C Programming - Command Line Arguments - Discussion

Discussion Forum : Command Line Arguments - Find Output of Program (Q.No. 1)
What will be the output of the program (myprog.c) given below if it is executed from the command line?
cmd> myprog one two three
/* myprog.c */

int main(int argc, char **argv)
    printf("%c\n", **++argv);
    return 0;
myprog one two three
myprog one
Answer: Option
No answer description is available. Let's discuss.
26 comments Page 1 of 3.

Sravya said:   4 years ago
Good explanation @Cherry.

Sanjana said:   5 years ago
But argv[0] must be a program name right?

Nikhil n said:   5 years ago
Perfect answer @Seenu.

GORIPARTHI. Udaya lakshmi said:   5 years ago
How the answer comes o?

Priyanka said:   6 years ago
Consider a pointer *p for storing single dimension array. *++p points to next element in the array.

In the case of 2-dimensional array **p is used and **++p points to next word.

Finally coming to answer **argv[] is a 2-dimensional representation of command line arguments and,

**++argv[] points to next word. %c is used for printing character at that position. So, the answer is "o".

Swapnil said:   7 years ago
Thanks @Cherry.

Akhilesh said:   9 years ago

So, ++argv[1]=*argv.

i.e, *(++(*argv)) = here value is incremented so it will Print next character provided %c should be there in printf.

Sunaina said:   9 years ago
Another question is :
What will be the output of the program (myprog.c) given below if it is executed from the command line?

cmd> myprog friday tuesday sunday

/* myprog.c */

int main(int argc, char *argv[])
printf("%c", *++argv[1]);
return 0;

For this answer is "r".

Can anyone explain difference in these ?

As the confusion is that in this question after ++ we move to next string and in another question as I mentioned above we move to one character.... why?

Shalini said:   9 years ago
**argv is pointer to array of pointers.
argv having base address of "myprog".
++argv having base address of "one".
*++argv(dereference) prints "one".
**++argv(two times dereference) prints 'o'.

Vineet said:   10 years ago
argv have address of argv[0];
and so on...

++argv means address of argv[1] and now value at argv[1] is address of o of(one). then again value at that address is o.

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