confusion while understanding parameters of main

Question

while understanding the parameters of main function i.e, int argc, char* argv[]

i wrote a piece of code to understand these parameters.

#include <stdio.h>
int main(int argc,char*argv[])
{
    printf("test\n");
    printf("%d %c",argc,*argv[argc-1]);
    return 0;
}

This prints

test

1 F

here I don't understand why there is F as output. I mean how this is executed to result in output as F ?

I read about these perameters and main function at here and here. But still I don't understand how these works.

please explain .

EDIT: as mentioned in comments if I change the code to

printf("%d %s",argc,argv[argc-1]);

Now i'm getting the whole path of the file F:\file path

so does it mean argv[0] is the location of the file in drive?

What parameters did you supply when running your program? And did you happen to name your program something starting with F ? — M.M, Jun 11 '16 at 03:43
@M.M I am running code::blocks and the program name is temp\main.c — mssirvi, Jun 11 '16 at 03:53
argc is not 0, please read my que again. output of argc is 1. — mssirvi, Jun 11 '16 at 03:57
@KenWhite Commonly, `argv[0]` is the name of the executable (often with full path), so it's probably not garbage. Perhaps the program is on `F:` drive. — M.M, Jun 11 '16 at 03:57
Perhaps http://stackoverflow.com/questions/3024197/what-does-int-argc-char-argv-mean will help — Ed Heal, Jun 11 '16 at 03:58
@M.M yeah the program is on F: drive . so it means argv[0] is always the drive name ? — mssirvi, Jun 11 '16 at 04:00
@mssirvi to see what `argv[0]` is, make the code change suggested by Ed Heal — M.M, Jun 11 '16 at 04:00
after changing the code suggested by Ed Heal, this prints the path of the file. — mssirvi, Jun 11 '16 at 04:03
It is the path of the executable. See http://stackoverflow.com/questions/2050961/is-argv0-name-of-executable-an-accepted-standard-or-just-a-common-conventi — Ed Heal, Jun 11 '16 at 04:14

score 1 · Answer 1 · answered Jun 11 '16 at 17:21

1

It is not defined in the C standard, but on Unix argv[0] is the name of the executable. Then argv[1] the first argument, etc. I think that this is also true, most of the time, on Microsoft's Dos and their Windowing OSes.

answered Jun 11 '16 at 17:21

ctrl-alt-delor

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score 0 · Answer 2 · answered Jun 11 '16 at 17:10

In simple words: When you give your commandline the instruction to run your program, you can append some text, which can be accessed in your program.

#include <stdio.h>
int main(int argc,char*argv[])
{
    printf("This is the path or name of your programm: ");
    printf("%s\n", argv[0]);

    if(argc > 1) {
        printf("This is the first argument you gave your programm: ");
        printf("%s\n", argv[1]);
    }

    if(argc > 2) {
        printf("This is the second argument you gave your programm: ");
        printf("%s\n", argv[2]);
    }
    return 0;
}

Try to run this example with:

<path_to_the_programm> Hallo Welt

You will see argc is an integer, which tells you how many arguments you gave the program. I hope it helped you.

confusion while understanding parameters of main

2 Answers2