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I observe some very weird performance for read and write access on Intel machine.
I wrote a C program that allocate an array first. The code of the program is at [1] ; You can compile it by running Make. (I don't use any compiling optimization.)
The sequence of the operations of the program is as follows:
allocate a char array;
init each element of array to be 1;
use clflush to flush the whole array from cache;
read each cache line of the array by using tmp = array[i];
(Do simple calculation after reading each cache line)
use clflush to flush the whole array from cache;
write each cache line of the array by using array[i] = tmp;
(Do the same simple calculation after reading each cache line)
I run the program on Intel(R) Xeon(R) CPU E5-1650 v2 # 3.50GHz (Haswell arch.) with turbo boost disabled.
The command I used to run the program is:
sudo ./rw-latency-test-compute 5210 10 1
I got the read latency for the whole array is 6670us, while the write latency for the whole array is 3518us.
The interesting part is
If I don't do any computation after I read/write a cache line, the read latency for the whole array is 2175us, while the write latency for the whole array is 3687us.
So doing computation seems speed up the execution... :-(
Do you have any suggestion/explanation on this weird performance?
The whole assembly code of the program can be found at [2].
The assembly code of the inner loop is as follows:
0000000000400898 <read_array>:
400898: 55 push %rbp
400899: 48 89 e5 mov %rsp,%rbp
40089c: 53 push %rbx
40089d: 48 83 ec 28 sub $0x28,%rsp
4008a1: 48 89 7d d8 mov %rdi,-0x28(%rbp)
4008a5: 48 89 75 d0 mov %rsi,-0x30(%rbp)
4008a9: c7 45 e8 00 00 00 00 movl $0x0,-0x18(%rbp)
4008b0: c7 45 e4 00 00 00 00 movl $0x0,-0x1c(%rbp)
4008b7: eb 58 jmp 400911 <read_array+0x79>
4008b9: b8 00 00 00 00 mov $0x0,%eax
4008be: e8 38 ff ff ff callq 4007fb <sw_barrier>
4008c3: 8b 45 e4 mov -0x1c(%rbp),%eax
4008c6: 48 98 cltq
4008c8: 48 03 45 d8 add -0x28(%rbp),%rax
4008cc: 0f b6 00 movzbl (%rax),%eax
4008cf: 88 45 ef mov %al,-0x11(%rbp)
4008d2: 0f be 45 ef movsbl -0x11(%rbp),%eax
4008d6: 89 c1 mov %eax,%ecx
4008d8: 03 4d e8 add -0x18(%rbp),%ecx
4008db: ba 01 80 00 80 mov $0x80008001,%edx
4008e0: 89 c8 mov %ecx,%eax
4008e2: f7 ea imul %edx
4008e4: 8d 04 0a lea (%rdx,%rcx,1),%eax
4008e7: 89 c2 mov %eax,%edx
4008e9: c1 fa 0f sar $0xf,%edx
4008ec: 89 c8 mov %ecx,%eax
4008ee: c1 f8 1f sar $0x1f,%eax
4008f1: 89 d3 mov %edx,%ebx
4008f3: 29 c3 sub %eax,%ebx
4008f5: 89 d8 mov %ebx,%eax
4008f7: 89 45 e8 mov %eax,-0x18(%rbp)
4008fa: 8b 55 e8 mov -0x18(%rbp),%edx
4008fd: 89 d0 mov %edx,%eax
4008ff: c1 e0 10 shl $0x10,%eax
400902: 29 d0 sub %edx,%eax
400904: 89 ca mov %ecx,%edx
400906: 29 c2 sub %eax,%edx
400908: 89 d0 mov %edx,%eax
40090a: 89 45 e8 mov %eax,-0x18(%rbp)
40090d: 83 45 e4 40 addl $0x40,-0x1c(%rbp)
400911: 8b 45 e4 mov -0x1c(%rbp),%eax
