如何在C ++中访问新的Main Function的参数? [英] How do I access arguments for a new Main Function in C++?

问题描述

我有一个新的主要功能，要求链接程序在Visual Studio中指向它.

I have a new main function that I asked the linker to point to in the Visual Studio.

我可以正确执行此功能.但是，我得到的命令行参数不正确.

I can correctly execute this function. However the command line arguments that I am getting is not correct.

当前这是我的签名

int NewMain(int argc, const char* argv[])
    {
    cout << "New Main" << endl;
    for (int i = 0; i < argc; ++i)
        {
        cout << argv[i] << "\n";
        }
    return 0;
    }

但是，当我将相同的函数与标准main一起使用时，我会得到所有参数.

However when I use the same function with the standard main, I am getting all the arguments.

推荐答案

当您指定链接器的入口点时，调用的函数不会传递 any 参数.它的签名是int entrypoint(void);

When you specify an entry point to the linker, the function that gets called is not passed any arguments. Its signature is int entrypoint(void);

如果需要命令行，可以使用GetCommandLine进行检索.如果要将其解析为参数，则可以自己进行解析，也可以使用CommandLineToArgvW，但请注意，宽字符版本仅提供 功能，因此如果需要标准命令行(使用char，而不是wchar_t)，您必须自己执行此操作.

If you want the command line, you can retrieve it with GetCommandLine. If you want to parse that into arguments, you can either do so on your own, or use CommandLineToArgvW -- but note that this is only available in a wide-character version, so if you want a standard command line (using chars, not wchar_ts) you have to do that on your own.

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