Use cdq / idiv for signed 32-bit / 32-bit => 32 bit division,xor edx,edx / div for unsigned. With the dividend in EAX to start with, and the divisor specified as an operand to DIV or IDIV. If you zero EDX/RDX instead of sign-extending into EDX:EAX before idiv, you can get a large positive result for -5 / 2, for example. Using the “full … Read more
I keep seeing this form often: div 3, 15 this is not any valid INTEL mneumonic! To divide 15 by 3: For the second error, you cannot move a 16 bit register into a 32 bit register like that. You need to use one of the following: Or:
For the 1st error you forgot to type a comma between the register and the immediate. For the 2nd and 3rd errors the CH and CL registers cannot be used for addressing memory. Use SI, DI, or BX instead. Since your array is defined as words you must treat it as such!Change into something like … Read more
It’s probably: This is a common idiom for zeroing a register on x86. This would correspond with i = 0 in the C code. If you are curious “but why ?” the short answer is that the xor instruction is fewer bytes than mov $0, %ebx. The long answer includes other subtle reasons. I am … Read more
There are a few operations you can only do with DI/SI (or their extended counterparts, if you didn’t learn ASM in 1985). Among these are Which are, respectively, operations for repeated (= mass) storing, loading and scanning. What you do is you set up SI and/or DI to point at one or both operands, perhaps put a … Read more
I was reading the following definition for syscall: I saw it explained that the line cmpq $-4095 %rax determines whether %rax contains a value between -1 and -4095. How does it do that? What exactly does the cmpq instruction do?
You have as input values. The instruction just subtracts 3(dec) in EDX from 100(hex) in EAX resulting in 100h-3h=0FDh.
As has already been stated, wrapping brackets around an operand means that that operand is to be dereferenced, as if it were a pointer in C. In other words, the brackets mean that you are reading a value from (or storing a value into) that memory location, rather than reading that value directly. So, this: … Read more
Maybe you’re wondering about this: These lines are known as the assembly function prologue. The first 2 instructions save the previous base pointer (ebp) and set EBP to point at that position on the stack (right below the return address). This sets up EBP as a frame pointer. The sub esp,12 line is saving space … Read more
when I have the the operation in assembly, then i have read that the that the value in edx:eax is divided by the operand ecx. I also know that the quotient is stored in eax and the remainder in edx. so but what exactly is the value in edx:eax ?? Can someone explain it to … Read more