默认INT主要论据在C / C ++ [英] Default int main arguments in C/C++

问题描述

我是用C / C ++项目乱搞，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;
}

所以我总是那种想知道这一点，究竟是什么做的那些默认参数用C做/ C ++下INT主？我知道，应用程序仍然会编译没有他们，但他们所服务的目的是什么？

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

他们认为传递到命令行程序的参数。举例来说，如果我有计划的a.out 和我正是如此调用它：

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 

的argv 的内容将是包含字符串数组

The contents of argv will be an array of strings containing

1. [0] 的a.out - 可执行文件的文件名始终是第一要素


2. [1] ARG1 - 其他参数


3. [2] ARG2 - 我通过

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 持有的argv 元素的数量（在C你需要另一个变量知道有多少元素有在一个数组，当传递给函数）。

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).

您可以用这个简单的程序自己尝试一下：

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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