Question

这是一个关于指向字符串的指针与指针（复数）到字符串数组的问题。这是代码 - 请查看问题的评论：

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

    // Here, we're just loading each string into an incrementing pointer and printing as we go. It's not an array of pointers. This works fine.

    char* dumb = NULL;
    cout << argc << endl;
    for (int i = 0; i < argc; i++) {
        dumb = argv[i];
        cout << dumb << endl;
        dumb++;
    }

    // Next we'll try to load the strings into an array of pointers, and print them out. This causes an error (see below).

    char** dumber = NULL;
    cout << argc << endl;
    for (int i = 0; i < argc; i++) {
        *(dumber + i) = argv[i];
        cout << dumber[i] << endl;
    }

    return 0
}

Testing.exe中0x001899F7处的未处理异常：0xC0000005：访问冲突写入位置0x00000000。

有人可以纠正我吗？

Answer 1

你的指针数组在哪里？

char** dumber =NULL;

只是指向char＆＃39;的指针。不是一个指针数组。你需要

char** dumber = new char*[argc];

Answer 2

将此行添加到代码中：

dumber = malloc(100); //   adjust the size as per your requirements

因为每次尝试时都没有为“dumber”分配任何空间写一个新字符串，你将它写在0x00000000位置，写在这个
location会导致访问冲突，因为不允许任何进程写入此地址该地址专门为空指针保留。

Answer 3

这就是我最终的结果。也许这会对其他新手有所帮助，所以我想我发布了。它通过释放分配给dumber数组的内存来完成，我理解这是一件好事。

#include <iostream>
using namespace std;

int main(int argc, char* argv[]) {
// int main(int argc, char** argv) { // An alternative.

    // Just loading each string into the same pointer and printing as we go.
    // It's not an array of pointers.
    char* dumb;
    cout << argc << endl;
    for (int i = 0; i < argc; i++) {
        dumb = argv[i];
        cout << dumb << endl;
    }

    // Next we'll load the strings into an array of pointers, and print them out.
    char** dumber = new char* [argc];
    cout << argc << endl;
    for (int i = 0; i < argc; i++) {
        *(dumber + i) = argv[i];
        cout << dumber[i] << endl;
    }

    // Deallocate the memory dynamically assigned by the 'new' operator for the
    // pointer array.
    delete[] dumber;
    dumber = 0;

    cin.get();
    return 0;
}

指针 - 单个字符串与字符串数组

3 个答案: