VSCode：clang：错误：链接器命令失败，退出代码为 1（使用 -v 查看调用）

Question

我已经为我所在的一个类编写了一个代码，它应该拆分一个字符串并显示第二个字符串中有多少个唯一字符。但是，每次我尝试构建代码时，我都会收到“clang: error: linker command failed with exit code 1 (use -v to see invocation)”。我查看了有关此错误的其他问题，但没有找到 VSCode 的解决方案（如果有任何区别，则在 MacBook Pro 上）。有人看到我的错误在哪里吗？还有一个子问题：我是否一定需要#include 字符串？我被告知 std::string 是 iostream 库的一部分。如果我不需要字符串库，我什么时候需要字符串？谢谢大家。

    #include <iostream>
    #include <string>

    int splitwords(std::string, char);
    int findnumchar(std::string);

    int main()
    {
        std::string txt1("ABCDEF,GHI,JKLMN,OP");
        std::string txt2("BACDGABCDAZ");
        int result;
        char delimiter = ',';
        result = splitwords(txt1, delimiter);

        result = findnumchar(txt2);


    }

    int splitwords(std::string txt, char delimiter)
    {
        int start, found, cnt = 0;
        std::string splitstr;

        start = 0;
        while ((found = txt.find(delimiter, start)) != std::string::npos)
        {
            splitstr = txt.substr(start, found - start);
            std::cout << "Split  Word" << splitstr << std::endl;
            start = found + 1;
            cnt += 1;
        }
        splitstr = txt.substr(start, txt.length() - start);
        std::cout << "Split  Word" << splitstr << std::endl;
        return cnt + 1;
    }

    int findnumchnar(std::string txt)
    {

        int uniquecnt = 0, index;
        int seen[26] = {0};
        std::string::iterator iter;
        for (iter = txt.begin(); iter < txt.end(); iter++)
        {
            index = *iter - 'A';
            if (seen[index] == 0)
            {
                seen[index] = 1;
                uniquecnt+=1;
            }
        }

        for (int i = 0; i <= 26; i++)
        {
            if (seen[i] == 1)
            {
                std::cout << static_cast<char>(i + 'A') << "\t";
            }
        }
        std::cout << std::endl;
        std::cout << "The number of unique characters: " << uniquecnt << std::endl;
        return uniquecnt;
    }

Answer 1

有一个简单的错字，这就是我收到错误的原因。对不起各位，不过还是谢谢你的回答！我将 findnumchar 定义为 findnumchnar。总是检查愚蠢的东西

Answer 2

我有 findnumchnar() 而不是声明的 findnumchar()

Answer 3

因为 std::string::npos 是 64 位常量，而 found 是 32 位整数，编译器会抱怨 while 循环内部的条件永远不会为真（因为理论上 found 永远不会变大足以等于 std::string::npos)。要解决此问题，只需将 int 替换为 unsigned long long int。完整代码见下面sn-p：

    #include <iostream>
    #include <string>

    int splitwords(std::string, char);
    int findnumchar(std::string);

    int main()
    {
        std::string txt1("ABCDEF,GHI,JKLMN,OP");
        std::string txt2("BACDGABCDAZ");
        int result;
        char delimiter = ',';
        result = splitwords(txt1, delimiter);

        result = findnumchar(txt2);


    }

    int splitwords(std::string txt, char delimiter)
    {
        unsigned long long start, found, cnt = 0;
        std::string splitstr;

        start = 0;
        while ((found = txt.find(delimiter, start)) != std::string::npos)
        {
            splitstr = txt.substr(start, found - start);
            std::cout << "Split  Word" << splitstr << std::endl;
            start = found + 1;
            cnt += 1;
        }
        splitstr = txt.substr(start, txt.length() - start);
        std::cout << "Split  Word" << splitstr << std::endl;
        return cnt + 1;
    }

    int findnumchnar(std::string txt)
    {

        int uniquecnt = 0, index;
        int seen[26] = {0};
        std::string::iterator iter;
        for (iter = txt.begin(); iter < txt.end(); iter++)
        {
            index = *iter - 'A';
            if (seen[index] == 0)
            {
                seen[index] = 1;
                uniquecnt+=1;
            }
        }

        for (int i = 0; i <= 26; i++)
        {
            if (seen[i] == 1)
            {
                std::cout << static_cast<char>(i + 'A') << "\t";
            }
        }
        std::cout << std::endl;
        std::cout << "The number of unique characters: " << uniquecnt << std::endl;
        return uniquecnt;
    }