访问导致运行时错误的动态数组

Question

我一直在搞乱动态内存，结果碰上了一堵巨大的墙。

我正在尝试创建一个程序，用户可以在其中输入任意数量的字符串，然后可以随时退出，但是在输入第二个字符串后，程序崩溃而没有给我任何特定的错误消息。

#include "stdafx.h"
#include "string.h"
#include "stdio.h"
#include "stdlib.h"
#include "new"

int _tmain(int argc, _TCHAR* argv[])
{
    //Variables
    int i=0,end=0,requiresSize=1;
    char ** temp;
    char  item[256]="a";
    char ** requires;


    //Initialize each element in requiers
    requires = new char * [requiresSize];
    for(int j=0;j<requiresSize*2;j++){
        requires[j]= new char[256];
    }


    while(strcmp(item,"q-")){
        end=0;
        printf("Enter h- for help.\nEnter q- to quit.\n");
        printf("Please enter a string\n");
        gets_s(item);
        if(!strcmp(item,"h-")){
            printf("Enter a string to add to the list.\nEnter p- to print the list.\n");
            end=1;
        }   
        if(!strcmp(item,"q-")){
            break;
        }
        if(!strcmp(item,"p-")){
            if(requires[0]!=NULL){
                for(int j=0;j<requiresSize;j++){
                    printf("%d. %s\n",j,requires[j]);
                }
            }
            end=1;
        }
        while(end==0){
                printf("check1:i=%d\n",i);
            //if search index is larger than size of the array,reallocate the array
            if(i>= requiresSize){
                temp = new char * [requiresSize*2];
                //Initialize each element in temp
                printf("check2:temp initalized\n");
                for(int j=0;j<requiresSize*2;j++){
                    temp[j]= new char[256];
                }
                printf("check3:temp itmes initialized\n");
                for(int j =0;j<requiresSize;j++){
                    //for each element in requires, copy that element to temp
                    temp[j]=requires[j];
                }
                printf("check4:copied requires into temp\n");
                delete * requires;
                requires = temp;
                printf("check5:deleted requires and set requires equal to temp\n");
                delete  temp;
                requiresSize = requiresSize *2;         
            }
                printf("check6:\n");
            //if the index at requires is not empty, check to see if it is the same as given item
            if(requires[i]!= NULL){
                printf("check8:index at requires is not empty\n");
                //I know the error occurs here, something to do with accessing requires[i]
                if(!strcmp( item,  requires[i])){
                printf("check9:index at requires is the same as item\n");
                    //if they are the same, break out of the loop, item is already included
                    break;
                }else{
                printf("check10:index at requires is different\n");
                    //otherwise, increase the index and check again (continue loop)
                    i++;
                    break;
                }
            }else{
                printf("check11:index at requires is null, item added\n");
                //if the index is empty, add the item to the list and break out of loop
                requires[i]=  item;
                break;
            }
                printf("check7\n");

        }
    }
    delete requires;
    return 0;
}

提前谢谢你。

Answer 1

你需要意识到像temp = requires这样的赋值语句（在这种情况下）只是复制指针，所以temp现在指向与requires相同的内存位置；它不会复制该内存。

这会导致两个问题：

1. 您正在为temp 的每个元素分配新的256 字符数组，然后重新分配temp 中的每个字符* 以指向不同的位置，从而泄漏所有内存；现在无法引用新分配的内存，因此您无法释放它。

1234563 ，同样，requires 现在也指向）。

另外，如果你使用new[] 分配一个数组，你必须使用delete[] 来释放它。所以requires = new char * [requiresSize]; 要求你在程序结束时使用delete [] requires;，而不仅仅是delete requires;。 requires 的每个 256 字符元素都相同。

因此，将temp[j]=requires[j]; 替换为对strcpy（或strncpy）的适当调用。并且不要删除temp；最后的 delete [] requires; 将处理这个问题，因为它现在指向那段内存。