Question

业余C ++学生赋值是创建一个函数，它接受两个字符串向量，并返回第三个向量，它以交替的形式组合前两个的内容，最后是剩余的任何剩余。
它编译并运行，但在显示内容之前要求输入，一次一行。请不要花哨，我只想知道我搞砸了哪里。谢谢！

vector<string> shuffle(vector<string> L1, vector<string> L2);

int main() {

    string s;
    vector<string> L1;
    vector<string> L2;

    cout << "Enter words for the first list\n";
    while (cin >> s && s != "END") {
        L1.push_back(s);
    }

    cout << "Enter words for the second list\n";
    while (cin >> s && s != "END") {
        L2.push_back(s);
    }   

    shuffle(L1, L2);
    for (size_t k = 0; k < L3.size(); k++) {
    cout << L3[k] << endl;
    }

    cout << endl;

    return 0;
}

vector<string> shuffle(vector<string> L1, vector<string> L2) {  
    vector<string> L3;
    string s;

    for (size_t i = 0; i < L1.size() && i < L2.size(); i++) {
        if (i < L1.size()) {
            cout << L1[i] << " ";
            cin >> s;
            L3.push_back(s);
        }   
        if (i < L2.size()) {
            cout << L2[i] << " ";
            cin >> s;
            L3.push_back(s);
        }
    }       

    if (L1.size() > L2.size()) {
        for (size_t j = 0; j < (L1.size()-L2.size()); j++) {
            cout << L1[j+L2.size()] << " ";
            cin >> s;
            L3.push_back(s);
        }
    }

    if (L2.size() > L1.size()) {
        for (size_t j = 0; j < (L2.size()-L1.size()); j++) {
            cout << L2[j+L1.size()] << " ";
            cin >> s;
            L3.push_back(s);
        }
    }

    return L3;
}

Answer 1

您的函数shuffle要求输入，因为您从中调用了cin >> s。 shuffle并没有真正使用作为参数传递的两个字符串向量。实际上，它忽略了传递给它的字符串的值，形成了新输入的字符串L3。

C ++函数返回向量组合两个向量参数

1 个答案: