0 as a pointer is the NULL pointer, and 3 isn't a valid pointer.

CodePudding user response：

Given

#include <stdio.h>
int main() {
    int *a = 0;
    int *b = 3;
    *a   = *b  ;
    printf("%d", a);
    printf("%d", b);
    return 0;
}

the printed values can not be predicted as the code invokes undefined behavior in multiple ways.

First, both *a and *b invoke undefined behavior by dereferencing invalid pointers - a is initialized to a null pointer value, and b is initialized to point to address 3, which is almost certainly invalid also.

Second,printf("%d", a); invokes undefined behavior by trying to print an int * variable with the %d format specifier for int. The proper code would be

printf("%p", ( void * ) a);

It's not clear what the currently-posted code is supposed to do.