We have Function callee is called by caller as below left and there assembly cod

Question

We have Function callee is called by caller as below left and there assembly code on the right: caller: push %ebp %esp, %ebp sex18,%esp mov sub mov xc(%ebp),%eax mov mov mov call 0x80483ed %eax,8x4(%esp) 0x8(%ebp),%eax %eax,(%esp) void caller(int x, int y) int z; z - callee(x,y); mov sh1l mov leave ret %eax,-9x4(%ebp) $0x2, -0x4(Kebp) -9x4(%ebp),%eax return; int callee(int x, int y) callee: if (x>y) push %ebp return x %esp, %ebp 0x8(%ebp),%eax mov mov return y; jle 8x80483fd(callee+16> mov 0x8(%ebp),%eax jmp 0x8048400 mov xc(%ebp),%eax pop ret %ebp When caller starts executing(before push %ebp), we have %esp: 0xbffff13c %ebp: Oxbffff1a8 What value does %esp get when callee starts executing?

Explanation / Answer

Answer is as follows :

According to given information, before the caller part, the %esp contians 0xbffff18c, in this first %esp is updated at instruction mov %esp,%ebp i.e. value of %ebp is transfer to %esp , means now %esp contains 0xbffff1a8.

After that %esp is updated at instruction sub $0x18,%esp, means it subtract the contents of %esp with 18 and update value of %esp. Here 0x18 is decimal value due to $ sign. So we convert it to hexadecimal i.e. 12 and 12 is equal to C.

0xbffff1a8 - 0xC = BFFFF196

in instruction mov 0xc(%ebp),eax , 0xbffff1a8 + c = 0xbffff19c i.e. value of eax

in instruction mov 0x8(%ebp),eax,  0xbffff1a8 + 8 = 0xbffff130

So after getting subtract we get 0xbffff19c - 0xbffff130 = 0xbffff16c

So when callee function starts execution the %esp contains BFFFF16C i.e. option 3

if there is any query please ask in comments..

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