Re: Strange code generated by compiler

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autumn48@xxxxxxxxx wrote:
why does the program randomly copy the value to be assigned, to
a register before moving it to the actual memory destination (like
when it assigns 22 to v[7], 55 to v[8] and 12 to v[10]). this doesn't
happen when using a normal c style array.

In case of a C array allocated on the stack:

int x[10];
x[0] = 22;

The compiler knows the address of the memory where the elements are located, so it can hard code it for you:

mov [esp+14h], 22

A C++ vector internally uses a C array too, but the memory is allocated dynamically (runtime) on the heap. The address of a vector object doesn't directly point to its internal array. A pointer must be dereferenced first:

vector<int> x(10);
x[0] = 22;

What happens here is that vector<T>::operator[] gets called, where an internal pointer gets dereferenced, like this:

x.internal_pointer[0] = 22;

The compiler can not hard code the address of anything allocated at runtime, it has to load it from a pointer into a register first, at least on the x86 processor family:

mov eax, [esp+14h] ; operator*: resolve pointer
mov [eax], 22 ; operator=: assign value

Try to compile the following piece of code:

int* ptr = (int*)malloc(10 * sizeof(int));
ptr[0] = 22;
ptr[1] = 55;
[...]
free(ptr);

The resulting assembly code should be very similar in nature to your vector example. In the code above, 'ptr' is not the direct address of your array, but a pointer that contains the address, so it essentially generates the same machine instructions as vector did (with physical memory offsets differring, of course).

Tom
.

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