C / C ++中的默认int main参数 [英] Default int main arguments in C/C++

问题描述

我在使用C / C ++进行项目，我注意到了这一点：

I was messing around with projects in C/C++ and I noticed this:

C ++

#include <iostream.h>

int main (int argc, const char * argv[]) {
    // insert code here...
    cout << "Hello, World!\n";
    return 0;
}

和

C

#include <stdio.h>

int main (int argc, const char * argv[]) {
    // insert code here...
    printf("Hello, World!\n");
    return 0;
}

所以我一直想知道这是什么，参数在C / C ++在int main？我知道应用程序仍然会编译没有他们，但他们服务的目的是什么？

So I've always sort of wondered about this, what exactly do those default arguments do in C/C++ under int main? I know that the application will still compile without them, but what purpose do they serve?

推荐答案

它们保存在命令行上传递给程序的参数。例如，如果我有程序 a.out ，然后调用它：

They hold the arguments passed to the program on the command line. For example, if I have program a.out and I invoke it thusly:

$ ./a.out arg1 arg2 

argv 将是一个包含

1. [0] - 可执行文件的文件名始终是第一个元素


2. [1] arg1参数


3. [2] arg2 - 我已通过

1. [0] "a.out" - The executable's file name is always the first element
2. [1] "arg1" - The other arguments
3. [2] "arg2" - that I passed

argc 包含 argv 中的元素数量（如C中所示，

argc holds the number of elements in argv (as in C you need another variable to know how many elements there are in an array, when passed to a function).

您可以使用这个简单的程序自己尝试：

You can try it yourself with this simple program:

#include <stdio.h>

int main(int argc, char ** argv){
    int i;
    for(i = 0; i < argc; i++){
        printf("Argument %i = %s\n", i, argv[i]);
    }
    return 0;
}

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