Question

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT. */

    string list[] = {"fiorello", "nonuterine", "asquint", "commodore", "semiprogressive",
                     "aviculturist", "brayley", "tendentious", "hungriness", "overbulkily",
                     "subfumigation", "praline", "fiorello", "presurvey", "unjealous",
                     "brayley", "unimpassionate", "welshman", "dcor", "traducianist"};

    int size = sizeof(list);
    for (int i = 0; i < size; i++) {
        cout << list[i] << endl; 
        // THIS IS WHERE I REALIZE NOTHING ELSE PRINTS AFTER THIS POINT.
    }
    cout << endl;

    int z = sizeof(list) / sizeof(list[0]);
    sort(list, list + z);
    for (int y = 0; y < z; y++) {
        cout << list[y] << endl;
    }

    return 0;
}

我没有强大的C ++背景，来自HTML，CSS等，所以试图解决这个问题。

我要完成的是打印出阵列，然后按字母顺序打印出来，然后找到重复项并删除并重新打印出来。最后，找出数组中每个单词的长度并打印出来。

Answer 1

正如评论中所述，您第一次使用sizeof时出错。一个很好的解决方案是根本不使用它，而是使用标准库算法，它将通过模板推导找到大小：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
    string list[]={"fiorello","nonuterine","asquint","commodore","semiprogressive","aviculturist","brayley","tendentious","hungriness","overbulkily","subfumigation","praline","fiorello","presurvey","unjealous","brayley","unimpassionate","welshman","dcor","traducianist"};

    // Operate on each item in list - don't need to mention count explicitly
    for ( auto&& s : list )
        cout << s << '\n'; 
    cout << endl;

    // Same as sort(list, list+z)
    sort( begin(list), end(list) );

    for ( auto&& s : list )
        cout << s << '\n'; 
    cout << endl;
}

您的评论建议您计划删除重复项，但仍希望使用C风格的数组。所以你可能会在列表计数中使用变量;你可以通过使用：

得到这个

size_t count = distance( begin(list), end(list) );

而不是使用sizeof事物。除了容易出错之外，即使您稍后将代码更改为使用容器而不是C样式数组，这也会继续工作。

C ++：为什么我的代码在第一个for循环后停止运行？

1 个答案: