wrapped1.cpp
来源: 探讨 C++ 新标准
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//wrapped1.cpp -- using a function wrapper as an argument
#include <iostream>
#include <math.h>
#include <functional>
using namespace std;
template <typename T, typename F>
T use_f(T v, F f)
{
static int count = 0;
count++;
cout << " use_f count = " << count << ", &count = " << &count << endl;
return f(v);
}
class Fp
{
private:
double z_;
public:
Fp(double z = 1.0) : z_(z) {}
double operator()(double p) { return z_*p; }
};
class Fq
{
private:
double z_;
public:
Fq(double z = 1.0) : z_(z) {}
double operator()(double q) { return z_+ q; }
};
double dub(double x) {return 2.0*x;}
int main()
{
double y = 1.21;
function<double(double)> ef1 = dub;
function<double(double)> ef2 = sqrt;
function<double(double)> ef3 = Fq(10.0);
function<double(double)> ef4 = Fp(10.0);
function<double(double)> ef5 = [](double u) {return u*u;};
function<double(double)> ef6 = [](double u) {return u+u/2.0;};
cout << "Function pointer dub:\n";
cout << " " << use_f(y, ef1) << endl;
cout << "Function pointer sqrt:\n";
cout << " " << use_f(y, ef2) << endl;
cout << "Function object Fp:\n";
cout << " " << use_f(y, ef3) << endl;
cout << "Function object Fq:\n";
cout << " " << use_f(y, ef4) << endl;
cout << "Lambda expression 1:\n";
cout << " " << use_f(y, ef5) << endl;
cout << "Lambda expression 2:\n";
cout << " " << use_f(y,ef6) << endl;
// cin.get();
return 0;
}
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