400914: 48 98 cltq
400916: 48 3b 45 d0 cmp -0x30(%rbp),%rax
40091a: 7c 9d jl 4008b9 <read_array+0x21>
40091c: b8 e1 0f 40 00 mov $0x400fe1,%eax
400921: 8b 55 e8 mov -0x18(%rbp),%edx
400924: 89 d6 mov %edx,%esi
400926: 48 89 c7 mov %rax,%rdi
400929: b8 00 00 00 00 mov $0x0,%eax
40092e: e8 3d fd ff ff callq 400670 <printf#plt>
400933: 48 83 c4 28 add $0x28,%rsp
400937: 5b pop %rbx
400938: 5d pop %rbp
400939: c3 retq
000000000040093a <write_array>:
40093a: 55 push %rbp
40093b: 48 89 e5 mov %rsp,%rbp
40093e: 53 push %rbx
40093f: 48 83 ec 28 sub $0x28,%rsp
400943: 48 89 7d d8 mov %rdi,-0x28(%rbp)
400947: 48 89 75 d0 mov %rsi,-0x30(%rbp)
40094b: c6 45 ef 01 movb $0x1,-0x11(%rbp)
40094f: c7 45 e8 00 00 00 00 movl $0x0,-0x18(%rbp)
400956: c7 45 e4 00 00 00 00 movl $0x0,-0x1c(%rbp)
40095d: eb 63 jmp 4009c2 <write_array+0x88>
40095f: b8 00 00 00 00 mov $0x0,%eax
400964: e8 92 fe ff ff callq 4007fb <sw_barrier>
400969: 8b 45 e4 mov -0x1c(%rbp),%eax
40096c: 48 98 cltq
40096e: 48 03 45 d8 add -0x28(%rbp),%rax
400972: 0f b6 55 ef movzbl -0x11(%rbp),%edx
400976: 88 10 mov %dl,(%rax)
400978: 8b 45 e4 mov -0x1c(%rbp),%eax
40097b: 48 98 cltq
40097d: 48 03 45 d8 add -0x28(%rbp),%rax
400981: 0f b6 00 movzbl (%rax),%eax
400984: 0f be c0 movsbl %al,%eax
400987: 89 c1 mov %eax,%ecx
400989: 03 4d e8 add -0x18(%rbp),%ecx
40098c: ba 01 80 00 80 mov $0x80008001,%edx
400991: 89 c8 mov %ecx,%eax
400993: f7 ea imul %edx
400995: 8d 04 0a lea (%rdx,%rcx,1),%eax
400998: 89 c2 mov %eax,%edx
40099a: c1 fa 0f sar $0xf,%edx
40099d: 89 c8 mov %ecx,%eax
40099f: c1 f8 1f sar $0x1f,%eax
4009a2: 89 d3 mov %edx,%ebx
4009a4: 29 c3 sub %eax,%ebx
4009a6: 89 d8 mov %ebx,%eax
4009a8: 89 45 e8 mov %eax,-0x18(%rbp)
4009ab: 8b 55 e8 mov -0x18(%rbp),%edx
4009ae: 89 d0 mov %edx,%eax
4009b0: c1 e0 10 shl $0x10,%eax
4009b3: 29 d0 sub %edx,%eax
4009b5: 89 ca mov %ecx,%edx
4009b7: 29 c2 sub %eax,%edx
4009b9: 89 d0 mov %edx,%eax
4009bb: 89 45 e8 mov %eax,-0x18(%rbp)
4009be: 83 45 e4 40 addl $0x40,-0x1c(%rbp)
4009c2: 8b 45 e4 mov -0x1c(%rbp),%eax
4009c5: 48 98 cltq
4009c7: 48 3b 45 d0 cmp -0x30(%rbp),%rax
4009cb: 7c 92 jl 40095f <write_array+0x25>
4009cd: b8 ee 0f 40 00 mov $0x400fee,%eax
4009d2: 8b 55 e8 mov -0x18(%rbp),%edx
4009d5: 89 d6 mov %edx,%esi
4009d7: 48 89 c7 mov %rax,%rdi
4009da: b8 00 00 00 00 mov $0x0,%eax
4009df: e8 8c fc ff ff callq 400670 <printf#plt>
4009e4: 48 83 c4 28 add $0x28,%rsp
4009e8: 5b pop %rbx
4009e9: 5d pop %rbp
4009ea: c3 retq
[1]https://github.com/PennPanda/rw-latency-test/blob/master/rw-latency-test-compute.c
[2] https://github.com/PennPanda/rw-latency-test/blob/2da88f1cccba40aba155317567199028b28bd250/rw-latency-test-compute.asm
Write is faster than read because if you read from RAM and use the value (that is, you don't just read and discard), the processor has to stall for the read at the point the value is used. However, write proceeds asynchronously and never stalls.
