What does the operation
movl (%esi, %ecx,4), %eax do?
To my understanding it would store in %eax the equivalent of %ecx * 4 + %esi.
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2 Answers
It's equivalent to the following in Intel syntax:
mov eax,[esi + ecx*4]
What it will do is read 32 bits from memory at the address formed by esi + ecx*4 and put that value in eax.
1 Comment
Jonathon Reinhart
Note that you can get
objdump to output Intel syntax with objdump -d -M intelYes, you are write.
It is called as the index addressing mode.
It's syntax is:
<constant1/label> (%reg1, %reg2, constant2)
i.e. either constant1 or label without < & >.
%reg1 and %reg2 must be GPRs.
<constant1/label> is optional.
%reg1 is optional.
It results in: constant1 + %reg1 + %reg2 * constant2
Generally, constant1 or label and %reg1 are used for the base address of an array. And %reg2 & constant2 are used for the index.
For example:
Let's say, you have a global array:
.section .data
.globl arr
.type arr, @object
.size arr, 20
.align 4
arr:
.long 10, 20, 30, 40, 50
Then you may write...
movl $2, %eax # Move the index into %eax
movl arr( , %eax, 4), %edx
This means: %edx = arr + %eax * 4. i.e. Base address + 8 bytes.
You may think of it like %edx = arr[2]
Generally,
If you have a global array having a lable, then you use label for the base address.
If you have a local array, then you use the %reg2 fot the base address.
If you have a global structure containing array, then you use both label of the structure + %reg2 containing the byte-offset of the member array.
That's what happens generally... but it depends on the situation...
