CS350 Intro Computer Systems Homework

Homework 5 on Control

1. Fill the table with 0's and 1's for four flags: CF (carry flag), SF (sign or negative number flag), ZF (zero flag), and OF (positive/negative overflow):

   	Number Type	%rdi/edi/di	%rsi/esi/si	Instruction	CF	SF	ZF	OF
   (a)	Unsigned	0xFFFE	0x4	addw %di, %si
   (b)	Unsigned	0xFFFE	0x4	addl %edi, %esi
   (c)	Signed two's complement	0xFFFE	0x2	addw %di, %si
   (d)	Signed two's complement	0xFFFE	0x2	addl %edi, %esi
   (e)	Signed two's complement	0xFFFFFFFF	0x80000000	addl %edi, %esi
   (f)	Signed two's complement	0xFFFF	-0xFFFF	subl %si, %di
   (g)	Signed two's complement	0xFFFFFFFE	0x7FFFFFFE	subl %esi, %edi
   (h)	Unsigned	0xF	0xFF	shlq 64, %rdi

2. In the assembly code below, jump targets are encoded in PC-relative form, in other words, target addresses are computed relative to the memory address of the next instruction. Following the Practice problem 3.15 in page 330, fill the table with target addresses by replacing aaaaaa-eeeeee with memory addresses:

   	Address	Instructions in Hexa	---	Assembly Instructions
   (a)	ab1234:	74 08		je aaaaaa
   	ab1236:	48 89 d0		mov %rdx,%rax
   (b)	abcdef:	7c 07		jl bbbbbb
   	abcdf1:	48 39 f7		cmp %rsi,%rdi
   (c)	cccccc:	7d 11		jge 0x123456
   	dddddd:	48 85 ab		test %rdi,%rdi
   (d)	ab01f0:	7f 2f ff ff		jg eeeeee
   	ab01f4:	48 39 d6		mov %rdx,%rsi

3. This problem checks if you followed the reverse logic usage for jumping to else part. Fill in the C code for the assembly. Again use the jump-to-else-part reverse logic we used in class and also by following the practice problem 3.18 of page 213.

   reverse_logic:
   	cmpq	%rsi, %rdi
   	jge	.L3
   	cmpq	%rdx, %rdi
   	jle	.L4
   	movq	%rdx, %rax
   	subq	%rdi, %rax
   	ret
   .L4:
   	leaq	(%rdi,%rdx), %rax
   	ret
   .L3:
   	cmpq	%rdx, %rsi
   	jle	.L7
   	movq	%rdx, %rax
   	subq	%rsi, %rax
   	ret
   .L7:
   	leaq	(%rsi,%rdx), %rax
   	ret
   
   
   long reverse_logic(long x, long y, long z)
   {
       long result;
       ...
       ...
       return result;
   }
     

4. Do problem 3.60.
5. This problem checks again if you followed the jump to middle logic used for a while loop. Fill in the C code for the assembly below. Again use the reverse logic we used in class and also following the practice problem 3.18 of page 213.

   
   loop_while_hw5:
   .LFB0:
   	movl	$1, %eax
   	jmp	.L2
   .L3:
   	movq	%rdi, %rdx
   	subq	%rsi, %rdx
   	addq	%rdx, %rax
   	addq	$1, %rsi
   .L2:
   	cmpq	%rdi, %rsi
   	jl	.L3
   	rep ret
   
   long loop_while_hw5(long a, long b)
   {
       long result = _____________;
       while (___________) {
   	result = _________________________;
   	b = ______________________________;
       }
       return result;
   }
     

6. For a C function switch_hw5 below and the assembly code along with a jump table, fill in the missing parts of the C code. See page 233 and Practice Problem 3.31 on how the labels in case statements and the Jump Table can be related.

   (a) C Code
   void switch_hw5(long a, long b, long c, long *dest)
   {
       long val;
       switch(a) {
       case _____:
   	val = _______________________;
   	break;
       case _____:
   	c = _______________________;
   	/* Fall through */
       case _____:
   	val = _______________________;
   	break;
       case _____:
       case _____:
   	val = _______________________;
   	break;
       default:
   	val = _______________________;
       }
       return val;
   }
       
   (b) Assembly Code
   switch_hw5:
   .L3:
   	movq	%rdx, %rax
   	subq	%rdi, %rax
   	ret
   .L5:
   	movq	%rsi, %rdx
   	salq	$4, %rdx
   	addq	%rsi, %rdx
   .L6:
   	movq	%rdx, %rax
   	xorb	$-1, %al
   	ret
   .L7:
   	leaq	(%rdx,%rsi), %rax
   	sarq	$4, %rax
   	ret
   .L2:
   	leaq	(%rdi,%rsi), %rax
   	ret
   (c) Jump Table
   .L4:
   	.quad	.L3
   	.quad	.L5
   	.quad	.L2
   	.quad	.L6
   	.quad	.L2
   	.quad	.L7
   	.quad	.L2
   	.quad	.L7