Related
This question already has answers here:
Binary Bomb - Phase 4
(1 answer)
having trouble with bomb lab phase 4 [closed]
(1 answer)
Closed 5 years ago.
I'm having a bit of trouble understanding the following assembly code for the bomb lab. Running through it so far, I've figured out that the answer is supposed to be two decimal values. If not it will explode the bomb. Then, function 4 is making sure that the first value inputted is between 0 and 30. (0x1e) Then it jumps to func4 where it does something to my number. I understand up to the
sub %esi,%eax
then I don't completely understand what is going on in the function. I've tried plugging in values and checking them in the registry but I still don't understand what is going in function 4.
My attempt at understanding it is that, first its setting registers equal to each other. Then its doing an arithmetic right shift by 31 (0x1f). Then its subtracting by 1 and adding it to eax.
So, from what I understand its taking a number. Subtracting it by 1 and then adding them together? Such that the simplified formula would be (x-1)*x ?
Func_4
00000000004010b2 <func4>:
4010b2: 55 push %rbp
4010b3: 48 89 e5 mov %rsp,%rbp
4010b6: 89 d0 mov %edx,%eax
4010b8: 29 f0 sub %esi,%eax
4010ba: 89 c1 mov %eax,%ecx
4010bc: c1 e9 1f shr $0x1f,%ecx
4010bf: 01 c8 add %ecx,%eax
4010c1: d1 f8 sar %eax
4010c3: 8d 0c 30 lea (%rax,%rsi,1),%ecx
4010c6: 39 f9 cmp %edi,%ecx
4010c8: 7e 0c jle 4010d6 <func4+0x24>
4010ca: 8d 51 ff lea -0x1(%rcx),%edx
4010cd: e8 e0 ff ff ff callq 4010b2 <func4>
4010d2: 01 c0 add %eax,%eax
4010d4: eb 15 jmp 4010eb <func4+0x39>
4010d6: b8 00 00 00 00 mov $0x0,%eax
4010db: 39 f9 cmp %edi,%ecx
4010dd: 7d 0c jge 4010eb <func4+0x39>
4010df: 8d 71 01 lea 0x1(%rcx),%esi
4010e2: e8 cb ff ff ff callq 4010b2 <func4>
4010e7: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax
4010eb: 5d pop %rbp
4010ec: c3 retq
Phase_4
00000000004010ed <phase_4>:
4010ed: 55 push %rbp
4010ee: 48 89 e5 mov %rsp,%rbp
4010f1: 48 83 ec 10 sub $0x10,%rsp
4010f5: 48 8d 4d fc lea -0x4(%rbp),%rcx
4010f9: 48 8d 55 f8 lea -0x8(%rbp),%rdx
4010fd: be 6d 2a 40 00 mov $0x402a6d,%esi
401102: b8 00 00 00 00 mov $0x0,%eax
401107: e8 a4 fb ff ff callq 400cb0 <__isoc99_sscanf#plt>
40110c: 83 f8 02 cmp $0x2,%eax
40110f: 75 0b jne 40111c <phase_4+0x2f>
401111: 8b 45 f8 mov -0x8(%rbp),%eax
401114: 83 e8 20 sub $0x20,%eax
401117: 83 f8 1e cmp $0x1e,%eax
40111a: 76 05 jbe 401121 <phase_4+0x34>
40111c: e8 b4 05 00 00 callq 4016d5 <explode_bomb>
401121: ba 3e 00 00 00 mov $0x3e,%edx
401126: be 20 00 00 00 mov $0x20,%esi
40112b: 8b 7d f8 mov -0x8(%rbp),%edi
40112e: e8 7f ff ff ff callq 4010b2 <func4>
401133: 83 f8 0e cmp $0xe,%eax
401136: 75 06 jne 40113e <phase_4+0x51>
401138: 83 7d fc 0e cmpl $0xe,-0x4(%rbp)
40113c: 74 05 je 401143 <phase_4+0x56>
40113e: e8 92 05 00 00 callq 4016d5 <explode_bomb>
401143: c9 leaveq
401144: c3 retq
I'm not understanding what the function below does. From what I gather, function 4 does something like (x+x)*2 or it does something like (high-low)/2 if a condition is reached. (I might be wrong on this). From reading the code, I also understood that in order to "defuse" the bomb. I need two decimal inputs, and the second one should be 14.
