C ++中的const运算符重载问题 [英] Const operator overloading problems in C++
本文介绍了C ++中的const运算符重载问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在使用const版本重载operator()时遇到麻烦:
I'm having trouble with overloading operator() with a const version:
#include <iostream>
#include <vector>
using namespace std;
class Matrix {
public:
Matrix(int m, int n) {
vector<double> tmp(m, 0.0);
data.resize(n, tmp);
}
~Matrix() { }
const double & operator()(int ii, int jj) const {
cout << " - const-version was called - ";
return data[ii][jj];
}
double & operator()(int ii, int jj) {
cout << " - NONconst-version was called - ";
if (ii!=1) {
throw "Error: you may only alter the first row of the matrix.";
}
return data[ii][jj];
}
protected:
vector< vector<double> > data;
};
int main() {
try {
Matrix A(10,10);
A(1,1) = 8.8;
cout << "A(1,1)=" << A(1,1) << endl;
cout << "A(2,2)=" << A(2,2) << endl;
double tmp = A(3,3);
} catch (const char* c) { cout << c << endl; }
}
这给了我以下输出:
- 调用NONconst版本--调用NONconst版本-A(1,1)= 8.8
- 调用了NONconst版本-错误:您只能更改矩阵的第一行。
如何实现C ++调用operator()的const版本?我正在使用GCC 4.4.0。
How can I achieve that C++ call the const-version of operator()? I am using GCC 4.4.0.
推荐答案
重载看起来不错,但是您永远不会在const对象上调用它。您可以尝试以下操作:
The overloading looks fine but you never call it on a const object. You can try this:
void foo(const Matrix& A) {
cout << "A(1,1)=" << A(1,1) << endl;
}
Matrix A(10,10);
foo(A);
这会给您:
- const-version was called - A(1,1)=0
这篇关于C ++中的const运算符重载问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文