提问者:小点点

重新生成数组时出错:写入到时缓冲区溢出


因此,我试图重新生成这个一维动态数组,但无法修复此错误:写入'NEW_ARR'时缓冲区溢出:可写大小为'NEWLENGTH*1'字节,但可能会写入2个字节

void regrow(char *&arr, int &length,int newLength) //Funcion to regrow an array
{
    char* new_arr = new char[newLength];
    for (int index = 0; index < length; index++)
    {
        new_arr[index] = arr[index];   //**Error occurs here** 
    }
    length = newLength;
    delete[] arr;
    arr = new_arr;
}

共1个答案

匿名用户

缓冲区溢出错误通常检测您是否尝试在未分配的空间中写入,这可能是由于newlength小于length本身造成的,这可以通过if-return检查来避免:

#include <iostream>

void regrow(char *&arr, int &length,int newLength) //Funcion to regrow an array
{
    if(length >= newLength){ //Check for correct input
        return;
    }else{

    char* new_arr = new char[newLength];
    for (int index = 0; index < length; index++)
    {
        new_arr[index] = arr[index];   //**Error occurs here**
    }
    length = newLength;
    delete[] arr;
    arr = new_arr;
    }
}

int main()
{
    int a = 5;
    int b = 8;
    char*array = new char[a]{'C','B','a','d','f'};

    regrow(array,a,b);


    for(int i = 0; i < a; ++i){
        std::cout << array[i] << std::endl;
    }

    return 0;
}