Question

For the C code 1 int loop while(int a, int b) int result 1; while (a < b) t result (atb); return result; Gcc generates the following assembly code: %ebp+8, mov1 movl movl %ebp+12 at 8(%ebp) ,%ecx 12(%ebp) ,%ebx $1,%eax a at b 5 jge L11 leal (%ebx, %ecx), Xodx 8 L12: inull %eax , %eax addl $1,%ecx addl $1, edx cmp1 %ecx , %ebx 9 10 12 13 J8 14 ·L11: .L12 In generating this code, occ makes an interesting transformation that, in effect, introduces a new program variable. A. Register %edx is initialized on line 6 and updated within the loop on line 11. Consider this to be a new program variable. Describe how it relates to the variables in the C code. B. Create a table of register usage for this function C. Annotate the assembly code to describe how it operates. D. Write a goto version of the function (in C) that mimics how the assembly code program operates.

Answer 1

Answer A: The %edx register initialized on line 6 is used to stor the sum of the inputs a and b, and it is updated on line 11 to increment the sum by 1 as the value of a is incremented through the loop.

Answer B & Answer C in table below:

				Register Table (input a's value at EBP + 8 and input b's value at EBP +12 )
Step no.	Operation	Operatnds	Description of operation	ECX	EBX	EAX	EDX
1	movl	8(%ebp), %ecx	/* Move value of 'a' into ECX. */	a
2	movl	12(%ebp), %ebx	/* Move value of 'b' into EBX. */	a	b
3	movl	$1, %eax	/* Move value 1 into EAX. */	a	b	1
4	cmpl	%ebx, %ecx	/* Compare the values in registrs EBX and ECX i.e. b and a respectively (a >= b) */	check(a > = b)
5	jge	.L11	/* If the contents of ECX are greater than or equal to the contents of EBX , jump to .L11. Else, continue next instruction. */	a	b	1
6	leal	(%ebx,%ecx), %edx	/* Add contents of EBX and ECX and store in EDX */	a	b	1	(a+b)
7	movl	$1, %eax	/* Move value 1 into EAX. */	a	b	1	(a+b)
8	.L12:		/* Label .L12 */	a	b	1	(a+b)
9	imull	%edx, %eax	/* Multiply contents of EDX with contents of EAX and store result in EAX */	a	b	(a+b+(n-1))^n	(a+b)
10	addl	$1, %ecx	/* Add 1 to contents of ECX and store in ECX */	a+n	b	(a+b+(n-1))^n	(a+b)
11	addl	$1, %edx	/* Add 1 to contents of EDX and store in EDX */	a+n	b	(a+b+(n-1))^n	(a+b)+n
12	cmpl	%ecx, %ebx	/* Compare the values in registrs ECX and EBX i.e. a and b respectively (b > a(incremented value)) */	check(b > a+n)
13	jg	.L12	/* If the contents of EBX are greater than the contents of ECX, jump to .L12. Else, continue next instruction i.e actually exit. */	a+n	b	(a+b+(n-1))^n	(a+b)+n
14	.L11:		/* Label .L11 */

Answer D:

int loop_while_mod(int a, int b)
{
   int result = 1;
   int temp = 0;
   if(a >= b) goto L11;
   temp = a+b;
   L12:
    result = temp*result;
    a++;
    temp++;
   if(b > a) goto L12;
    L11:
    return result;
}