Question

在遍历此向量时我没有得到正确的输出，我不确定为什么

vector<string> words = {"once","i","was","seven"};
for(int i=0;i<=words.size();i++){
    cout<<words[i]<<endl;
    for(int j =0; j<=words.size();j++){
        cout<<words[j]<<endl;
    }
}

此代码应打印：once once i was seven i once i was seven was once i was seven seven once i was seven，但它打印：once once i was seven，我该如何解决

Answer 1

打印“一次，我7岁”后，由于单词size为4，并且您正在循环5个元素（从i = 0到4，而word [4]将是未定义的行为），它将崩溃。只需从条件中删除等号，您就可以开始了。

vector<string> words = {"once","i","was","seven"};
for(int i = 0; i < words.size(); i++){
    cout << words[i] << endl;
    for(int j = 0; j < words.size(); j++){
        cout << words[j] << endl;
    }
}

Answer 2

vector<string> words = {"once","i","was","seven"};

words[0] == "once"
words[1] == "i"
words[2] == "was"
words[3] == "seven"

<= words.size()使循环带您进入

words[4] == <undefined behaviour>

因此，最大索引为size()-1。您通常可以通过使用基于范围的for循环来避免麻烦：

for(const auto& outer_word : words) {
    cout<< outer_word << endl;
    for(const auto& word : words) {
        cout << word << "\n";
    }
}

遍历向量时如何获得正确的输出？

2 个答案: