C Programming - Command Line Arguments - Discussion

Discussion Forum : Command Line Arguments - Find Output of Program (Q.No. 6)
6.
What will be the output of the program (sample.c) given below if it is executed from the command line?
cmd> sample "*.c"
/* sample.c */
#include<stdio.h>

int main(int argc, int *argv)
{
    int i;
    for(i=1; i<argc; i++)
        printf("%s\n", argv[i]);
    return 0;
}
*.c
"*.c"
sample *.c
List of all files and folders in the current directory
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
22 comments Page 2 of 3.

Prashant Kumar said:   8 years ago
I am getting the output as list of all .c file in the current directory? why it's so?

Kavithha said:   1 decade ago
Since argv[] is a string, it print ly the string pointed by the argv[1] i.e. *.c .

Shail said:   1 decade ago
Warning: format \'%s\' expects type \'char *\', but argument 2 has type \'int\'.

Mangusta said:   1 decade ago
Hey, But argument is char * argv, not char* *argv or char* argv[].

Anu said:   1 decade ago
Can someone explain about the output of the above program ?

AppiReddy said:   1 decade ago
Any one can explain the output of the above program please.

Anish said:   1 decade ago
argc=2
argv[0]=sample
argv[1]=*.c

So it prints *.c

Varun Gupta said:   1 decade ago
@Chandrasekar not a relevant explanation

Tanya said:   7 years ago
Please explain the correct answer.

Shamshad said:   1 decade ago
Why don't the output be "*.c" ?.


Post your comments here:

Your comments will be displayed after verification.