字符串查找函数返回奇数

string.find function returning odd numbers

本文关键字：返回 函数 查找 字符串      更新时间：2023-10-16

我正试图找到找到字符的位置。

const char* normalize(std::string path) 
{
    std::cout << "executed   " << path << std::endl;
    //"foo//\bar////bar2///../.bar2" -- foo/bar/bar2
    std::size_t found;
    std::size_t found2;
    std::size_t curchar = 0;
    std::string final;
    std::string buffer;
    bool notdone = true;
    while (notdone) {
        //std::cout << "loop" << std::endl;
        //find the current element
        // can be / or 
        found = path.find("/", curchar);
        found2 = path.find("\",curchar);
        std::cout << found << std::endl;
        SDL_Delay(2000);
        if (found != std::string::npos && found2 != std::string::npos) {
            if (found < found2){
                //read from the curchar to the slash
                if (curchar-found > 1){
                    buffer = path.substr(curchar,found-curchar-1);
                    //add to path
                    final = final + "/" + buffer;
                }
                curchar = found+1;
                //buffer will be the file/component
            }else{
                if (curchar-found2 > 1){
                    buffer = path.substr(curchar,found2-curchar-1);
                    //add to path
                    final = final + "/" + buffer;
                }
                curchar = found2+1;
            }
        }else if(found != std::string::npos){
            //std::cout << "loop2" << found == std::string::npos << std::endl;
            //std::cout << "loop2   " << path.substr(curchar, 1) << std::endl;
            if (curchar-found > 1){//
                buffer = path.substr(curchar,found-curchar-1);
                //add to path
                final = final + "/" + buffer;
            }
            curchar = found+1;
        }else if(found2 != std::string::npos){
            std::cout << "loop3" << std::endl;
            if (curchar-found2 > 1){
                buffer = path.substr(curchar,found2-curchar-1);
                //add to path
                final = final + "/" + buffer;
            }
            curchar = found2+1;
        }else{
            std::cout << "finishing" << std::endl;
            final = final + "/" + path.substr(curchar,path.size()-curchar);
            notdone = false;
        }
    }
    return final.c_str();
}
normalize("test/");

这个代码应该打印出"4"，但它却打印出了18。它在无限循环中打印出18个。但是，如果我使用std::cout << path.find("/", curchar) << std::endl，它确实会打印4。一开始我以为它实际上并没有返回std::size_t，但我检查了一下，它确实返回了。

您的下一行正在创建问题

  //find the current element
  // can be / or 
  found = path.find("/", curchar);

我在我的linux终端上运行，GCC被视为下一行，作为上一行注释的延续。

basic.cpp:18:9: warning: multi-line comment [-Wcomment]
         // can be / or 
         ^
basic.cpp: In function ‘const char* normalize(std::string)’:
basic.cpp:21:22: warning: ‘found’ may be used uninitialized in this function [-Wmaybe-uninitialized]
         std::cout << found << std::endl;
                      ^

现在，由于上面的注释样式，您的下一行（代码）被视为注释。由于发现未初始化，因此它有垃圾值，这使您的逻辑出错，因为它没有进入您重置标志未完成的路径。

然而，GCC或任何其他编译器都应该发出警告（使用未初始化变量），如果我们仔细阅读，我们可能会进行回溯并理解问题。

解决方案是将注释样式更改为

/* // can be / or  */
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