Simple y86 do-nothing program halt

0
[terminal]$ ./yis a.yo
Stopped in 11 steps at PC = 0x2c.  Status 'HLT', CC Z=1 S=0 O=0
Changes to registers:
%eax:   0x00000000  0x00000004
%ebx:   0x00000000  0x00000005
%esp:   0x00000000  0x00000023
%ebp:   0x00000000  0x0000002f

Changes to memory:
0x0020: 0x2c803fa0  0x27803fa0
0x0024: 0x20000000  0x05000000
0x0028: 0x905fb054  0x04000000
0x002c: 0x45205fa0  0x37000000
0x0030: 0x32200120  0x0d000000
0x0034: 0x00905fb0  0x00000000
[terminal]$ cat a.yo
  0x000:              |     .pos 0
  0x000:              | init:
  0x000: 30f437000000 |     irmovl Stack, %esp
  0x006: 2045         |     rrmovl %esp, %ebp
  0x008: 800e000000   |     call Main
  0x00d: 00           |     halt
                      | 
  0x00e:              | Main:
  0x00e: a05f         |     pushl %ebp
  0x010: 2045         |     rrmovl %esp,%ebp
                      | 
  0x012: 30f004000000 |     irmovl $4,%eax
  0x018: a00f         |     pushl %eax
  0x01a: 30f305000000 |     irmovl $5,%ebx
  0x020: a03f         |     pushl %ebx
  0x022: 802c000000   |     call Sum
                      | 
  0x027: 2054         |     rrmovl %ebp,%esp
  0x029: b05f         |     popl %ebp
  0x02b: 90           |     ret
                      | 
  0x02c:              | Sum:
  0x02c: a05f         |     pushl %ebp        #right here
  0x02e: 2045         |     rrmovl %esp,%ebp
                      | 
  0x030: 2001         |     rrmovl %eax,%ecx
  0x032: 2032         |     rrmovl %ebx,%edx
                      | 
  0x034: b05f         |     popl %ebp
  0x036: 90           |     ret
  0x037:              | Stack:

I'm using the yas simulator to compile and run my y86 assembly. I'm trying to understand why the program would halt at 0x2c, It does nothing except send 2 constants into a function (which aren't even used) that just moves the values that would be parameter into other registers.

assembly
y86
asked on Stack Overflow Oct 14, 2013 by SGM1 • edited Oct 14, 2013 by SGM1

1 Answer

0

You're overwriting parts of the code with stack contents.

You initialize esp to Stack, which is 0x37. By the time you reach Sum you'll have 5 DWORDs on the stack (3 * pushl, and 2 * call). Five DWORDs is 20 bytes (0x14), and 0x37 - 0x14 is 0x23 (remember, the stack grows backwards in memory). You can see this is the "Changes to registers" list: %esp: 0x00000000 0x00000023.

As you can see in the "Changes to memory" list, the DWORD at 0x2C (which is where Sum starts) has changed from 0x45205fa0 to 0x37000000. Assuming little-endian, that means the byte at address 0x2C is 0x00, which equals HALT.

answered on Stack Overflow Oct 14, 2013 by Michael

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