将参数传递给main [英] passing arguments to main

问题描述

我知道这是很基本的，但是我仍然坚持. 所以我有一个需要带一个变量n的函数，所以这是我的主要功能

I know this is fairly basic, but I'm still stuck. So I have a function that needs to take in a variable n, so this is my main function

int main(int argc, char* argv){
  sort(argv[1]);
    }

我正在这样调用程序:

    ./sort 4 <text.txt

但是数字4不会被识别或传递给函数.我究竟做错了什么?我知道argv [0]应该包含程序本身的名称，并且每个之后的名称都应该包含参数.

But the number 4 doesnt get recognized or passed into the function. What am I doing wrong? I know that argv[0] should hold the name of program itself and each one from there on should hold the arguments.

推荐答案

您应尝试全部打印它们.

You should try to print them all.

#include <stdio.h>

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

使用./a.out 4 < /somefile运行该代码会给我:

Running that code with ./a.out 4 < /somefile gives me:

argv[0] = './a.out'
argv[1] = '4'

最终，您必须记住"4"是指向字符数组的指针，并且可能必须将其解析为整数.

Eventually you'll have to remember that the '4' is a pointer to an array of characters, and you might have to parse it into an integer.

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