Standard Input for C program using Unix Command Line

Question

I have the following section of code in C:

int main(int argc, char *argv[]) 
{
    char *input;
    int lines;
    int p;


    input = (char*)malloc(10);
    input = argv[0];
    for(p=0;p<10;p++)
    {
        printf("%c",input[p]);
    }

On my Unix system I make the following call:

./program_name.exe < inputfile

where inputfile is a file that contains the following: 000000010Z

The output I receive to the previous commands is:

./program_

What am I missing?

If the goal is to read stdin, you may want to read about getc(), fgets() or scanf(). — Charlie Burns
– Charlie Burns, Commented Sep 19, 2013 at 23:19
the malloc is useless, you aren't using that allocated buffer, the < inputfile isn't a parameter to your program, its redirecting the stdin into your program — Keith Nicholas
– Keith Nicholas, Commented Sep 19, 2013 at 23:20
argv[0] is the program name. argv[1] is the first command line parameter. If you redirect a file into stdin you either need to open stdin as a file inside your program and read from it (fopen, fgets, fclose) or read ketstrokes (getchar, gets) which will return the same thing. — Jerry Jeremiah
– Jerry Jeremiah, Commented Sep 19, 2013 at 23:20
It looks like you're missing "name.exe". Try looping to 18 instead of 10! — William Pursell
– William Pursell, Commented Sep 19, 2013 at 23:27

zubergu · Accepted Answer · 2013-09-19 23:33:48Z

1

argv[0] holds name of program being executed. And that's the only thing you print.

What you're trying to do is to read from your file as it was stdin. But you still have to read it. Use getchar(), or any other function that reads input.

edited Sep 19, 2013 at 23:33

answered Sep 19, 2013 at 23:20

zubergu

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1 Comment

PumpkinPie Over a year ago

Yep, thats it. Thank you for the help!

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