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Hard to understand GCC assembly instructions to calculate conditional jump


Hard to understand GCC assembly instructions to calculate conditional jump

By : user2949264
Date : November 16 2020, 06:23 AM
I hope this helps . It's jge because the compiler swapped both the operands and the condition. It's doing global_count >= a. Notice that 0x10(%ebp) is a and %eax is global_count. To complicate matters further, at&t syntax itself has operands reversed.
code :


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Conditional Jump instructions?

Conditional Jump instructions?


By : Jim Lu
Date : March 29 2020, 07:55 AM
hope this fix your issue
Many processors have conditional jump instructions only for conditional jumps with a limited range. For example, the target of the jump may not be further than 128 bytes away from the current program counter.
code :
jnz distant_label
[other instructions]
distant_label:
jz nearby_label
jmp distant_label
nearby_label:
[other instructions]
distant_label:
What are the conditional jump instructions for Go's assembler?

What are the conditional jump instructions for Go's assembler?


By : user3616206
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I'm answering this so I don't lose the information, and so other people don't have to go through the same sleuthing game as me. Looking at optab.c and the x86 jumps we can match up the instruction encodings to solve the puzzle.
code :
JCC     JAE
JCS     JB
JCXZL   JECXZ
JEQ     JE,JZ
JGE     JGE
JGT     JG
JHI     JA
JLE     JLE
JLS     JBE
JLT     JL
JMI     JS
JNE     JNE, JNZ
JOC     JNO
JOS     JO
JPC     JNP, JPO
JPL     JNS
JPS     JP, JPE
x86 Assembly two jump instructions with the same target

x86 Assembly two jump instructions with the same target


By : Jacob
Date : March 29 2020, 07:55 AM
I wish did fix the issue. The result of assembly code compilation is a piece of native code, more specifically a sequence of bytes, where different parts of this sequence correspond to instructions from the original assembly. Intel x86 processors have a so-called CISC instruction set, which basically means that instruction length in bytes can vary from 1 to some 12, and that's without considering instruction set extensions that are available nowadays. The technique presented makes use of this fact.
The whole point is to mislead the person with a potential mischief in their mind. Code of the second example (with pop eax and retn) is what we want to execute and what will be executed. However, because we inserted a fake byte in front of the pop eax instruction, disassembled code will without paying close attention appear as if there was a call to some location in memory, since most of the less bright disassemblers automatically assume that machine code of a binary has no gaps in it.
difference between conditional instructions (cmov) and jump instructions

difference between conditional instructions (cmov) and jump instructions


By : ap-2
Date : March 29 2020, 07:55 AM
Hope that helps movcc is a so-called predicated instruction. That's fancy-speak for "this instruction executes under a condition (predicate)".
Many processors, including the x86, after doing an arithmetic operation (especially compare instructions), sets the condition code bits to indicate the status of the result of the operation.
How do we use jump in assembly using these instructions?

How do we use jump in assembly using these instructions?


By : somil lohani
Date : March 29 2020, 07:55 AM
help you fix your problem Branching in assembly is flexible, in a sense that you can arrange the code blocks in any order you want. You can also merge the branches by jumping to the same block. To archive this the instructions must to able to jump both forward and backward. The following example shows a use of backward jumping.
code :
073000:  bf 08 00 00 00          mov    edi, 0x8
073005:  31 c0                   xor    eax, eax
073007 <loop>:
073007:  01 f8                   add    eax, edi
073009:  83 ef 01                sub    edi, 0x1
07300c:  75 f9                   jne    073007 <loop>
07300e:
Unsigned numbers:
0x1234 =  1*16^3 +  2*16^2 + 3*16^1 +  4 = 4660 
0x3420 =  3*16^3 +  4*16^2 + 2*16^1 +  0 = 13344 
0x0A0B =  0*16^3 + 10*16^2 + 0*16^1 + 11 = 2571   (A is 10; B is 11)
0x8000 =  8*16^3 +  0*16^2 + 0*16^1 +  0 = 37268
0xFF7A = 15*16^3 + 15*16^2 + 7*16^1 + 10 = 65402  (F is 15)

Signed numbers:
0x1234 =   ; positive because 1 is smaller than 8
0x3420 =   ; positive because 3 is smaller than 8 
0x0A0B =   ; positive because 0 is smaller than 8
0x8000 =   ; negative because 8 is NOT smaller than 8
0xFF7A =   ; negative because F (15) is NOT smaller than 8

Signed numbers:
0x1234 = 4460  ; positive numbers are calculated like unsigned number
0x3420 = 13344 ; positive numbers are calculated like unsigned number
0x0A0B = 2571  ; positive numbers are calculated like unsigned number
0x8000 = 32768 - 65536 = -32768 ; a simple way to calculate negative number
0xFF7A = 65402 - 65536 = -134 ; a simple way to calculate negative number
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