为什么此代码无法检测到<<运算符？

Question

请向我解释一下为什么这没有检测到<<操作符。

我尽了最大的努力，甚至试图在两个类上重载<<(这是不必要的)。

#include<iostream>
using namespace std;
const int MAX = 10;
class Complex;
template<class t>
class stack {
private:
t stk[MAX];
int top;
public:
stack() { top = -1; }
void push(t data) {
if (top == MAX - 1)
cout << "Stack is full.";
else
stk[++top] = data;
}
t pop() {
if (top == -1) {
cout << "Stack is empty.";
return NULL;
}
else {
//return stk[top--];
t data = stk[top];
top--;
return data;
}
}
};  
class Complex {
private:
float real, imag;
public:
Complex(float r = 0.0, float i = 0.0) { real = r; imag = i; }
friend ostream& operator << (ostream& s, Complex& c);
};
ostream& operator << (ostream& s, Complex& c) {
s << "(" << c.real << "," << c.imag << ")";
return s;
}
int main() {
stack<int> s1;
s1.push(10);
s1.push(20);
s1.push(30);
s1.push(40);
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
stack<float> s2;
s2.push(3.14f);
s2.push(4.14f);
s2.push(5.14f);
s2.push(6.14f);
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
Complex c1(1.5f, 2.5f), c2(1.5f, 2.5f), c3(1.5f, 2.5f), c4(1.5f, 2.5f);
//cout<<c1;
stack<Complex> s3;
s3.push(c1);
s3.push(c2);
s3.push(c3);
s3.push(c4);
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
return 0;
}

Answer 1

函数签名应该是这样的:

std::ostream& operator<<(std::ostream& os, const T& obj)
{
// write obj to stream
return os;
}

函数pop不能返回NULL，因为它的类型和t不一样。修复代码可能像

#include <iostream>
using namespace std;
const int MAX = 10;
class Complex;
template <class t>
class stack {
private:
t stk[MAX];
int top;
public:
stack() { top = -1; }
void push(t data) {
if (top == MAX - 1)
cout << "Stack is full.";
else
stk[++top] = data;
}
t pop() {
if (top == -1) {
cout << "Stack is empty.";
return {}; // May throw, or return std::optional here
} else {
// return stk[top--];
t data = stk[top];
top--;
return data;
}
}
};
class Complex {
private:
float real, imag;
public:
Complex(float r = 0.0, float i = 0.0) {
real = r;
imag = i;
}
friend ostream &operator<<(ostream &s, const Complex &c);
};
ostream &operator<<(ostream &s, const Complex &c) {
s << "(" << c.real << "," << c.imag << ")";
return s;
}
int main() {
stack<int> s1;
s1.push(10);
s1.push(20);
s1.push(30);
s1.push(40);
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
stack<float> s2;
s2.push(3.14f);
s2.push(4.14f);
s2.push(5.14f);
s2.push(6.14f);
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
Complex c1(1.5f, 2.5f), c2(1.5f, 2.5f), c3(1.5f, 2.5f), c4(1.5f, 2.5f);
// cout<<c1;
stack<Complex> s3;
s3.push(c1);
s3.push(c2);
s3.push(c3);
s3.push(c4);
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
return 0;
}

在线演示。

相关问题:操作符重载的基本规则和习惯用法是什么?

Answer 2

它不起作用，因为第二个参数应该是const引用

class Complex{
private:
float real,imag;
public:
Complex(float r=0.0,float i=0.0){
real=r;
imag=i;
}
friend ostream& operator<<(ostream& s, const Complex& c);
};
ostream& operator<<(ostream& s, const Complex& c) {
s<<"("<<c.real<<","<<c.imag<<")";
return s;
}

Answer 3

原因:pop函数返回一个r值。您可以将r值视为内存中的无名对象(当然，还有更多)。不能将r值放入l值引用中。但是对于const引用，这无关紧要，因为您不打算修改它。

解决方案1:

friend ostream &operator<<(ostream &s, Complex &c); // accepts lvalue - c refers to some external Complex object.
friend ostream &operator<<(ostream &s, Complex &&c); // accepts rvalue - c becomes a normal local variable whose value is the passed rvalue.

解决方案2:

friend ostream &operator<<(ostream &s, const Complex &c); // accepts both