Is there any overhead to declaring a variable within a loop? (C++)

Question

Stack space for local variables is usually allocated in function scope. So no stack pointer adjustment happens inside the loop, just assigning 4 to var. Therefore these two snippets have the same overhead.

For primitive types and POD types, it makes no difference. The compiler will allocate the stack space for the variable at the beginning of the function and deallocate it when the function returns in both cases.

For non-POD class types that have non-trivial constructors, it WILL make a difference -- in that case, putting the variable outside the loop will only call the constructor and destructor once and the assignment operator each iteration, whereas putting it inside the loop will call the constructor and destructor for every iteration of the loop. Depending on what the class' constructor, destructor, and assignment operator do, this may or may not be desirable.

They are both the same, and here's how you can find out, by looking at what the compiler does (even without optimisation set to high):

Look at what the compiler (gcc 4.0) does to your simple examples:

1.c:

main(){ int var; while(int i < 100) { var = 4; } }

gcc -S 1.c

1.s:

_main:
    pushl   %ebp
    movl    %esp, %ebp
    subl    $24, %esp
    movl    $0, -16(%ebp)
    jmp L2
L3:
    movl    $4, -12(%ebp)
L2:
    cmpl    $99, -16(%ebp)
    jle L3
    leave
    ret

2.c

main() { while(int i < 100) { int var = 4; } }

gcc -S 2.c

2.s:

_main:
        pushl   %ebp
        movl    %esp, %ebp
        subl    $24, %esp
        movl    $0, -16(%ebp)
        jmp     L2
L3:
        movl    $4, -12(%ebp)
L2:
        cmpl    $99, -16(%ebp)
        jle     L3
        leave
        ret

From these, you can see two things: firstly, the code is the same in both.

Secondly, the storage for var is allocated outside the loop:

         subl    $24, %esp

And finally the only thing in the loop is the assignment and condition check:

L3:
        movl    $4, -12(%ebp)
L2:
        cmpl    $99, -16(%ebp)
        jle     L3

Which is about as efficient as you can be without removing the loop entirely.

These days it is better to declare it inside the loop unless it is a constant as the compiler will be able to better optimize the code (reducing variable scope).

EDIT: This answer is mostly obsolete now. With the rise of post-classical compilers, the cases where the compiler can't figure it out are getting rare. I can still construct them but most people would classify the construction as bad code.

Most modern compilers will optimize this for you. That being said I would use your first example as I find it more readable.