C Programming - Command Line Arguments - Discussion

@Vidhya Balan

Assume your current working directory in DOS is "C:\TURBOC" and your C program name is SAMPLE.C (which is in c:\turboc directory).

If you compile the program successfully, an EXE file SAMPLE.EXE will be created in your current directory i.e "c:\turboc".

You can run it by simply typing its name as given below.

C:\TURBOC>sample.exe Good Morning <press-enter-key>

(or)

C:\TURBOC>SAMPLE Good Morning <press-enter-key>

Here "sample" is the 'exe-file-name-or-command' and "Good Morning" is the arguments supplied to the exe-file-or-command called "sample" which is created by compiling the C program "sample.c".

Hope this will help you. Have a nice day!

argc which count number of word or characters separated by space and it holds int value. In this example three words are their including file name.Hence argc print 3.

argv[] this is argument vector array which holes all arguments in string format and store from starting index with 0.
argv[0]=sample
argv[1]=Good.....

Sample Good Morning are three words, so argc will total words like sample = 1 good = 2 and morning = 3 that is total 3 counts and argv contains argv[0] = sample, argv[1] = good, and argv[2] = morning.

So output will be at printf ("%d %s", argc, argv[1]); 3 count at argc, and good at argv[1] that is 3 Good.

The given answer is correct. I have tested it.

argc - will contain 3 (count of arguments argv[])

argv[0] = C:\TURBOC\SAMPLE.EXE
argv[1] = Good
argv[2] = Morning

Therefore, answer is "3 Good"

argc=argument count= sample,good,morning=3
argv[0]=by default it stores base address= it stores file name=sample
argv[1]=good
argv[2]=morning

%d argc=3
%s argv[1]=good

But sample is file name right arguments are given after file name is specified then good =0,morning =1 .. I don't understand if sample is argument here.

The 'argc' contains count value and argv[1] contains "Good".

On printing argc,argv,its o/p will be 3 Good.

Because count value is 3 here.

argv[0]=Base address=Sample
argv[1]=Good
argv[2]=Morning


%d argc=3 because (argv[0],argv[1],argv[2] only three arguments argv[])

%s argv[1]=Good

But I have a doubt guys. Usually the count value will start with 0, but how here started with 1. Is that the rule for argc[]. Please explain.

In linux,
argc = 3 because no of arguments that are passed is 3 in the program.

arv[0] = ./a.out,
argv[1] = Good,
argv[2] = Morning.