Is this claim correct: "Anywhere you pass in a function, you can pass in a functor"?
for example:Code:bool cmp_function(int a, int b){ return a<b; } struct cmp_functor{ bool operator()(int a, int b){ return a<b; } }; std::sort(vec.begin(), vec.end(), cmp_function); std::sort(vec.begin(), vec.end(), cmp_functor);
no, that is incorrect. but you can often structure your code in a way that allows either to be used. for example:
Code:template <typename Function> void foo(Function callback, int data) { callback(data); } void bar(int) { } struct baz { void operator () (int) { } }; foo(bar, 1); foo(baz(), 1);
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Thank you, I am curious why "baz()" should carry "()". Isn't "baz" enough?
Also, in your example, what would the 2 signatures be if the foo only accepts function or funcotr?
Last edited by meili100; 11-21-2007 at 07:21 PM.
> I am curious why "baz()" should carry "()". Isn't "baz" enough?
Because baz is an object that needs to be instantiated, just like other classes or structs in C++.
> what would the 2 signatures be if the foo only accepts function or funcotr?
The simplest of the possibilities is probably
void foo( void (*callback )( int ), int data );
void foo( baz callback, int data );
