[SOLVED] c program and bash arguments

marbangens · 10-14-2012, 06:55 AM

I in doing the gnu c tutorial and stuck on the cow program

#include <stdio.h>

int
main()
{
int cows = 6;

if (cows > 1)
printf("We have cows\n");

if (cows > 10)
printf("loads of them!\n");

return 0;
}

This is no problem as it is here. But I want to make it take arguments
about how many cows I have. The code look like this now.

#include <stdio.h>

int main(int argc, char **argv)
{
int cows;

if (argc != 2)
{
printf("error - you must give an int argument\n");
return 1;
}

if (argc >= 3)
{
printf("error - only one argument is taken");
return 1;
}

cows = *argv;

if (cows > 1 || cows < 9)
printf("We have cows\n");

if (cows > 10)
printf("loads of them!\n");

return 0;
}

The problem is that I always get different output no mater what argument
I give. Like

./a.out 0
you have cows

./a.out 3
you have cows
loads of them!

and sometimes no output...

Snark1994 · 10-14-2012, 07:06 AM

When you post code, use [CODE][/CODE] tags around it to make it more legible...

Your problem is that '*argv' is a 'char*' (i.e. argv is a pointer to a pointer to a char, as 'char**' suggests). So you need to convert the numbers written as text in '*argv' into an int. To do this, just use the 'atoi' function:

Code:

cows = atoi(argv[1]);

Also, your second test on argc will never be triggered - you test 'argc != 2' and then 'argc >= 3'. If you wanted to detect the two conditions separately, you should use 'argc < 2' and then 'argc >= 3'.

Hope this helps,

H_TeXMeX_H · 10-14-2012, 07:08 AM

Make sure to put the code in code tags.

Code:

#include <stdio.h>

int main(int argc, char **argv)
{
int cows;

if (argc != 2)
{
printf("error - you must give an int argument\n");
return 1;
}

// this statement is redundant
if (argc >= 3)
{
printf("error - only one argument is taken");
return 1;
}

cows = *argv;

if (cows > 1 || cows < 9)
printf("We have cows\n");

if (cows > 10)
printf("loads of them!\n");

return 0;
}

You can't do 'cows = *argv;' it doesn't make sense, I'm sure the compiler complained with 'assignment makes integer from pointer without a cast'. Instead use sscanf to scan argv for an integer like:

Code:

#include <stdio.h>

int main(int argc, char * argv[])
{
	int cows;

	if (2 != argc)
	{
		fprintf(stderr, "ERROR: Need exactly one argument\n");
		return 1;
	}

	// get cows
	sscanf(argv[1], "%d", &cows);

	if (cows > 1)
	{
		printf("We have cows\n");
	}

	if (cows > 10)
	{
		printf("loads of them!\n");
	}

	return 0;
}

EDIT: or you can use atoi like Snark1994 suggests.

marbangens · 10-15-2012, 02:42 AM

Thanks for the help.
and I will use "["CODE]"["/CODE] next time.

Snark1994 · 10-16-2012, 03:43 AM

Did that solve your problem? If so, can you click the "Mark this thread as SOLVED" link at the top of the page, please? That way people know the thread contains a solution.

Thanks,

marbangens · 10-24-2012, 12:24 PM

sorry snark hop you can forgive me for the delay.
I have been busy.

Snark1994 · 10-25-2012, 03:42 AM

Aha, no problem at all - it's just a bit of etiquette really.

Have a nice day,