CODEWITHSUNDEEP
c++optimizationpow
Qix - MONICA WAS MISTREATED
Qix - MONICA WAS MISTREATED

Reputation: 15113

C++: pow() optimizations with two constants

Simple question: When the compiler faces a call to, say, pow() with two constants (i.e. values from macros), is it optimized by evaluating it at compile time, or is it still calculated at run-time?

Example:

#define V_BITMEM_GRID 3
#define V_BITMEM_TOTAL pow(V_BITMEM_GRID,2)

Thanks!

EDIT If not, is there a way to calculate the square/cube of a macro as another macro (like I'm attempting above) at compile-time?

Upvotes: 0

Views: 511

Answers (3)

mikithskegg
mikithskegg

Reputation: 816

gcc counts such expressions at compilation time, e.g.

Upvotes: 0

rasmus
rasmus

Reputation: 3216

You shouldn't depend on it. A macro based approach is:

#define POW1(x) (x)
#define POW2(x) ((x)*(x))
#define POW3(x) (POW2(x)*POW1(x))
...
#define POW(x, y) POW##y(x)

Upvotes: 1

Luchian Grigore
Luchian Grigore

Reputation: 258618

It can be both. It depends on how intrusive the compiler is, whether it has access to the function implementation and can correctly evaluate it. There's no rule that specifies how it's supposed to be, as long as observed behavior is the same.

For example, I got the following:

#define X 1
#define Y 2
int foo(int x, int y)
{
    return x + y;
}

int main(int argc, char* argv[])
{
    cout << foo(X,Y);
00BE1000  mov         ecx,dword ptr [__imp_std::cout (0BE203Ch)] 
00BE1006  push        3    
00BE1008  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (0BE2038h)] 
}

The function, as you can see, isn't even called. So it is possible that the call is eliminated for good.

Upvotes: 2

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