c++ - 在 C++ 中查找精确的字符串匹配

Question

这是我用来检测 txt 文件中一行字符串的代码:

int main()
{
    std::ifstream file( "C:\\log.txt" );

    std::string line;
    while(!file.eof())
    {
        while( std::getline( file, line ) )   
        {
            int found = -1;
            if((found = line.find("GetSA"))>-1)
                std::cout<<"We found GetSA."<<std::endl;
            else if ((found = line.find("GetVol"))>-1)
                std::cout<<"We found GetVol."<<std::endl;
            else if ((found = line.find("GetSphereSAandVol"))>-1)
                std::cout<<"We found GetSphereSAandVol."<<std::endl;
            else
                std::cout<<"We found nothing!"<<std::endl;

        }
    }
    std::cin.get();
}

这是我的日志文件:

GetSA (3.000000)

GetVol (3.000000)

GetSphereSAandVol (3.000000)

GetVol (3.000000)

GetSphereSAandVol (3.000000)

GetSA (3.00000)

错误是，程序不会去寻找“GetSphereSAandVol”，因为它停在了“GetSA”处。显然，程序认为“GetSphereSAandVol”中包含“GetSA”，所以会执行:

if(found = line.find("GetSA"))
    std::cout<<"We found GetSA."<<std::endl;

这不是我想要的，因为我希望程序执行:

else if (found = line.find("GetSphereSAandVol"))
    std::cout<<"We found GetSphereSAandVol."<<std::endl;

那么，无论如何我都可以避免这种情况？得到我真正想要的？非常感谢。

Answer 1

您误解了find 的工作原理。阅读 documentation .

条件应该是这样的:

if ((found = line.find("xyz")) != line.npos) { /* found "xyz" */ }

---

我会这样写你的整个程序:

int main(int argc, char * argv[])
{
    if (argc != 2) { std::cout << "Bad invocation\n"; return 0; }

    std::ifstream infile(argv[1]);

    if (!infile) { std::cout << "Bad filename '" << argv[1] << "'\n"; return 0; }

    for (std::string line; std::getline(infile, line); )
    {
        int pos;

        if ((pos = line.find("abc")) != line.npos)
        {
            std::cout << "Found line 'abc'\n";
            continue;
        }

        if ((pos = line.find("xyz")) != line.npos)
        {
            std::cout << "Found line 'xyz'\n";
            continue;
        }

        // ...

        std::cout << "Line '" << line << "' did not match anything.\n";
    }
}