c++ - 在 C++ 中输出时出现段错误

Question

我正在尝试打印出一个整数数组。当我尝试如下打印时出现段错误。如果我取消注释“In for loop”，它将打印除数组最后一项之外的所有内容，并且它仍然有段错误。当我取消注释这两个注释(或只是“使用 for 循环完成”)时，一切都正常打印。为什么会发生这种情况，我该如何解决？

for( int i = 0; i < l.usedLength; i++ )
{
    //cout << "**********In for loop" << endl;
    cout << l.largeInt[ i ];
}
//cout << "**********done with for loop" << endl;

这是全类:

#include "LargeInt.h"
#include <ctype.h>

LargeInt::LargeInt()
{
    usedLength = 0;
    totalLength = 50;

    largeInt = new int[totalLength];
    for( int i=0; i<totalLength; i++ )
    {
        largeInt[i] = 0;
    }
}

LargeInt LargeInt::operator+(const LargeInt &l) const
{}

LargeInt LargeInt::operator-(const LargeInt &l) const
{}

LargeInt LargeInt::operator*(const LargeInt &l) const
{}

LargeInt LargeInt::operator/(const LargeInt &l) const
{}

bool LargeInt::operator==(const LargeInt &l) const
{}

ostream& operator<<(ostream &out, const LargeInt &l)
{
    cout << "In output" << endl;

    if( l.usedLength == 0 )
    {
        cout << 0;
    }
    else
    {
        cout << "In else... NON 0" << endl;

        for( int i = 0; i < l.usedLength; i++ )
        {
            cout << "In for loop" << endl;
            cout << l.largeInt[ i ];
        }
        //cout << "done with for loop" << endl;
    }
    //cout << "after the if.... all done with output" << endl;
}

istream& operator>>(istream &in, LargeInt &l)
{
    char x;
    while (std::cin.get(x) && x >= '0' && x <= '9')
    {
        l.largeInt[ l.usedLength ] = x-48;
        l.usedLength++;
        //need to check array length and make bigger if needed
    }

}

主要内容:

#include <stdlib.h>
#include <iostream>

#include "LargeInt.h"

int main(int argc, char** argv) {

    cout << "\nJosh Curren's Assignment #5 - Large Integer\n" << endl;

    LargeInt lint;

    cout << "Enter a large int: ";
    cin >> lint;

    cout << "\nYou entered: " << endl;
    cout << lint << endl;
    cout << endl;


    return (EXIT_SUCCESS);
}

Answer 1

您忘记了 ostream& operator<<(ostream &out, const LargeInt &l) 中的最后一行:

return out;

这条线完美地工作。