使用 gcc 编译代码后，std::out_of_range 异常字符串不会打印

Question

我正在尝试测试以下异常处理代码：

#include "Array_Template.h"
using namespace std;

int main()
{
    Array<int> intArr1;
    //attempt to use out-of-range subscript
    try{
        cout << "\nAttempt to assign 1000 to intArr1[6]" << endl;
        intArr1[6] = 1000;
    } //end try
    catch (const out_of_range &ex){
        cout << "An exception occurred: " << ex.what() << endl;
    } //end catch
    return 0;
} //end main

在我的 Array 类模板中：

#ifndef ARRAY_H
#define ARRAY_H

#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <stdexcept>

//forward declarations of friend functions as specializations of template
template<typename T> class Array;

template<typename T>
std::istream &operator>>(std::istream &, Array<T> &);
template<typename T>
std::ostream &operator<<(std::ostream &, const Array<T> &);

//Array class template
template<typename T>
class Array
{
    friend std::istream &operator>> <>(std::istream &, Array<T> &);
    friend std::ostream &operator<< <>(std::ostream &, const Array<T> &);
public:
    Array(int = 5); //default constructor of array size 5
    Array(const Array<T> &); //copy constructor
    ~Array(); //destructor
    int getSize() const; //return size of array
    const Array<T> &operator=(const Array<T> &); //overloaded assignment operator
    bool operator==(const Array<T> &) const; //overloaded equality operator
    bool operator!=(const Array<T> &r) const //overloaded inequality operator, inline definition
    {
        return !(*this == r); //invokes Array<T>::operator==
    }
    T &operator[](int); //overloaded subscript operator for non-const objects
    T operator[](int) const; //overloaded subscript for const objects
private:
    int size; //pointer-based array size
    T *arrPtr; //pointer to first element of array
}; //end class Array

template<typename T>
Array<T>::Array(int s)
    :size(s > 0 ? s : 5), arrPtr(new T[size])
{
    for (int i = 0; i < size; ++i)
        arrPtr[i] = 0;
} //end Array constructor

template<typename T>
//ref return creates a modifiable lvalue
T& Array<T>::operator[](int index)
{
    if (index < 0 || index >= size) //check if out of bounds
        throw std::out_of_range::out_of_range("index out of range");
    return arrPtr[index]; //ref return
} //end operator[]

template<typename T>
//const ref return creates an rvalue
T Array<T>::operator[](int index) const
{
    if (index < 0 || index >= size) //check if out of bounds
        throw std::out_of_range::out_of_range("index out of range");
    return arrPtr[index]; //returns copy of this element
} //end operator[]

#endif

程序在 Visual Studio 中编译并执行良好，显示正确的一行错误消息：

但是当我用 gcc 编译同一个程序并运行它时，我得到以下输出：

* `./arrays_template' 中的错误：free()：无效大小：0x0000000001426030 * ======= 回溯：========= /lib64/libc.so.6(+0x7364f)[0x7fd7c2e3d64f] /lib64/libc.so.6(+0x78eae)[0x7fd7c2e42eae] /lib64/libc.so.6(+0x79b87)[0x7fd7c2e43b87] ./arrays_template[0x4011a1] ./arrays_template[0x400fc7] /lib64/libc.so.6(__libc_start_main+0xf5)[0x7fd7c2debbe5] ./arrays_template[0x400c29]

之后是冗长的内存映射跟踪。 有人可以解释这里到底发生了什么以及如何解决它吗？通过修复，我的意思是如何获得与 catch 块中指定的相同的错误消息。

Answer 1

以下代码错误：

throw std::out_of_range::out_of_range("index out of range");
//       ^^^^^^^^^^^^^^

应该改写为：

throw std::out_of_range("index out of range");

调用constructor的std::out_of_range

---

还要确保修改operator[](int) const的返回类型。

Answer 2

在两个运算符中更改此语句

throw std::out_of_range::out_of_range("index out of range");

到

throw std::out_of_range("index out of range");