从终端读取 input.txt 文件和 output.bmp 文件(C 编程) [英] Read input.txt file and also output.bmp file from terminal (C-programming)

问题描述

我必须做一个作业，我必须编写一个 C 程序，它从控制台获取输入文件名作为命令行参数.
它应该将数据从 input.txt 文件(输入文件具有 bmp 文件的信息 - 颜色等)移动到生成的 output.png 文件.20 20 参数代表 output.png 图像的宽度和高度.

I have to do an assignment where I have to write a C-Programm, where it gets the input-file-name from the console as command line parameter.
It should move the data from the input.txt file (the input file has the information for the bmp file - color etc.) to the generated output.png file. The 20 20 parameters stand for width and height for the output.png image.

例如控制台请求(在 Linux 上测试)将如下所示:

So the console-request for example (tested on Linux) will look like this:

./main input.txt output.bmp 20 20

我知道这段代码会读取一个 input.txt 文件并将其显示在屏幕上.

I know that this code reads an input.txt File and puts it on the screen.

FILE *input;
int ch;
input = fopen("input.txt","r");
ch = fgetc(input);
while(!feof(input)) {
    putchar(ch);
    ch = fgetc(input);
}
fclose(input);

这会(例如)将其写入 output.png 文件.

And this would (for example) write it to the output.png file.

FILE *output;
int i;
     output = fopen("ass2_everyinformationin.bmp", "wb+"); 
 for( i = 0; i < 55; i++)               
 {
     fputc(rectangle_bmp[i], output);
 }
 fclose(output);

但此代码仅适用，如果我直接在代码中硬编码名称，而不是使用命令行参数.
我没有任何线索，如何实施，我也没有在互联网上找到任何有用的信息，也许有人可以帮助我.

But this code works only, if I hard-code the name directly in the code, not by using a command line parameters.
I don't have any clue, how to implement that and I also didn't find any helpful information in the internet, maybe someone can help me.

问候

推荐答案

标准 main() 的完整原型是

The full prototype for a standard main() is

int main(int argc, char* argv[]);

你得到一个带有参数数量的整数，argc 和
字符串"列表(只要它们在 C 中存在)，argv.

You get an int with the number of arguments, argc and
a list of "strings" (as far as they exist in C), argv.

例如，您可以使用

#include "stdio.h"
int main(int argc, char* argv[])
{

    printf("Number: %d
", argc);
    printf("0: %s
", argv[0]);
    if (1<argc)
    {
        printf("1: %s
", argv[1]);
    }
}

开始玩弄参数.

请注意，这只是一个使用命令行参数的基本示例，故意不实现任何内容.这符合公认的 StackOverflow 政策，即提供作业帮助，而无需解决这些问题.

Note that this is intentionally not implementing anything but a basic example of using command line parameters. This matches an accpeted StackOverflow policy of providing help with assignments, without going anywhere near solving them.

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