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如何处理命令行参数?

[英]How to process command line arguments?

Yesterday I made a simple program in c++ that uses the arguments passed through command line. 昨天我用c ++编写了一个简单的程序,该程序使用了通过命令行传递的参数。

E.G. myDrive:\myPath\myProgram.exe firstWord secondWord

The program run fine and do what it has to, but there's a little curiosity I have: I had to write argc --; 该程序可以正常运行并执行所需的操作,但是我有一点好奇:我必须写argc --; before I could use it well, otherwise I have a run-time crash [The compiler won't speak!]. 在我不能好好使用它之前,否则会发生运行时崩溃[编译器不会讲话!]。

In particular argc gives me a bad time when I don't give any word as argument to the program when I run it... 特别是当我在运行程序时不给任何单词作为程序的参数时, argc给我带来了糟糕的时光...

Now it works, so isn't bad at all, but I wonder why this is happening! 现在它可以工作了,一点也不坏,但是我想知道为什么会这样! [PS making argc --; [PS制作argc --; and printing it, it gives 0 as value!] 并打印出来,它的值为0 !]

EDIT: Here all the istructions that use argc 编辑:这里所有使用argc的指令

int main(int argc, char *argv[]) {
    [...]
    argc --;
    if(argc > 0){
        if(firstArg.find_last_of(".txt") != string::npos){
            reading.open(argv[1], ios::binary);
            [...]
        }
    }
    if ((!(firstArg.find_last_of(".txt") != string::npos)) && argc > 0){
    [...]
        for(int i = 1; i <= argc; i ++){
        [...]
        toTranslate = argv[i][j];
        [...]
        toTranslate = argv[i][j];
        }
    }
}

The arguments include the name of the program itself as well, so argc is always at least 1. 参数还包括程序本身的名称,因此argc始终至少为1。

Here's the typical loop: 这是典型的循环:

int main(int argc, char * argv[])
{
    for (int i = 0; i != argc; ++i)
    {
        std::cout << "Argument #" << i << ": " << argv[i] << "\n";
    }
}

Alternatively you can print backwards: 或者,您可以向后打印:

while (argc--)
{
    std::cout << argv[argc] << "\n";
}

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