I'm stuck trying to figure out the first value, and trying to identify the correct formula to use in order to figure out the first value.
Function_4
00000000004010b2 <func4>:
4010b2: 55 push %rbp
4010b3: 48 89 e5 mov %rsp,%rbp
4010b6: 89 d0 mov %edx,%eax
4010b8: 29 f0 sub %esi,%eax
4010ba: 89 c1 mov %eax,%ecx
4010bc: c1 e9 1f shr $0x1f,%ecx
4010bf: 01 c8 add %ecx,%eax
4010c1: d1 f8 sar %eax
4010c3: 8d 0c 30 lea (%rax,%rsi,1),%ecx
4010c6: 39 f9 cmp %edi,%ecx
4010c8: 7e 0c jle 4010d6 <func4+0x24>
4010ca: 8d 51 ff lea -0x1(%rcx),%edx
4010cd: e8 e0 ff ff ff callq 4010b2 <func4>
4010d2: 01 c0 add %eax,%eax
4010d4: eb 15 jmp 4010eb <func4+0x39>
4010d6: b8 00 00 00 00 mov $0x0,%eax
4010db: 39 f9 cmp %edi,%ecx
4010dd: 7d 0c jge 4010eb <func4+0x39>
4010df: 8d 71 01 lea 0x1(%rcx),%esi
4010e2: e8 cb ff ff ff callq 4010b2 <func4>
4010e7: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax
4010eb: 5d pop %rbp
4010ec: c3 retq
Phase_4
00000000004010ed <phase_4>:
4010ed: 55 push %rbp
4010ee: 48 89 e5 mov %rsp,%rbp
4010f1: 48 83 ec 10 sub $0x10,%rsp
4010f5: 48 8d 4d fc lea -0x4(%rbp),%rcx
4010f9: 48 8d 55 f8 lea -0x8(%rbp),%rdx
4010fd: be 6d 2a 40 00 mov $0x402a6d,%esi
401102: b8 00 00 00 00 mov $0x0,%eax
401107: e8 a4 fb ff ff callq 400cb0 <__isoc99_sscanf#plt>
40110c: 83 f8 02 cmp $0x2,%eax
40110f: 75 0b jne 40111c <phase_4+0x2f>
401111: 8b 45 f8 mov -0x8(%rbp),%eax
401114: 83 e8 20 sub $0x20,%eax
401117: 83 f8 1e cmp $0x1e,%eax
40111a: 76 05 jbe 401121 <phase_4+0x34>
40111c: e8 b4 05 00 00 callq 4016d5 <explode_bomb>
401121: ba 3e 00 00 00 mov $0x3e,%edx
401126: be 20 00 00 00 mov $0x20,%esi
40112b: 8b 7d f8 mov -0x8(%rbp),%edi
40112e: e8 7f ff ff ff callq 4010b2 <func4>
401133: 83 f8 0e cmp $0xe,%eax
401136: 75 06 jne 40113e <phase_4+0x51>
401138: 83 7d fc 0e cmpl $0xe,-0x4(%rbp)
40113c: 74 05 je 401143 <phase_4+0x56>
40113e: e8 92 05 00 00 callq 4016d5 <explode_bomb>
401143: c9 leaveq
401144: c3 retq
I try to achieve stack smashing when I have only the executable file .
I use the objdump to get the assembly code for this source code :
#include<stdio.h>
#include<string.h>
void func(char *str) {
char buffer[24];
int *ret;
strcpy(buffer,str);
}
int main(int argc, char **argv) {
int x;
x = 0;
func(argv[1]);
x = 1;
printf("%d\n”,x);
}
at run time ./a,out (value)....I need to insert the (value ) in such away I insert the NOP in stack location and that last part of (value) is the address of my next instruction.
I have 40 byte before reaching the location that contain the return address of the fun() .
