清除cout缓冲区(c)

你好我正在写一个简短的程序来实现一个 shell,我遇到了一个不寻常的问题.由于某种原因,我无法清除std :: cout缓冲区.该程序不会打印出消息.我理解一个简单的解决方案是切换到std :: cerr,但有没有办法用cout打印消息？

我尝试过的事情：

> std :: cout.flush()

>在将任何内容写入标准输出后插入std :: endl.

>将std :: flush插入输出流

> std :: cout.setf(std :: ios :: unitbuf);我发现这应该是非缓冲输出.

任何帮助非常感谢这里是我的代码：

int main()
{
    //Tryed this to unbuffer cout, no luck.
    std::cout.setf(std::ios::unitbuf);

    std::string input;

    //Print out shell prompt and read in input from keyboard.
    std::cout << "myshell> ";   
    std::getline(std::cin, input);

    //**********************************************************************
    //Step 1) Read in string and parse into tokens.
    //**********************************************************************

    char * buf = new char[input.length() + 1];
    strcpy(buf, input.c_str());

    int index = 0;
    char * command[256]; 

    command[index] = std::strtok(buf, " ");    //Get first token.
    std::cout << command[index] << std::endl;

    while (command[index] != NULL)
    {
        ++index;
        command[index] = std::strtok(NULL," ");    //Get remaining tokens.
        std::cout << command[index] << std::endl;
    }   

    std::cout.flush(); //No luck here either

    //HERE IS WHERE MY PROBLEM IS.
    std::cout << index << " items were added to the command array" << std::endl;

    delete[] buf;

    return 0;   
}

问题是你在while循环的最后一次迭代中向cout发送NULL,这导致UB,在你的情况下是干扰cout.在向cout发送任何内容之前检查NULL,你没事：

if (command[index] != NULL) {
    std::cout << command[index] << std::endl;
}