Question

2) Draw a line between each of the IA32 assembler routines on the left and its equivalent C function on the right. (If there is no matching C function, do not draw a line.):

pushl %ebp movl %esp, %ebp movl 8(%ebp), %edx movl 12(%ebp), %eax cmpl %edx, %eax jle .L1 movl %edx, %eax .L1: leave ret pushl %ebp movl %esp, %ebp movl 8(%ebp), %edx movl (%edx), %edx movl 12(%ebp), %eax movl (%eax), %eax cmpl %edx, %eax jle .L1 movl %edx, %eax .L1: leave ret pushl %ebp movl %esp, %ebp movl 8(%ebp), %eax movl 12(%ebp), %edx cmpl %edx, %eax jge .L1 subl %edx, %eax .L1: leave ret pushl %ebp movl %esp, %ebp movl 8(%ebp), %edx movl 12(%ebp), %eax cmpl %edx, %eax jge .L1 movl %edx, %eax .L1: leave ret

int fun1(int a, int b) { if (a < b) return a-b; else return a; } int fun2(int a, int b) { if (a > b) return a; else return b; } int fun3(int *a, int *b) { if (*a < *b) return *a; else return *b; } int fun4(int a, int b) { if (a < b) return a; else return b; }

(b) Consider the following IA32 assembly language code fragment:

       .data
       .align 4
A:     .long 10, 20, 30, 40, 50
       .text
main: <code>

Determine the decimal value stored in register %eax if <code> in the above code fragment is replaced by each of the following:

i) movl $A, %ebx
   movl 4(%ebx), %eax

ii) movl $2, %ecx
   movl A(,%ecx, 4), %eax

iii)  movl $24, %eax
   sarl $2, %eax

iv) movl $4, %ecx
   leal 4(%ecx,%ecx,4), %eax

Answer 1

SOLUTION:

2)

(b)

(i)

movl $A, %ebx ;Move the address of A into register %eax

movl 4(%ebx), %eax ;Add 4 byte (one memory unit in 'Long' type ) to address stored in %ebx and move the contents of this address into %eax.

Value stored in %eax is 20

(ii)

movl $2, %ecx ;Move number 2 into register %ecx

movl A( , %ecx, 4), %eax ;Multiply content on %ecx with 4. Add the result to the address of A (struct_base) This gives the resultant address and move the contents of this address into %eax.

Value stored in %eax is 30

(iii)

movl $24, %eax ;Move number 24 into register %eax

sarl $2, %eax ;Arithmetic right shift to content in %eax for 2 times.

ie, (24)10 = (0001 1000)2

First shift -> 00001100

Second shift -> (00000110)2 = (6)10

Value stored in %eax is 6

(iv)

movl $4, %ecx ;Move number 4 into register %ecx

leal 4( %ecx, %ecx, 4), %eax ;Multiply content on %ecx with 4, add this with content on %ecx, and add 4 to it. This gives the resultant address

Resultant Address = (4 * %ecx) + %ecx + 4 = (4 * 4) + 4 + 4

Value stored in %eax is 24

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