08048444 <func>:
8048444: 55 push %ebp
8048445: 89 e5 mov %esp,%ebp
8048447: 83 ec 48 sub $0x48,%esp
804844a: 8b 45 08 mov 0x8(%ebp),%eax
804844d: 89 45 d4 mov %eax,-0x2c(%ebp)
8048450: 65 a1 14 00 00 00 mov %gs:0x14,%eax
8048456: 89 45 f4 mov %eax,-0xc(%ebp)
8048459: 31 c0 xor %eax,%eax
804845b: 8b 45 d4 mov -0x2c(%ebp),%eax
804845e: 89 44 24 04 mov %eax,0x4(%esp)
8048462: 8d 45 dc lea -0x24(%ebp),%eax
8048465: 89 04 24 mov %eax,(%esp)
8048468: e8 eb fe ff ff call 8048358 <strcpy#plt>
804846d: 8b 45 f4 mov -0xc(%ebp),%eax
8048470: 65 33 05 14 00 00 00 xor %gs:0x14,%eax
8048477: 74 05 je 804847e <func+0x3a>
8048479: e8 fa fe ff ff call 8048378 <__stack_chk_fail#plt>
804847e: c9 leave
804847f: c3 ret
08048480 <main>:
8048480: 55 push %ebp
8048481: 89 e5 mov %esp,%ebp
8048483: 83 e4 f0 and $0xfffffff0,%esp
8048486: 83 ec 20 sub $0x20,%esp
8048489: c7 44 24 1c 00 00 00 movl $0x0,0x1c(%esp)
8048490: 00
8048491: 8b 45 0c mov 0xc(%ebp),%eax
8048494: 83 c0 04 add $0x4,%eax
8048497: 8b 00 mov (%eax),%eax
8048499: 89 04 24 mov %eax,(%esp)
804849c: e8 a3 ff ff ff call 8048444 <func>
80484a1: c7 44 24 1c 01 00 00 movl $0x1,0x1c(%esp)
80484a8: 00
80484a9: b8 90 85 04 08 mov $0x8048590,%eax
80484ae: 8b 54 24 1c mov 0x1c(%esp),%edx
80484b2: 89 54 24 04 mov %edx,0x4(%esp)
80484b6: 89 04 24 mov %eax,(%esp)
80484b9: e8 aa fe ff ff call 8048368 <printf#plt>
80484be: b8 00 00 00 00 mov $0x0,%eax
80484c3: c9 leave
80484c4: c3 ret
80484c5: 90 nop
80484c6: 90 nop
problem if I insert 00 its consider as (31) ASCII .How I can insert hex values.
... I hope the Que is clear
objdump -w -Mintel :
08048444 <func>:
8048444: 55 push ebp
8048445: 89 e5 mov ebp,esp
8048447: 83 ec 48 sub esp,0x48
804844a: 8b 45 08 mov eax,DWORD PTR [ebp+0x8]
804844d: 89 45 d4 mov DWORD PTR [ebp-0x2c],eax
8048450: 65 a1 14 00 00 00 mov eax,gs:0x14
8048456: 89 45 f4 mov DWORD PTR [ebp-0xc],eax
8048459: 31 c0 xor eax,eax
804845b: 8b 45 d4 mov eax,DWORD PTR [ebp-0x2c]
804845e: 89 44 24 04 mov DWORD PTR [esp+0x4],eax
8048462: 8d 45 dc lea eax,[ebp-0x24]
8048465: 89 04 24 mov DWORD PTR [esp],eax
8048468: e8 eb fe ff ff call 8048358 <strcpy#plt>
804846d: 8b 45 f4 mov eax,DWORD PTR [ebp-0xc]
8048470: 65 33 05 14 00 00 00 xor eax,DWORD PTR gs:0x14
8048477: 74 05 je 804847e <func+0x3a>
8048479: e8 fa fe ff ff call 8048378 <__stack_chk_fail#plt>
804847e: c9 leave
804847f: c3 ret
08048480 <main>:
8048480: 55 push ebp
8048481: 89 e5 mov ebp,esp
8048483: 83 e4 f0 and esp,0xfffffff0
8048486: 83 ec 20 sub esp,0x20
8048489: c7 44 24 1c 00 00 00 00 mov DWORD PTR [esp+0x1c],0x0
8048491: 8b 45 0c mov eax,DWORD PTR [ebp+0xc]
8048494: 83 c0 04 add eax,0x4
8048497: 8b 00 mov eax,DWORD PTR [eax]
8048499: 89 04 24 mov DWORD PTR [esp],eax
804849c: e8 a3 ff ff ff call 8048444 <func>
80484a1: c7 44 24 1c 01 00 00 00 mov DWORD PTR [esp+0x1c],0x1
80484a9: b8 90 85 04 08 mov eax,0x8048590
80484ae: 8b 54 24 1c mov edx,DWORD PTR [esp+0x1c]
80484b2: 89 54 24 04 mov DWORD PTR [esp+0x4],edx
80484b6: 89 04 24 mov DWORD PTR [esp],eax
80484b9: e8 aa fe ff ff call 8048368 <printf#plt>
80484be: b8 00 00 00 00 mov eax,0x0`
You could use ./a.out $(perl -e "print '\x97';") and replace \x97 by the hex you want to use.
