Thread: Scaning chars while initializing program?

Hello. I was wondering if anyone knew how to scan in a string if you place it right after how to initiate the program.
i.e. a.out ContentsOfString

and also if anyone could help out... i need to scan the line below it into a string also.
so...

a.out ContentsOfString
String2contets

any advice could help.

Thanks, Curtis

Is your requirement parsing the command line arguments?

i need to be able to save both the strings to use them in a cypher. But i also must run the program as
$ a.out key <---The key used for the cypher

The following program is reading the keys passed to the program and print them.For example,I simply printed the arguments.If you have any argument,you use those strings in your cypher.

Code:

#include	<stdio.h>
	int
main ( int argc, char *argv[] )
{
	int i=0;
	if(argc==1)
		printf ( "Please Enter the Command line arguments\n"  );
	else
	{
		for(i=1;i<argc;i++)
			printf ( "Arguments Passed Are:%s\n", argv[i] );
	}
	return 0;
}

Creedy, the C program can be passed, at start up, many parameters - that's just no problem at all.

They are loaded into a char array with multiple rows, with one row for each word.

I guess you wanna give a string with multiple lines in a command line argument and parsing it ?
If yes
give the string in a command line enclosed in ""
$ a.out string1 "string2
continuation of string 2
....... etc "

Thank you vivekraj, Adak, and Alexander jack. I understand it better now, it all just came together. I think i should be able to finish the program now. Thanks.

Dangit. One more issue. If i want the first letter of the key then i would printf argv[ i ][ 0 ] ?

when i try to print it i get an error:
format ‘%s’ expects type ‘char *’, but argument 2 has type ‘int’

While printing the first character alone use %c instead of %s

Code:

                        printf ( "First char :%c\n", argv[i][0] );

Okay cool. Thanks everyone.