My Computer Systems course uses Bryant and O’Hallaron’s Computer Systems: A Programmer’s Perspective. The following is a fun homework assignment from our introduction to Assembly. The language and example assembly is copied directly from the text.
int decode2(int x, int y, int z);
is compiled into IA32 assembly code. The body of the code is as follows:
x at %ebp+8, y at %ebp+12, z at %ebp+16
1 movl 12(%ebp), %edx
2 subl 16(%ebp), %edx
3 movl %edx, %eax
4 sall $31, %eax
5 sarl $31, %eax
6 imull 8(%ebp), %edx
7 xorl %edx, %eax
Parameters x, y, and z are stored at memory locations with offsets 8, 12, and 16 relative to the address in register %ebp. The code stores the return value in register %eax.
Write C code for decode2 that will have an effect equivalent to our assembly code:
int decode2(int x, int y, int z)
{
int n = y - z; // 1-2
int m = (n << 31) >> 31; // 3-5
return (n * x) ^ m; // 6-7
}
Compiling the above code with gcc -O1 -c decode2.c and then running objdump -d decode2.o yields:
decode2.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <decode2>:
0: 29 d6 sub %edx,%esi
2: 89 f0 mov %esi,%eax
4: c1 e0 1f shl $0x1f,%eax
7: c1 f8 1f sar $0x1f,%eax
a: 0f af f7 imul %edi,%esi
d: 31 f0 xor %esi,%eax
f: c3 retq
For the following assembly, which has the exact same structure as exercise 3.54:
1 movl 16(%ebp), %edx
2 movl 12(%ebp), %eax
3 addl 8(%ebp), %eax
4 addl %eax, %eax
5 leal (%edx,%edx,2), %edx
6 cmpl %edx, %eax
7 setg %al
8 movzbl %al, %eax
The equivalent C code as a function, foo:
int foo(int x, int y, int z)
{
y += x + y; // 2-4
x *= 3; // 5
return y > x; // 6-8
}
Compiling the above code with gcc -01 -c foo.c on and then running objdump -d foo.o yields:
foo.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <foo>:
0: 8d 34 77 lea (%rdi,%rsi,2),%esi
3: 8d 3c 7f lea (%rdi,%rdi,2),%edi
6: 39 fe cmp %edi,%esi
8: 0f 9f c0 setg %al
b: 0f b6 c0 movzbl %al,%eax
e: c3 retq