If C, the end of string character is 0x00 (or '\0' if you prefer). So if you make your string exactly 39 characters long, then the 40th character will be the zero - and it will be in exactly the right place. There is no way to copy more than one zero in a C string - unless you use a function other than strcpy (for example, memcpy). But if you are relying on the argv[1] to be the source of your zero, then this is the only way. You could of course subtract something from the string before processing it - if you want, you could do
L = strlen(argv[1]);
for(int ii = 0; ii < L; ii++) if(argv[1][ii] == '0') argv[1][ii] = '\0';
This would turn every '0' into '\0'. But then you can't do a simple strcpy, you would have to do memcpy.
And you have to hope that you don't get a segfault for writing to memory you don't own…
I have the following from an objdump. This was C code compiled by gcc for an IA32.
08048e9a <my_func>:
8048e9a: 55 push %ebp
8048e9b: 89 e5 mov %esp,%ebp
8048e9d: 83 ec 48 sub $0x48,%esp
8048ea0: 89 5d f4 mov %ebx,-0xc(%ebp)
8048ea3: 89 75 f8 mov %esi,-0x8(%ebp)
8048ea6: 89 7d fc mov %edi,-0x4(%ebp)
8048ea9: 8d 5d d0 lea -0x30(%ebp),%ebx
8048eac: 89 5c 24 04 mov %ebx,0x4(%esp)
8048eb0: 8b 45 08 mov 0x8(%ebp),%eax
8048eb3: 89 04 24 mov %eax,(%esp)
8048eb6: e8 52 04 00 00 call 804930d <read_num>
8048ebb: 8d 7d dc lea -0x24(%ebp),%edi
8048ebe: be 00 00 00 00 mov $0x0,%esi
8048ec3: 8b 03 mov (%ebx),%eax
8048ec5: 3b 43 0c cmp 0xc(%ebx),%eax
8048ec8: 74 05 je 8048ecf <my_func+0x35>
8048eca: e8 fc 03 00 00 call 80492cb <other_func>
8048ecf: 03 33 add (%ebx),%esi
I am interested in finding out the values being compared on line 8048ec5 In gdb I can step to this line and I can read %eax just fine from info registers but how can I read 0xc(%ebx)? This means 0xc offset from %ebx or 0xc + %ebx?
It refers to the 32-bit value at the address %ebx + 0xc in memory.
