Question

我是C ++的新手，并且正在努力使以下代码工作：

struct Keywords
{
     const char* const* keys;
     int count;
}

Keywords GetKeywords()
{
     Keywords keywords;
     keywords.count = 10;
     string keys = "a b c d e";
     keywords.keys = keys???;
     return keywords;
}

int main(int argc, char *argv[])
{
     Keywords keywords = GetKeywords();
     cout<<keywords.count<<endl; //prints 10
     cout<<keywords.keys<<endl;  //prints a memory address like 0030F838
}

如何让keys指向a b c d e？到目前为止，我已经尝试了这个

char * writable = new char[keys.size() + 1];
strcpy(writable, keys.c_str());
writable[keys.size()] = '\0';

const char* a = writable;
const char** ptr = &a;
results.keys = ptr;

调试时我可以看到results.keys显示a b c d e，但只要GetKeywords()返回，results.keys就会指向不可读的内存位置。

我无法对struct Keywords和GetKeywords()方法签名进行任何更改。

Answer 1

考虑到Some programmer dude，Miles Budnek和其他人提供的提示，我能够解决问题，如下所述：

在我原来的问题中，密钥由string keys = "a b c d e";表示。但是，我在a b c d e上工作的作业是通过迭代set<string>项来构建的。所以这是如何完成的：

struct Keywords
{
    const char* const* keys;
    int count;
}

Keywords GetKeywords()
{
    Keywords keywords;
    keywords.count = 5;

    set<string> keys;
    GetKeys(keys);//returns keys = {"a", "b", "c", "d", "e"}

    char** charKeysPtr = new char *[keys.size() + 1];
    set<string>::iterator it;
    int index = 0;

    //Iterate the set and copy each key
    for (it = keys.begin(); it != keys.end(); it++)
    {
        charKeysPtr[index] = new char[it->size() + 1];
        strcpy(charKeysPtr[index], it->c_str());
        charKeysPtr[index][it->size()] = '\0';
        index++;
    }

    keywords.Keys = charKeysPtr;

    return keywords;
}

int main(int argc, char *argv[])
{
     Keywords keywords = GetKeywords();

     //Print keys
     for (int i = 0; i < keywords.count; ++i)
        cout << keywords.keys[i] << " ";

     //Output is "a b c d e"

     //Memory cleanup and etc.
}

谢谢你们！

为指向const char的const指针的指针赋值（const char * const * p）

1 个答案: