The difference between cmpl and cmp

Question

I am trying to understand assembly to be able to solve a puzzle. However I encountered the following instructions:

0x0000000000401136 <+44>:    cmpl   $0x7,         


        

Answer 1

Just to add one extra detail for clarification to @glglgl's excellent answer is that the "l" in "cmpl" is an operation suffix indicating that the operation is being done on a long number (32 bit integer or 64-bit floating point).

Answer 2

According to my understanding cmpl compares unsigned.

It does both, in a way.

The difference in signed vs. unsigned is here the usage of the jump instructions.

For >, there is ja for unsigned and jg for signed (jump if above and jump if greater).

For <, there is jb for unsigned and jl for signed (jump if below and jump if less).

To be exact, here is the meaning of several jump commands:

For unsigned comparisons:

JB/JNAE (CF = 1)           : Jump if below/not above or equal
JAE/JNB (CF = 0)           : Jump if above or equal/not below
JBE/JNA (CF = 1 or ZF = 1) : Jump if below or equal/not above
JA/JNBE (CF = 0 and ZF = 0): Jump if above/not below or equal

For signed comparisons:

JL/JNGE (SF <> OF)          : Jump if less/not greater or equal
JGE/JNL (SF = OF)           : Jump if greater or equal/not less
JLE/JNG (ZF = 1 or SF <> OF): Jump if less or equal/not greater
JG/JNLE (ZF = 0 and SF = OF): Jump if greater/not less or equal

Answer 3

I don't think x86 actually has an instruction called CMPL. It's probably part of your assembler syntax to give hints on operands or something else (like JZ and JE being the same).

From the intel manual on what it is doing:

Compares the first source operand with the second source operand and sets the status flags in the EFLAGS register according to the results. The comparison is performed by subtracting the second operand from the first operand and then setting the status flags in the same manner as the SUB instruction. When an immediate value is used as an operand, it is sign-extended to the length of the first operand.

Sign-ness is given implicitly, because of the two's complement representation of numbers.

How to manipulate the jump? If you are sure that the jump should do the exact opposite than what it is doing, you just have to change JA to JBE.