How to write a BASH command in C which use stdin and stdout

Question

I want to write a programme where I compile a C programme, and save it where BASH recognises programmes are (/usr/bin or somewhere).

For usage I want to run in the terminal

$ c_programme <file_input> | <programme_to_pipe_stdout>

I can see that with the popen() function I can use ffmpeg to process video:

FILE *pipein = popen("ffmpeg -i teapot.mp4 -f image2pipe -vcodec rawvideo -pix_fmt rgb24 -", "r");
FILE *pipeout = popen("ffmpeg -y -f rawvideo -vcodec rawvideo -pix_fmt rgb24 -s 1280x720 -r 25 -i - -f mp4 -q:v 50 -an -vcodec mpeg4 output.mp4", "w");

But in the above teapot.mp4 was already specified in the code, not input into the C programme when the programme was run in BASH. How can I make a variable in the C code which will read from the shell?

Similarly, I don't want to output to a file, I just want to output to stdout, so I can pipe it. How would I do this?

Thanks!

Answer 1

How can I make a variable in the C code which will read from the shell?

A C program can get the arguments passed to it on the command line by declaring main as int main(int argc, char *argv[]), after which argv[0] refers to the name of the program, argv[1] is the first argument, argv[2] is the second, and so on up to argv[argc-1]. The end is also indicate by argv[argc] being a null pointer.

For example, the output of this program:

#include <stdio.h>


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

when executed as foo Argument1 Argument2 is:

argv[0] = foo.
argv[1] = Argument1.
argv[2] = Argument2.

Similarly, I don't want to output to a file, I just want to output to stdout, so I can pipe it. How would I do this?

Use printf, puts, or putchar.