x=*ptr ; is equivalent to x=*ptr; ptr=ptr 1;?-CodePudding

In the below code

ptr1--->ptr--->&c=4

when i print *ptr and **ptr1 both prints value of c i.e 4 everything is fine here, but after initialization of x with *ptr , i read somewhere

that x=*ptr is equivalent to x=*ptr; ptr= ptr 1;

if it's true that means pointer ptr is pointing to a new address but still *ptr and **ptr1 prints the value of c why?

int main(void)
{
    int x, c=4, *ptr=&c,**ptr1=&ptr;
    printf("\nBefore incrementation ptr points to c\n%p == %p", ptr,&c);
    x=*ptr  ;
    printf("\n%d,%d,%d,%p,%p,%p\n\n", x,*ptr,**ptr1,&c,ptr,ptr1); 
    return 0;
}

CodePudding user response：

DO NOT DO THIS under normal circumstances... This 'happily' exploits "undefined behaviour" with the code generated by my C compiler. This is BAD PRACTICE, but useful to understand the behaviour you experienced.

int main() {
    int x = 123;
    int y = 456;
    int z = 789;
    int *p = &y; // 'p' points at stack variable 'y'...

    printf( "%d %d %d\n", p[-1], p[0], p[1] );

    return 0;
}

Output:

789 456 123

I repeat: This is for educational purposes only.

CodePudding user response：

My environment : WSL2 with gcc9.3 I add two value to printf to show the address of local variables:

#include <stdio.h>

int main(void)
{
        int x, c=4, *ptr=&c,**ptr1=&ptr;
        printf("Before incementation ptr points to c\n%p == %p\n", ptr,&c);
        x=*ptr  ;
        printf("ptr:%p\n", ptr);
        printf("x:%d\n*ptr:%d\n**ptr1:%d\n&c:%p\nptr:%p\nptr1:%p\n&ptr:%p\n&ptr1:%p\n",
                        x, *ptr, **ptr1, &c, ptr, ptr1, &ptr, &ptr1);
        return 0;
}

And the output is:

Before incementation ptr points to c
0x7fffe4ffe0d0 == 0x7fffe4ffe0d0
ptr:0x7fffe4ffe0d4
x:4
*ptr:4
**ptr1:4
&c:0x7fffe4ffe0d0
ptr:0x7fffe4ffe0d4
ptr1:0x7fffe4ffe0d8
&ptr:0x7fffe4ffe0d8
&ptr1:0x7fffe4ffe0e0

Note that? The "ptr" is adjacent to "c" on stack. So after "x=*ptr ", "ptr" pointing to itself.

if it's true that means pointer ptr is pointing to a new address but still *ptr and **ptr1 prints the value of c why?

So you can see that "ptr" does not pointing to c.

The next problem is when does "*ptr" changed, I'am single-instruction tracing the code, I found it is changed in a piece of pre-call asm-code:

119b:       48 8b 45 e8             mov    -0x18(%rbp),%rax
119f:       48 8d 55 e0             lea    -0x20(%rbp),%rdx
11a3:       48 89 c6                mov    %rax,%rsi
11a6:       48 8d 3d 5b 0e 00 00    lea    0xe5b(%rip),%rdi        # 2008 <_IO_stdin_used 0x8>
11ad:       b8 00 00 00 00          mov    $0x0,