C++ 四舍五入保留有效数字

#include <iostream>#include <cmath>double _f(double t, int n) {double m = pow(10, n);double result = floor(t * m + 0.5) / m;return result;}//保留4位有效数字double f4(double n) {return _f(n, 4);}double f3(double n) {return _f(n, 3);}double f2(double n) {return _f(n, 2);}double f1(double n) {return _f(n, 1);}int main (){    double num = f4(3.1415926);    std::cout<<num<<std::endl;    return 0;}