No - a while loop is not known to always compare. It is only known to always require an expression. But it's up to the compiler to decide what code to generate if that expression can be computed at compile time.



No - the C compiler likes early out, so you can write code like:

if ((num_elements != 0) && ((total_bytes / num_elements) > 5)) {
    ...
}

Let's look at some code generated by gcc:

int test(int a,int b,int c) {
    if (a && !b && !c) {
        return 1;
    }
    return 0;
}

Produces the following code for a x86 processor:

.globl test
        .type   test, @function
test:
        pushl   %ebp
        movl    %esp, %ebp
        subl    $4, %esp
        cmpl    $0, 8(%ebp)
        je      .L2                <= instant out if a is zero
        cmpl    $0, 12(%ebp)
        jne     .L2                <= instant out if b is non-zero
        cmpl    $0, 16(%ebp)
        jne     .L2                <= instant out if c is non-zero
        movl    $1, -4(%ebp)       <= Return value if "if" statement did match
        jmp     .L3
.L2:
        movl    $0, -4(%ebp)       <= Return value if "if" statement didn't match
.L3:
        movl    -4(%ebp), %eax
        leave
        ret
        .size   test, .-test

And above example shows you wrong. I'll leave it up to you to take a closer took at the C standard and figure out the relevant paragraphs.