why is argc returning 6 as opposed to 3?

Question

I have following program and i am passing it 2 arguments at the command line as shown below. I was expecting the argc to be 3, but it prints it as 6. why?

#include <iostream>
#include <string>
#include <stdlib.h>

using namespace std;
void usage();

int main(int argc, char *argv[])
{
    cout << argc << endl;
    if (argc != 3)
            usage();

    string regex = argv[1];
    string searchString = argv[2];

    cout << "regex: " << regex << " . searchString: " << searchString << endl;

    return 0;
}

void usage()
{
    cout << "Usage: ./stringmatch <regex> <searchString>" << endl;
    exit(1);
}

Command line:

[jim@cola c++]$ ./stringmatch [yyyy\d\d\d]* yyyy1234
 6
 Usage: ./stringmatch <regex> <searchString>

score 7 · Accepted Answer · answered Dec 01 '12 at 22:11

7

Your shell is expanding the glob pattern [yyyy\d\d\d]* so the actual number of arguments this results in depends on the contents of the current directory!

The [yyyy\d\d\d] becomes a character class matching the characters y and d, and the * matches anything that follows, so I'm guessing your current directory has 4 files that start with y or d. To see what it expands to, use echo:

$ echo [yyyy\d\d\d]*

To make it work as intended, quote the argument:

$ ./stringmatch '[yyyy\d\d\d]*' yyyy1234

answered Dec 01 '12 at 22:11

Thomas

174,939
50
355
478

1

How is shell treating [yyyy\d\d\d]* to? can you please explain more? – Jimm Dec 01 '12 at 22:13
@Jimm try printing out argv and see what you get. – John Carter Dec 01 '12 at 22:14
2

@Jimm It's called 'shell globbing'. – bames53 Dec 01 '12 at 22:14
but \d is digit, so why is shell treating it as character 'd'? – Jimm Dec 01 '12 at 22:17
1

A glob is not a regexp, and \ merely does escaping, nothing magical. Since there's no point escaping the letter `d`, `\d` is simply interpreted as `d`. – Thomas Dec 01 '12 at 22:19

score 2 · Answer 2 · answered Dec 01 '12 at 22:11

2

[yyyy\d\d\d]* is expanded by the shell into multiple file names that match the pattern. Quote it to get 3.

answered Dec 01 '12 at 22:11

bmargulies

97,814
39
186
310

Olaf Dietsche · Answer 3 · 2012-12-01T22:26:50.550

2

The shell resolves your command line to

./stringmatch [yd]* yyyy1234

which gives all files starting with either y or d plus yyyy1234. So if you have 4 files starting with either y and d plus yyyy1234 plus ./stringmatch will give exactly 6.

See Filename Expansion and Pattern Matching for more information.

If you want just two arguments you must quote the first one with single or double quotes

./stringmatch '[yyyy\d\d\d]*' yyyy1234

edited Dec 01 '12 at 22:26

answered Dec 01 '12 at 22:13

Olaf Dietsche

72,253
8
102
198

this behavior of shell, is it because of presence of []? – Jimm Dec 01 '12 at 22:15
The reason for this is because of character set `[]` and wildcard `*`, see updated answer. – Olaf Dietsche Dec 01 '12 at 22:24

score 1 · Answer 4 · answered Dec 01 '12 at 22:12

1

Because your shell is expanding the arguments - wrap them in '' to prevent this.

answered Dec 01 '12 at 22:12

John Carter

53,924
26
111
144

why is argc returning 6 as opposed to 3?

4 Answers4