Hicham, visitation is not done at compile-time. Doc states as far as I remember that the compiler might (with high probability) inline the calls, so that they have similar performance as a switch statement. Switch usually equals to a jump in assembler. But there must be runtime overhead, since you put at runtime object instances into the variant. Greetings, Ovanes On Tue, Apr 21, 2009 at 4:09 PM, Hicham Mouline <hicham@mouline.org> wrote:

Following my prev message, I wrote the following test on a win platform, compiled with VS2005 /O2 in release mode

#include <iostream> #include <boost/variant/variant.hpp> #include <windows.h>

typedef boost::variant<int, double, char> VT;

struct print_type : public boost::static_visitor<const char*> { const char* operator()( int ) { return "int"; } const char* operator()( double ) { return "double"; } const char* operator()( char ) { return "char"; } };

int main() { const VT v('c'); LARGE_INTEGER usec0, usec1; const char* volatile res;

QueryPerformanceCounter( &usec0 ); for(size_t s=0; s<2000000; ++s) res = v.apply_visitor( print_type() ); QueryPerformanceCounter( &usec1 ); std::cout<< (usec1.QuadPart - usec0.QuadPart) <<std::endl;

QueryPerformanceCounter( &usec0 ); for(size_t s=0; s<2000000; ++s) res = "char"; QueryPerformanceCounter( &usec1 ); std::cout<< (usec1.QuadPart - usec0.QuadPart) <<std::endl; }

There is no statistically significant difference between the 2 cases.

Can I deduce that at least on this platform, for this code, the visitation is done at compile-time?

Rds,

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