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
08048544 <compare_password>:
8048544: 55 push %ebp
8048545: 89 e5 mov %esp,%ebp
8048547: 83 ec 38 sub $0x38,%esp
804854a: 8b 45 0c mov 0xc(%ebp),%eax
804854d: 89 45 d4 mov %eax,-0x2c(%ebp)
8048550: 65 a1 14 00 00 00 mov %gs:0x14,%eax
8048556: 89 45 f4 mov %eax,-0xc(%ebp)
8048559: 31 c0 xor %eax,%eax
804855b: c7 45 e4 00 00 00 00 movl $0x0,-0x1c(%ebp)
8048562: c7 45 e0 00 00 00 00 movl $0x0,-0x20(%ebp)
8048569: eb 22 jmp 804858d <compare_password+0x49>
804856b: 8b 45 e0 mov -0x20(%ebp),%eax
804856e: 03 45 d4 add -0x2c(%ebp),%eax
8048571: 0f b6 10 movzbl (%eax),%edx
8048574: 8b 45 e0 mov -0x20(%ebp),%eax
8048577: 05 44 a1 04 08 add $0x804a144,%eax
804857c: 0f b6 00 movzbl (%eax),%eax
804857f: 31 c2 xor %eax,%edx
8048581: 8d 45 ea lea -0x16(%ebp),%eax
8048584: 03 45 e0 add -0x20(%ebp),%eax
8048587: 88 10 mov %dl,(%eax)
8048589: 83 45 e0 01 addl $0x1,-0x20(%ebp)
804858d: 83 7d e0 09 cmpl $0x9,-0x20(%ebp)
8048591: 7e d8 jle 804856b <compare_password+0x27>
8048593: c7 45 e0 00 00 00 00 movl $0x0,-0x20(%ebp)
804859a: eb 2c jmp 80485c8 <compare_password+0x84>
804859c: 8b 55 08 mov 0x8(%ebp),%edx
804859f: 89 d0 mov %edx,%eax
80485a1: c1 e0 02 shl $0x2,%eax
80485a4: 01 d0 add %edx,%eax
80485a6: 01 c0 add %eax,%eax
80485a8: 03 45 e0 add -0x20(%ebp),%eax
80485ab: 05 e0 a0 04 08 add $0x804a0e0,%eax
80485b0: 0f b6 10 movzbl (%eax),%edx
80485b3: 8d 45 ea lea -0x16(%ebp),%eax
80485b6: 03 45 e0 add -0x20(%ebp),%eax
80485b9: 0f b6 00 movzbl (%eax),%eax
80485bc: 38 c2 cmp %al,%dl
80485be: 75 04 jne 80485c4 <compare_password+0x80>
80485c0: 83 45 e4 01 addl $0x1,-0x1c(%ebp)
80485c4: 83 45 e0 01 addl $0x1,-0x20(%ebp)
80485c8: 83 7d e0 09 cmpl $0x9,-0x20(%ebp)
80485cc: 7e ce jle 804859c <compare_password+0x58>
80485ce: 83 7d e4 08 cmpl $0x8,-0x1c(%ebp)
80485d2: 7e 07 jle 80485db <compare_password+0x97>
80485d4: b8 01 00 00 00 mov $0x1,%eax
80485d9: eb 05 jmp 80485e0 <compare_password+0x9c>
80485db: b8 00 00 00 00 mov $0x0,%eax
80485e0: 8b 55 f4 mov -0xc(%ebp),%edx
80485e3: 65 33 15 14 00 00 00 xor %gs:0x14,%edx
80485ea: 74 05 je 80485f1 <compare_password+0xad>
80485ec: e8 2f fe ff ff call 8048420 <__stack_chk_fail#plt>
80485f1: c9 leave
80485f2: c3 ret
080485f3 <main>:
80485f3: 55 push %ebp
80485f4: 89 e5 mov %esp,%ebp
80485f6: 83 e4 f0 and $0xfffffff0,%esp
80485f9: 83 ec 30 sub $0x30,%esp
80485fc: 65 a1 14 00 00 00 mov %gs:0x14,%eax
8048602: 89 44 24 2c mov %eax,0x2c(%esp)
8048606: 31 c0 xor %eax,%eax
8048608: c7 44 24 04 00 00 00 movl $0x0,0x4(%esp)
804860f: 00
8048610: 8d 44 24 10 lea 0x10(%esp),%eax
8048614: 89 04 24 mov %eax,(%esp)
8048617: e8 f4 fd ff ff call 8048410 <gettimeofday#plt>
804861c: 8b 54 24 10 mov 0x10(%esp),%edx
8048620: 8b 44 24 14 mov 0x14(%esp),%eax
8048624: 0f af c2 imul %edx,%eax
8048627: 89 04 24 mov %eax,(%esp)
804862a: e8 21 fe ff ff call 8048450 <srand#plt>
804862f: e8 3c fe ff ff call 8048470 <rand#plt>
8048634: 89 44 24 18 mov %eax,0x18(%esp)
8048638: 8b 4c 24 18 mov 0x18(%esp),%ecx
804863c: ba 67 66 66 66 mov $0x66666667,%edx
8048641: 89 c8 mov %ecx,%eax
8048643: f7 ea imul %edx
8048645: c1 fa 02 sar $0x2,%edx
8048648: 89 c8 mov %ecx,%eax
804864a: c1 f8 1f sar $0x1f,%eax
804864d: 29 c2 sub %eax,%edx
804864f: 89 d0 mov %edx,%eax
8048651: c1 e0 02 shl $0x2,%eax
