How exactly does argc work when I don't use quotation marks?

Question

My code is very simple :

#include <stdio.h>

int main(int argc, char **argv)
{
    printf("%d\n", argc);
    return 0;
}

It prints 17 after I input

./a.out 1 2 3 + *

Shouldn't it print 6 instead? The program's behavior is as expected if I put '' around every argument except ./a.out. This is strange, because I can use some programs with flags without using quotation marks('', "").

Linux deals with the `*` argument in its own way. It does not reach the program. I suggest you print out all the 17 arguments given. — Weather Vane, May 19 '20 at 20:10
`+` isn't special to the shell. The only specials that need to be quoted at listed in 2.2 here: https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html — R.. GitHub STOP HELPING ICE, May 19 '20 at 20:15
Replace `./a.out` with `echo` and you'll see what the arguments are. — Barmar, May 19 '20 at 20:17

score 4 · Accepted Answer · answered May 19 '20 at 20:20

4

If you don't quote or escape *, it's expanded as a wildcard by the shell when forming the program arguments. It's replaced with all the names in the current directory (except the ones that begin with .).

You must have had 12 files and subdirectories in your directory when you ran this.

answered May 19 '20 at 20:20

Barmar

741,623
53
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612

Might have had fewer than 12 if some had spaces in their names – stark May 19 '20 at 20:31
2

@stark No, the expansion of a wildcard does not undergo word splitting. – Barmar May 19 '20 at 20:32

How exactly does argc work when I don't use quotation marks?

1 Answers1