为什么在重载<<中字符串会改变后面的数字大小

#include 
using namespace std;

class rational {
private:
    long denom, den;   // denom is the numerator 
    //and den is the denominator
public:
    rational(int up = 0, int down = 1);
    bool operator<(rational r) const;
    bool operator<=(rational r) const;
    bool operator == (rational r) const;
    bool operator!=(rational r) const;
    rational& operator-() ;
    friend ostream& operator<<(ostream& out, const rational& r);// Add output operator << overload here
};
rational::rational(int up, int down) {
    denom = up;
    den = down;
}
bool rational::operator<(rational r) const {
    return (this->denom / this->den < r.denom / r.den);
}
bool rational::operator<=(rational r) const {
    return (this->denom / this->den <= r.denom / r.den);
}
bool rational::operator==(rational r) const {
    return (this->denom / this->den == r.denom / r.den);
}
bool rational::operator!=(rational r) const {
    return (this->denom / this->den != r.denom / r.den);
}
rational& rational::operator-()  {
    rational a(-denom, den);
    return a;
}
ostream& operator<<(ostream& out, const rational& r) {
    out << r.denom << "/" << r.den;
    return out;
}
int main() {
    rational r1(1, 2);
    rational r2(1, 3);
    cout << r1 << " " << r2 << endl;
    cout << (r1 < r2) << endl;
    cout << (r1 <= r2) << endl;
    cout << (r1 != r2) << endl;
    cout << -r1 << endl;
}

如图，运行到第38行时结果与预期不符

把“/”改成‘/’时就好了

后面又试了一下，发现只要是一个字符串都会让后面的数字结果出错

这是为什么

rational& rational::operator-()

不要返回堆栈上的临时变量，而是这么写

rational& rational::operator-() {
    rational *a = new rational(-denom, den);
    return *a;
}

• 你可以参考下这个问题的回答, 看看是否对你有帮助, 链接: https://ask.csdn.net/questions/7721770