Question

string word;
int l,eFound,xFound;
l = word.size();
cout <<"Enter a word: ";
cin >> word;

for (l>0 ; word.at(l)!='x' || word.at(l)!='e'; l--)
    if (word.at(l) == 'e'){
        eFound = true;
    }
    else if (word.at(l) == 'x'){
        xFound = true;
    }

if (eFound == true && xFound == true){
    cout << "Your word, "<<word<<", contains the character 'e'"<<"\n";
    cout << "Your word, "<<word<<", contains the character 'x'";
}
if (eFound == true && xFound != true){
    cout << "Your word, "<<word<<", contains the character 'e'";
}
if (xFound == true && eFound != true){
    cout << "Your word, "<<word<<", contains the character 'x'";
}

我不确定发生了什么，我正在尝试使用for循环来检测某个单词的输入中的e或x。我单击了具有相同错误的其他页面，但是它们具有不同的代码，并且我不太了解所解释的内容。那么，是什么导致此错误？我上第一门编程课已经两周了，如果我问一个愚蠢的问题，对不起。

Answer 1

问题是std::string的索引从零开始。不是来自1。因此，如果word.at(l)，l = word.size();将崩溃。

您应将语句更改为：l = word.size() - 1;。

此外，将循环条件更改为for (; l >= 0 ; l--)

建议：

请使用库功能：

赞：

#include <iostream>
#include <string>

using namespace std;

int main() {
    string word;
    cout <<"Enter a word: ";
    cin >> word;

    bool eFound = word.find('e') != string::npos;
    bool xFound = word.find('x') != string::npos;

    if (eFound) {
        cout << "Your word, "<<word<<", contains the character 'e'" << "\n";
    }

    if (xFound) {
        cout << "Your word, "<<word<<", contains the character 'x'" << "\n";
    }

    return 0;
}

引发'std :: out_of_range'what（）实例后调用终止方法what（）：basic_string

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