C ++：附加到向量字符串

Question

我正在编写一个“猪拉丁”程序；读取用户的输入（名字和姓氏），使输入变为小写并根据名称中的内容更改名称。如果第一个字母（名字和姓氏）都是元音，我们应该在其末尾添加“ way”。

如果第一个字母是辅音，我们将把第一个字母移到字符串的末尾，并在其末尾添加“ ay”。

尝试在字符串末尾添加文本时，我的代码给我错误。它说不能将字符串转换为字符，我不确定这意味着什么。它还说我不能将输出操作数<<用于字符串，即使我以前使用过它也是如此。

错误出现在“ strcpy”和我输出名称的最终代码中。

37：错误：无法转换'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >'到'char*'的参数'1'来'char* strcpy(char*, const char*)'

47：错误：无法转换'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >'到'char*'的参数'1'来'char* strcpy(char*, const char*)'

54：错误：不对应的'operator<<'在'std::cout << first'

我只需要一些帮助来修复错误并查看我哪里出错了。包括完整的代码。

#include <iostream>
#include <vector>
#include <cstring>
#include <string>
using namespace std;
int main()
{
    int q, s;
    char shea[] = "way";
    char gavin_stop_looking_at_ponies[] = "ay";
    vector <string> first;
    vector <string> last;
    cout << "Please enter your first name." << endl;
    for (int i = 0; i < first.size(); i++)
    {
        getline (cin, first[i]);
        string nfirst = first[i];
        while (nfirst[q])
        {
            nfirst[q] = tolower(nfirst[q]);
        }
        first[i] = nfirst;
    }
    cout << "Please enter your last name." << endl;
    for (int j = 0; j < last.size(); j++)
    {
        getline (cin, last[j]);
        string nlast = last[j];
        while (nlast[s])
        {
            nlast[s] = tolower(nlast[s]);
        }
        last[j] = nlast;
        }
    if ( (first[0] == "a") ||( first [0] == "e") || (first [0] == "i") || (first [0] == "o")     || (first [0] == "u"))
    {
        strcpy (first, "way");
    }
    else
    {
        first[first.size()] = first[0] + "ay";
    }

    if ( (last[0] == "a") ||( last [0] == "e") || (last [0] == "i") || (last [0] == "o") || (last [0] == "u"))
    {
        strcpy (last, "way");
    }
    else
    {
        last[last.size()] = last[0] + "ay";
    }
    cout << first << last << endl;
    return 0;
}

Answer 1

我已经在您的代码中添加了一些问题的解释和解决方案的建议。如果您不了解任何内容，请发表评论，我会尽力澄清。

#include <iostream>

// You don't need 'vector' for this.
#include <vector>

// You won’t often need the C string header in C++.
#include <cstring>

#include <string>
using namespace std;
int main()
{
    // These variables are unused.
    int q, s;
    char shea[] = "way";
    char gavin_stop_looking_at_ponies[] = "ay";

    // 'first' and 'last' are both names, not collections
    // of names.
    string first;
    string last;
    vector <string> first;
    vector <string> last;

    // 'endl' is unnecessary here; it outputs a newline and
    // flushes the stream, but standard output is usually
    // line-buffered, meaning that newline flushes the
    // stream regardless.
    cout << "Please enter your first name.\n"
    cout << "Please enter your first name." << endl;

    // If you just want to get one name, 'getline' is perfect.
    getline(cin, first);

    // This loop would run 0 times because 'first' is an
    // empty vector.
    for (int i = 0; i < first.size(); i++)
    {
        getline (cin, first[i]);
        string nfirst = first[i];
        while (nfirst[q])
        {
            nfirst[q] = tolower(nfirst[q]);
        }
        first[i] = nfirst;
    }

    // To make a string lowercase, use 'tolower' on each character.
    // Here's one way to do it:
    for (string::size_type i = 0; i < first.size(); ++i)
        first[i] = tolower(first[i]);

    // Here's another, with C++11 enabled:
    for (auto& c : first)
        c = tolower(c);

    cout << "Please enter your last name.\n";
    cout << "Please enter your last name." << endl;

    // Same thing.
    getline(cin, last);
    for (int j = 0; j < last.size(); j++)
    {
        getline (cin, last[j]);
        string nlast = last[j];
        while (nlast[s])
        {
            nlast[s] = tolower(nlast[s]);
        }
        last[j] = nlast;
    }

    // Now 'first' is a string, and 'first[0]' is a 'char'.
    // "a" is a string literal; 'a' is a character literal.
    // You can compare each character individually:
    if (first[0] == 'a' || first[0] == 'e' || first[0] == 'i' || first[0] == 'o' || first[0] == 'u')

    // Or you can say "if the character was found in this
    // set of vowels".
    if (string("aeiou").find(first[0]) != string::npos)

    if ( (first[0] == "a") ||( first [0] == "e") || (first [0] == "i") || (first [0] == "o")     || (first [0] == "u"))
    {
        // This would try to copy "way" into 'first':
        // formerly a vector of string objects, now just a
        // string object. 'strcpy' wants a character buffer,
        // and will overwrite characters in that buffer—
        // probably not what you want:
        //
        // "aaron" => "wayon"
        // 
        strcpy (first, "way");

        // Instead, just append "way":
        first += "way";
    }
    else
    {
        // This says "take the first first character of the
        // string, add the value of that character to a
        // pointer to a buffer containing "ay", then try to
        // copy the resulting pointer past the end of the
        // string. Again, not quite what you intended!
        first[first.size()] = first[0] + "ay";

        // Think of it instead like this: take everything
        // after the first character, add a string consisting
        // of the first character back onto the end, then add
        // "ay" after that.
        first = first.substr(1) + string(1, first[0]) + "ay";
    }

    // Duplicated code! You could move the above logic into
    // a function to avoid this duplication. Then you only
    // have to work on it in one place. :)
    if ( (last[0] == "a") ||( last [0] == "e") || (last [0] == "i") || (last [0] == "o") || (last [0] == "u"))
    {
        strcpy (last, "way");
    }
    else
    {
        last[last.size()] = last[0] + "ay";
    }

    // I need a space between my first and last names!
    cout << first << ' ' << last << '\n';
    cout << first << last << endl;
    return 0;
}