Question

Assembly

Please answer the following above, compile and single step through the program, writing down the value of the destination for each instruction.

1. Using Visual Studio on the system, create a project using the code given below unsigned char short int int gArray 0x09, 0xFA, 0x5A, 0x18, 0x48, 0xAC, 0xD4, 0x71 ; cAr rays 1 [ ] = { 0:09, 0xfa, Ox5A, Ox18, 0x48, OxAC, OxD4, 0x71 }; gArray! [ ] = { Ox09, OxFA, Ox5A, Ox18, 0x48, OxAC, OxD4, 0x71 }; int test ) asm mov al, gArray lea esi, gArray mov dl, byte ptr [esi] mov eax, dword ptr [esi+2] mov al,gArray[5] esir eax *1 hop mov ax, gArraySI lea esi, gArraysI mov dl, byte ptr [esi] mov edx, dword ptr [esi+ mov ax,gArraySI[5] esi= edx- *2 hop mov eax, gArrayI lea esi, gArrayI mov dl, byte ptr [esi] mov eax, dword ptr [esi+2] mov eax, gArrayll eax= esi= eax *3 nop return 0 int main( int argc, char *argv) test )

Answer 1

mov al,gArray lea esi, gArray mov dl, byte ptr esi] al - s 2924604 eds, drord pts fea4t213568048218.. *1 212 mov al,gArrayt5] al - nop mov ax,gArraysI lea esi,gArraysI mov dl, byte ptr esi] mov eax, dword ptr [e51+2] mov ax, gArraySI[5] ах 2924612 esi= edx 2* 6144 ax- nop mov eax, gArrayl lea esi,gArrayI mov dl, byte ptr esi] mov eax, dword ptr [esi+2] mov eax, gArrayII5] eax 2924628 esi edx x 1569949440 nop return 0;

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