Command Line Arguments in C – Don’t be Confused be Practical!

DataFlair Team

3 years ago

So far, we have seen that no arguments were passed in the main function. But the C programming language gives the programmer the provision to add parameters or arguments inside the main function to reduce the length of the code. These arguments are called command line arguments in C.

In this tutorial, we will discuss:

Command Line arguments
Components of command line arguments
C program to understand command line arguments

What are Command Line Arguments in C?

Command line arguments are nothing but simply arguments that are specified after the name of the program in the system’s command line, and these argument values are passed on to your program during program execution.

Components of Line Arguments

In order to implement command line arguments, generally, 2 parameters are passed into the main function:

Number of command line arguments
The list of command line arguments

The basic syntax is:

int main( int argc, char *argv[] )
{
.
.
// BODY OF THE MAIN FUNCTION
.
.
}

Another way to implement command line arguments is:

int main( int argc, char **argv[] )
{
.
.
// BODY OF THE MAIN FUNCTION
.
.
}

Now, let us discuss the components of this code in detail:

argc: It refers to “argument count”. It is the first parameter that we use to store the number of command line arguments. It is important to note that the value of argc should be greater than or equal to 0.
agrv: It refers to “argument vector”. It is basically an array of character pointer which we use to list all the command line arguments.

C Program to Understand the Command Line Arguments in C

Here is a code in C that illustrates the use of command line arguments.

// The program name is cl.c
#include<stdio.h>
int main(int argc, char** argv)
{

printf("Welcome to DataFlair tutorials!\n\n");

int i;
printf("The number of arguments are: %d\n",argc);
printf("The arguments are:");

for ( i = 0; i < argc; i++)
{
printf("%s\n", argv[i]);
}
return 0;
}

If you’re still having any doubts related to the above syntax then don’t forget to check this

Quiz on Command Line Arguments in C

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15

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Answered

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Question 1 of 15

1. Question

#include <iostream>

using namespace std;

int main(int argc)

{

cout <<argc<< “\n”;

return 0;

}

warning with output 1

Compilation Error

Program terminates without any output

Correct

Incorrect

Question 2 of 15

2. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout <<argc<< “\n”;

return 0;

}

warning with output 1

Compilation Error

Program terminates without any output

1

Correct

Incorrect

Question 3 of 15

3. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout <<argv<< “\n”;

return 0;

}

Compilation Error

0x7ff65

1

Correct

Incorrect

Question 4 of 15

4. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout <<argv[0]<< “\n”;

return 0;

}

.a/.out

0x7ff65

1

Correct

Incorrect

Question 5 of 15

5. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout <<“argument:”<<argv[1]<< “\n”;

return 0;

}

argument: 1

argument: 2

argument: .a/.out

argument:

Correct

Incorrect

Question 6 of 15

6. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout <<“argument:”<<argv[argc]<< “\n”;

return 0;

}

argument:1

argument:0

argument: .a/.out

argument:

Correct

Incorrect

Question 7 of 15

7. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout << argc<< ” “;

for (int i = 0; i < argc; ++i)

cout << argv[i];

return 0;

}

1

.a/.out

1 .a/.out

Correct

Incorrect

Question 8 of 15

8. Question

#include <iostream>

using namespace std;

int main(int argc, char** argv)

{

cout <<argv[2]<< “\n”;

return 0;

}

1

.a/.out

HOSTNAME =Check

Correct

Incorrect

Question 9 of 15

9. Question

#include <iostream>

using namespace std;

int main()

{

cout <<argc<< “\n”;

return 0;

}

1

Compilation Error

.a/.out

Correct

Incorrect

Question 10 of 15

10. Question

#include <iostream>

using namespace std;

int main(char** argv)

{

cout <<argv[0]<< “\n”;

return 0;

}

Program stops with warning

Compilation Error

.a/.out

Correct

Incorrect

Question 11 of 15

11. Question

what is the first argument in command line arguments?

list of CLA

number of CLA

argument vector

none of the above

Correct

Incorrect

Question 12 of 15

12. Question

what is an argument vector?

array of pointer to CLA

Number of CLA passed

Name of program

none of the above

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Question 13 of 15

13. Question

what is an argument count?

array of pointer to CLA

Number of CLA passed

Name of program

none of the above

Correct

Incorrect

Question 14 of 15

14. Question

what does argv[0] holds?

Null pointer

name of program

First CLA

none of the above

Correct

Incorrect

Question 15 of 15

15. Question

what does argv[1] holds?

Null pointer

name of program

First CLA

none of the above

Correct

Incorrect

Summary

In this tutorial, we understood the basic meaning behind command line arguments and its 2 components.

We inferred that one is used to find the number of command line arguments and the other is used to find the list of command line arguments such as the program name.

Feel free to leave a comment below so that we could better understand your responses.