The code in c: 1 int swap_n_add(int *xp, int *yp) { 2 int x = *xp; 3 int y = *yp

Question

The code in c:

1 int swap_n_add(int *xp, int *yp) {

  2     int x = *xp;

  3     int y = *yp;

  4

  5     *xp = y;

  6     *yp = x;

  7     return x + y;

  8 }

The same code in assembly

Group the assembly code instructions with their corresponding statements in c

above to the best of your ability.

db-pedaS disas ump of assembler code for function swap-n-add: 0x080483ed : push %ebp 0x080483ee : mov xesp.xebp 0x080483f0 : sub $0x10,%esp 0x080483f3 (+6): mov 0x8(%ebp),%eax 0x080483f6 : mov (zeax mov [%eax),%eax 0x080483f8(+11> : mov %eax,-0x8(%ebp) 0x080483fb(+14) : mov 0xcC%ebp),%eax 0x080483fe(+17>: mov [%eax),%eax 0x08048400(+19>: mov %eax,-0x4(%ebp) 0x08048403(+22> : mov 0x6(%ebp),%eax m0 U 0x08048406(+25> : mov -0x4(%ebp),%eax m0U 0x08048409(+28>: mov %eax,(xeax) 0x0804840b(+30 > : mov 0xcC%ebp),%eax 0x0804840e(+33) : mov -0x8(%ebp),%eax 0x08048411(+36 > : mov %edx,(zeax) m0U > 0x08048413(+38>: mov -0x4(%ebp),%eax 0x08048416(+41> : mov -0x6(%ebp),%edx 0x08048419(+44) : add xedx,xeax 0x0804841b(+46): leave 0x0804841c: ret nd of assembler dump. >>>>>14792580368-467d r01369-11122233334444 0 0 0 1 1 1 1 1 1 44444 a 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 s p0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Explanation / Answer

Matching the assembly code with C statements.

Push will put the value in stack and then it will move to register

So                              int swap_n_add(int *xp, int *yp) {   =      push   %ebp
                                                                                            mov    %esp, %esp
                                                                                            sub     $0x10, %esp

Then                           int x = *xp;       mov 0x8(%ebp),%eax
                                                         mov (%eax), %eax
                                                         mov %eax, -0x8(%ebp)
                                                         mov 0xc(%ebp), %eax

    int y = *yp;       (%eax), %eax
                                                         %eax,-0x4(%ebp)
                                                         0x8(%ebp), %eax
                                                         -0x4(%ebp), %edx

                                  *xp = y;           mov %edx, (%eax)
                                                        -0x(%ebp),%edx
                                                          %edx, (%eax)

    *yp = x;             mov %edx, (%eax)
                                                          -0x4(%ebp), %eax
                                                           -0x8(%ebp),%edx

       return x + y;                                   add %edx, %eax
                                                            leave
                                                            ret

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