8048654: 01 d0 add %edx,%eax
8048656: 01 c0 add %eax,%eax
8048658: 89 ca mov %ecx,%edx
804865a: 29 c2 sub %eax,%edx
804865c: 89 d0 mov %edx,%eax
804865e: 89 44 24 18 mov %eax,0x18(%esp)
8048662: 8b 54 24 18 mov 0x18(%esp),%edx
8048666: 89 d0 mov %edx,%eax
8048668: c1 e0 02 shl $0x2,%eax
804866b: 01 d0 add %edx,%eax
804866d: 01 c0 add %eax,%eax
804866f: 8d 90 60 a0 04 08 lea 0x804a060(%eax),%edx
8048675: b8 c0 87 04 08 mov $0x80487c0,%eax
804867a: 89 54 24 04 mov %edx,0x4(%esp)
804867e: 89 04 24 mov %eax,(%esp)
8048681: e8 7a fd ff ff call 8048400 <printf#plt>
8048686: b8 da 87 04 08 mov $0x80487da,%eax
804868b: 8d 54 24 22 lea 0x22(%esp),%edx
804868f: 89 54 24 04 mov %edx,0x4(%esp)
8048693: 89 04 24 mov %eax,(%esp)
8048696: e8 e5 fd ff ff call 8048480 <__isoc99_scanf#plt>
804869b: 8d 44 24 22 lea 0x22(%esp),%eax
804869f: 89 44 24 04 mov %eax,0x4(%esp)
80486a3: 8b 44 24 18 mov 0x18(%esp),%eax
80486a7: 89 04 24 mov %eax,(%esp)
80486aa: e8 95 fe ff ff call 8048544 <compare_password>
80486af: 89 44 24 1c mov %eax,0x1c(%esp)
80486b3: 83 7c 24 1c 01 cmpl $0x1,0x1c(%esp)
80486b8: 75 0e jne 80486c8 <main+0xd5>
80486ba: c7 04 24 dd 87 04 08 movl $0x80487dd,(%esp)
80486c1: e8 6a fd ff ff call 8048430 <puts#plt>
80486c6: eb 0c jmp 80486d4 <main+0xe1>
80486c8: c7 04 24 f2 87 04 08 movl $0x80487f2,(%esp)
80486cf: e8 5c fd ff ff call 8048430 <puts#plt>
80486d4: 8b 54 24 2c mov 0x2c(%esp),%edx
80486d8: 65 33 15 14 00 00 00 xor %gs:0x14,%edx
80486df: 74 05 je 80486e6 <main+0xf3>
80486e1: e8 3a fd ff ff call 8048420 <__stack_chk_fail#plt>
80486e6: c9 leave
80486e7: c3 ret
80486e8: 90 nop
80486e9: 90 nop
80486ea: 90 nop
80486eb: 90 nop
80486ec: 90 nop
80486ed: 90 nop
80486ee: 90 nop
80486ef: 90 nop
Ok, learning assembly code from scratch will take some time and effort, but there's no harm in getting the basics.
Each line of this output contains three parts:
The offset in the file where that piece of code is (in hex)
The bytes that make up that piece of code (each in hex, again)
The assembly language form of that code (basically reverse-translated from the bytes).
You can generally read the flow of the program through the last column. Instructions like JMPs will refer to other locations, which may or may not be nearby in the code. They may be presented in a labelled form like:
jmp 804858d <compare_password+0x49>
That says, jump to offset 0x804858d, so you can find that value in the first column. The label says that this is offset 0x49 after compare_password.
If you don't know what most of the instructions do, well, they mostly move, combine and compare individual words of memory and register. Even when you learn what each code does, understanding what it does in the context of this particular program can be hard. And you generally need to know the location of other important pieces of data when the program will be running to know what the effect will be.
There are lots of resources for learning computer programming at the level of debugging, assembly language and dissassembly, but I will leave it to others to refer you. If you really want to learn, a good way is to write your own simple program in C, and compile it to assembly. Then compare the C and assembly output side-by-side, figuring out how the C statements have been translated into instructions.