Question

This is some partial code , after a call to a function .

movl   12(%ebp),%ecx  
movl   8(%ebp),%esi    
movl   (%esi,%ebx,4),%edx  

This is the corresponds C function :

void  foo(MyType_t A[], int n);

So basically , %esi is A[] , and %ecx is n

What does the 3rd line means ? please note that the left operand has 3 sub-operands .

%edx = A[4 * %ebx + %esi] ? 

Regards

Answer 1

movl   (%esi,%ebx,4),%edx

is AT&T syntax for “scale, index, base” (SIB) addressing. %esi is the base address, %ebx is the index, and 4 is the “scale” (the multiplier applied to the index to generate a byte offset to the base address).

(The equivalent Intel syntax equivalent is more explicit: mov edx, dword ptr [esi+ebx*4].)

%edx = A[4 * %ebx + %esi] ?

Probably not, because the effective address calculation in this addressing mode works in units of bytes, whereas C array indexing works in units of the size of an array element.

If %esi is the base address of A in memory, and %ebx is n, and the size of a single element of A (i.e. sizeof(MyType_t)) is 4, then %esi + %ebx * 4 gives the address of A[n]; so the instruction would mean %edx = A[n] in that case.