How to be sure that a parameter will be treated as a constant during the compilation in C++?

Question

I wonder if the two following implementations will produce exactly the same thing with the same performances whatever the compiler I use :

template<class T, unsigned int TSIZE> MyClass1
{
    static const unsigned int size_const = 0;
    public:
        inline void Loop()
        {
            for(unsigned int i = 0; i < TSIZE; ++i) {
                /* DO SOMETHING ON DATA */
            }
        }
        T _data[TSIZE];
};

template<class T, unsigned int TSIZE> MyClass2
{
    static const unsigned int size_const = TSIZE;
    public:
        inline void Loop()
        {
            for(unsigned int i = 0; i < size_const; ++i) {
                /* DO SOMETHING ON DATA */
            }
        }
        T _data[size_const];
};

In the first one, as TSIZE used in the loop is a template parameter, it is almost guaranteed that the compiler will unroll the loop if needed. If the loop is unrolled in the first case, would it be unrolled in the second case (the only difference is that TSIZE is stored in a static const) ?

Thank you very much.

You should file a bug against the compiler vendor if it does not optimize the two versions you have in exactly the same way. — jxh, Aug 03 '12 at 20:49
Both constants are Integral Constant Expressions. They don't get any more const than that. — MSalters, Aug 04 '12 at 00:29

score 5 · Accepted Answer · answered Aug 03 '12 at 20:50

Whether the compiler will perform the optimization or not is different from whether it will treat the value as a compile time constant. In your particular example, and because the static const has not been defined anywhere, if the linker did not complain it means that the compiler only used it as a const-expression (compile time constant). Also note that if the compiler was not considering size_const as a const-expression, then the line T _data[size_const] (I am assuming you lost the T on the copy) would not compile.

Any odr-use (use other than as a compile time constant) of the static member would require a definition.

score 0 · Answer 2 · answered Aug 03 '12 at 20:46

0

Logically a compiler does have enough information to do the optimization you are describing however whether it will actually do it is extremely implementation-dependent and I wouldn't expect it to be universally supported

answered Aug 03 '12 at 20:46

YePhIcK

5,816
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Except that this is constant propogation, one of the first things any optimizing compilter will implement. :) – GManNickG Aug 03 '12 at 20:50

How to be sure that a parameter will be treated as a constant during the compilation in C++?

2 Answers2