在Linux上将C程序转换为Assembly（x86_64）

Question

I'm a newbie to assembly programming (especially x86_64).I have translated the following c program to assembly.

I understood most of the code but I'm not able to interpret the 2 Instructions I've commented out .

* Isn't the stmnt 1 corrupting contents of rax register by copying 42 to its lower word ?

note: as per the logic stmnt 1 should increment 42 in location whose address is stored in rax

addl $1,(%rax) 

could some one explain those 2 instructions .

temp.c

    #include<stdio.h>

    int main()
    {
        int a=42,*ptr;

        ptr = &a;
        (*ptr)++;

        return 0;
    }

temp.s

        .file   "temp.c"
        .text
        .globl  main
        .type   main, @function
    main:
    .LFB0:
        .cfi_startproc
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq    %rsp, %rbp
        .cfi_def_cfa_register 6
        subq    $32, %rsp
        movq    %fs:40, %rax
        movq    %rax, -8(%rbp)
        xorl    %eax, %eax
        movl    $42, -20(%rbp)
        leaq    -20(%rbp), %rax
        movq    %rax, -16(%rbp)
        movq    -16(%rbp), %rax  
        movl    (%rax), %eax         ;  stmt 1
        leal    1(%rax), %edx        ;  stmt 2
        movq    -16(%rbp), %rax
        movl    %edx, (%rax)
        movl    $0, %eax
        movq    -8(%rbp), %rcx
        xorq    %fs:40, %rcx
        je  .L3
        call    __stack_chk_fail
    .L3:
        leave
        .cfi_def_cfa 7, 8
        ret
        .cfi_endproc
    .LFE0:
        .size   main, .-main
        .ident  "GCC: (Ubuntu 5.3.1-14ubuntu2.1) 5.3.1 20160413"
        .section    .note.GNU-stack,"",@progbits

Answer 1

Yes, stmt1 is overwriting rax , but that is not a problem. stmt2 is then doing edx = eax + 1 in a tricky way using lea . The incremented value in edx is written back into memory at movl %edx, (%rax) . This means the (*ptr)++; has been split into multiple steps like this:

rax = ptr;     /* load pointer */
eax = *rax;    /* fetch current value */
edx = eax + 1; /* calculate new value */
rax = ptr;     /* load pointer again */
*rax = edx;    /* write new value */

Yes, addl $1,(%rax) could have been used. You probably don't have optimization enabled which is why you see inefficient code.

With optimization enabled, the following code:

void foo(int* ptr)
{
    (*ptr)++;
}

produces this assembly:

    addl    $1, (%rdi)
    ret

And this:

void foo(int* ptr)
{
    *ptr = 42;
    (*ptr)++;
}

gives:

    movl    $43, (%rdi)
    ret