Question

我正在尝试在数组中搜索值。阵列是否需要一个确定的数字？我该如何搜索这个号码？谢谢你的帮助！

#include <stdio.h>
#include <conio.h>
#include <iostream>
#include "windows.h"


using namespace std;
int main() {
    int userinput;

    int arrayofnumber[10] = { 5, 3, 77, 43, 6, 22, 7, 9, 84, 26 };

    int NumberInArray = sizeof(arrayofnumber) / sizeof(arrayofnumber[0]);

    cout << "What Number Would You Like To Search For? \n";
    cout << "Number of Values in Array = " << NumberInArray << endl;
    cin >> userinput;

    for (int i = 0; i >= NumberInArray; i++) {

        if (userinput == NumberInArray[i]) {
            cout << "We Found IT! It = " << NumberInArray[i] << endl;
        }

    }

    return 0;
}

Answer 1

for (int i = 0; i < NumberInArray; i++) {
    if (userinput == arrayofnumber[i]) {
        cout << "We Found IT! It = " << arrayofnumber[i] << endl;
        break;
    }
}

您的for循环条件i >= NumberInArray错误。它应该是i < NumberInArray。

NumberInArray[i]是非法的。你应该arrayofnumber[i]。

添加了break，以便在找到号码时，您可以退出循环。

Answer 2

该标准为我们提供find来执行此操作。所以我建议使用C ++的实现而不是自己编写它，这样做也会消除你对NumberInArray的需求：

if(cend(arrayofnumber) != find(cbegin(arrayofnumber), cend(arrayofnumber), userinput)) cout << "We Found IT! It = " << userinput << endl;

Live Example

如何在for循环中使用array [i]

2 个答案: