Question

这是我的计划。输出应该是'a'字符的序列，但由于某种原因，它不是。为什么呢？

#include <iostream>

using namespace std;

const int NAME_LENGTH = 16;

struct Record {
    char hotel_name[NAME_LENGTH];
};

int main() {
    int amount = 5;
    for (int i = 0; i < amount; i++) {
        Record * elementToBeAdded = new Record;
        for (int j = 0; j < NAME_LENGTH; j++)     
            elementToBeAdded->hotel_name[i] = 'a';
        elementToBeAdded->hotel_name[NAME_LENGTH-1] = '\0';
        cout << "string-" << elementToBeAdded->hotel_name << "-\n\n";
    }
}

Answer 1

因为你有错字

        elementToBeAdded->hotel_name[i] = 'a';

你的意思是j，而不是我。

Answer 2

当肯定打算使用i时，您的最内层循环正在使用j：

elementToBeAdded->hotel_name[i] = 'a';

因此，您永远不会设置Record＃1的char[]的第0个元素，也不会设置Record＃2的第1个元素等。这意味着每个通过最外层循环的时间除了第一个，酒店名称中的第一个char将保持未初始化状态，很可能是\0。

此外，您正在创建的Record对象永远不会被删除，因此这会在每次迭代时泄漏内存。

Answer 3

elementToBeAdded->hotel_name[i] = 'a';

应该是

elementToBeAdded->hotel_name[j] = 'a';

Answer 4

因为上面的源代码中存在拼写错误：

 for (int j = 0; j < NAME_LENGTH; j++)     
  elementToBeAdded->hotel_name[i] = 'a';

索引应该是[j]。

Answer 5

您i和j混淆了：

for (int j = 0; j < NAME_LENGTH; j++)     
    elementToBeAdded->hotel_name[i] = 'a'; //<-- should be j here

你也在泄漏内存，你应该在外循环结束时delete elementToBeAdded;。

Answer 6

现在您已经知道错误，请使用std::fill_n代替算法标题中的内容。

#include <algorithm>

// .....

for (int i = 0; i < amount; i++) {         

   Record * elementToBeAdded = new Record;
   std::fill_n( elementToBeAdded->hotel_name, NAME_LENGTH-2, 'a' );
   elementToBeAdded->hotel_name[NAME_LENGTH-1] = '\0';         

   cout << "string-" << elementToBeAdded->hotel_name << "-\n\n";
   delete  elementToBeAdded;
}

字符串初始化失败

6 个答案: