Can gcc/g++ tell me when it ignores my register?

Question

When compiling C/C++ codes using gcc/g++, if it ignores my register, can it tell me? For example, in this code

int main()
{
    register int j;
    int k;
    for(k = 0; k < 1000; k++)
        for(j = 0; j < 32000; j++)
            ;
    return 0;
}

j will be used as register, but in this code

int main()
{
    register int j;
    int k;
    for(k = 0; k < 1000; k++)
        for(j = 0; j < 32000; j++)
            ;
    int * a = &j;
    return 0;
}

j will be a normal variable. Can it tell me whether a variable I used register is really stored in a CPU register?

Answer 1

You can fairly assume that GCC ignores the register keyword except perhaps at -O0. However, it shouldn't make a difference one way or another, and if you are in such depth, you should already be reading the assembly code.

Google turned up an informative thread on this topic: http://gcc.gnu.org/ml/gcc/2010-05/msg00098.html

Answer 2

With respect to modern compilation and optimization techniques, the register annotation does not make any sense at all. In your second program you take the address of j, and registers do not have addresses, but one same local or static variable could perfectly well be stored in two different memory locations during its lifetime, or sometimes in memory and sometimes in a register, or not exist at all. Indeed, an optimizing compiler would compile your nested loops as nothing, because they do not have any effects, and simply assign their final values to k and j. And then omit these assignments because the remaining code does not use these values.

Answer 3

You can't get the address of a register in C, plus the compiler can totally ignore you; C99 standard, section 6.7.1 (pdf):

The implementation may treat any register declaration simply as an auto declaration. However, whether or not addressable storage is actually used, the address of any part of an object declared with storage-class specifier register cannot be computed, either explicitly (by use of the unary & operator as discussed in 6.5.3.2) or implicitly (by converting an array name to a pointer as discussed in 6.3.2.1). Thus, the only operator that can be applied to an array declared with storage-class specifier register is sizeof.

Unless you're mucking around on 8-bit AVRs or PICs, the compiler will probably laugh at you thinking you know best and ignore your pleas. Even on them, I've thought I knew better a couple times and found ways to trick the compiler (with some inline asm), but my code exploded because it had to massage a bunch of other data to work around my stubbornness.

Answer 4

Your second example is invalid in C. So you see well that the register keyword changes something (in C). It is just there for this purpose, to inhibit the taking of an address of a variable. So just don't take its name "register" verbally, it is a misnomer, but stick to its definition.

That C++ seems to ignore register, well they must have their reason for that, but I find it kind of sad to again find one of these subtle difference where valid code for one is invalid for the other.