大家帮忙看一下这份C++代码为什么通不过啊
#include<iostream>
#include<cmath>
using namespace std;

class Point{
public:
Point(int x=0,int y=0):x(x),y(y){}
int getX(){return x;}
int getY(){return y;}
friend double distant(Point &p1,Point &p2);
private:
int x,y;
};
double distant(Point &p1,Point &p2)
{
double x=p1.x-p1.x;
double y=p2.y-p2.y;
return static_cast<double>(sqrt(x*x,y*y));
}
int main()
{
Point myp1(1,1),myp2(4,5);
cout<<"The distance is: ";
cout<<distant(myp1,myp2);
system("pause");
return 0;
}
我用的是DevC++编译的，编译出现错误，错误在sqrt（）。错误如下：
F:\C语言及专业课课件\cpp类\5_6111.cpp In function `double distant(Point&, Point&)':
18 F:\C语言及专业课课件\cpp类\5_6111.cpp no matching function for call to `sqrt(double, double)'
note D:\Program Files\Dev-Cpp\include\math.h:151 candidates are: double sqrt(double)
note D:\Program Files\Dev-Cpp\include\math.h:151 long double std::sqrt(long double)
note D:\Program Files\Dev-Cpp\include\math.h:151 float std::sqrt(float)
但是你如果将那个distant函数中的double全改为float的话就可以。
不明白问题出在哪里？

[解决办法]
楼主写错了吧： return static_cast<double>(sqrt(x*x,y*y)); 应该是x*x+y*y吧
sqrt函数只接受一个参数，并且计算距离公式不对。这样：

C/C++ code

#include<iostream>#include<cmath>using namespace std;class Point{public:    Point(int x=0,int y=0):x(x),y(y){}    int getX(){return x;}    int getY(){return y;}    friend double distant(Point &p1,Point &p2);private:    int x,y;};double distant(Point &p1,Point &p2){    double x=p1.x-p2.x;    double y=p1.y-p2.y;       return static_cast<float>(sqrt(x*x+y*y));     }int main(){    Point myp1(1,1),myp2(4,5);       cout<<"The distance is: ";    cout<<distant(myp1,myp2);    system("pause");    return 0;}
[解决办法]
return static_cast<double>(sqrt(x*x,y*y));
应该是加法吧，逗号表达式等于sqrt(y*y),
而且sqrt返回值就是double，不用转换
[解决办法]
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楼主写错了吧： return static_cast<double>(sqrt(x*x,y*y));  应该是x*x+y*y吧
sqrt函数只接受一个参数，并且计算距离公式不对。这样：
C/C++ code

#include<iostream>
#include<cmath>
using namespace std;

class Point
{
public:
    Point(in……