请诸位看一下为什么会这样!
exc_mean.hpp
exc_mean.hpp
#include<iostream>
using namespace std;
class bad_hmean
{
private:
double v1;
double v2;
public:
bad_hmean(int a=0,int b=0):v1(a),v2(b){}
void mesg();
};
inline void bad_hmean::mesg()
{
cout<<"heman("<< v1 <<" , "<<v2<<"): "
<<"invalid arguments a=-b \n";
}
class bad_gmean
{
public:
double v1;
double v2;
bad_gmean(double a=0,double b=0):v1(a),v2(b){}
const char* mesg();
};
inline const char* bad_gmean::mesg()
{
return "gmean () srguments should be >0 \n";
}
error4.cpp
#include<iostream>
#include<cmath>
#include"exc_mean.hpp"
using namespace std;
double hmean(double a,double b);
double gmean(double a,double b);
int main()
{
double x,y,z;
cout<<"Enter two numbers: ";
while(cin>>x>>y)
{
try
{
z=hmean(x,y);
cout<<"Harmonic mean of "<<x<<" and "<<y<<" is "<<z<<endl;
cout<<"Geometric mean of "<<x<<" and "<<y<<" is "<<gmean(x,y)<<endl;
cout<<"enter next set of numbers <q to quit> : ";
}
catch(bad_hmean &bh)
{
bh.mesg();
cout<<"Try again! \n";
continue;
}
catch(bad_gmean &bg)
{
cout<<bg.mesg();
cout<<"Values uesd "<<bg.v1<<" , "<<bg.v2<<endl;
cout<<"Sorry ,you don"t play again. \n";
break;
}
}
cout<<"Bye! \n";
return 0;
}
double hmean(double a,double b)
{
if(a==-b)
//原句如下,编译运行没问题
//将此句替换成右边的两句,编译错误 // bad_hmean A=bad_hmean(a,b);
throw bad_hmean(a,b); //throw A;
return 2.0*a*b/(a+b);
}
double gmean(double a, double b)
{
if(a<0||b<0)
throw bad_gmean(a,b);
/* bad_gmean B=bad_gmean(a,b);//
throw B;*/
return sqrt(a*b);
}
报错如下:
error4.cpp:50:11: error: use of undeclared identifier “A”
throw A;
^
error4.cpp:57:11: error: use of undeclared identifier “B”
throw B;
^
2 errors generated.
怎么还说使用未声明的标示符 A和B
本人不是已经创建了A B这两个对象了吗?
到底怎么回事?
求指导
解决方案
2
哪个编译器这么蛋疼?
2
不是吧,什么版本?本人试了gcc没有这问题啊.规范定义的应该是throw expression;没有问题
14
if(a==-b) bad_hmean A=bad_hmean(a,b); throw A;
这种代码不报错才是怪事
起码加上大括号吧
if(a==-b){
bad_hmean A=bad_hmean(a,b);
throw A;
}
4
if 后假如不加括号,只对后一条语句有效,这样你的定义是在if里,但throw不是在if里,故找不到,加上大括号就好了