How to get command-line arguments in C
Command line arguments are referred to parameters that are passed to a program when it is to be executed via the command line/terminal.
We can pass command line arguments to a C program by adding individual arguments separated by spaces after the command to execute the binary file.
Syntax
Below is the syntax that shows how to pass command line arguments to a C program.
./<programName> <argument1> <argument2> ...
The programName is the name of the binary executable file, while argument1 and argument2 refer to the two arguments we pass to the program.
Example
Let's say we have a C program named passArguments. We will use the command ./passArguments in the command line to run the program.
If we want to pass two arguments, arg1 and arg2, to the program, we will add them after the program's name, separated by spaces.
So our shell command would be:
./passArguments arg1 arg2
How to retrieve the arguments
We will now examine how we can access those arguments within the program. To retrieve the command line arguments, we have two parameters that we can pass to the main() function.
argc: This integer variable refers to the number of arguments passed to the program via the command line.argv: This refers to the argument vector, an array of strings containing the arguments passed to the program.
Syntax
Below, we can see how to initialize these parameters in the main() function.
int main(int argc, char* argv[]){// write code here}
All arguments will be stored in the double-pointer char array named argv, with the number of arguments inside argc.
Example
We will now look at the C program below, which prints all the arguments passed to the program via the command line.
#include <stdio.h>
#include <unistd.h>
int main(int argc, char* argv[]){
for(int i = 0; i < argc; i++){
printf("\nArgument %d = argv[%d] = %s\n", i , i, argv[i]);
}
return 0;
}Note: To pass custom arguments to the program, click the "Run" button and in the terminal, use the command
./args arg1 arg2 ...wherearg1andarg2are the command line arguments.
Code explanation
Lines 4–11: We declare the
mainfunction, passingargc, andargvto access the command line arguments.Lines 6–8: We use a
forloop that starts from 0 and goes on till the number of arguments given to the program accessed via theargcvariable.Line 7: We print each command line argument by accessing each element from the
argvarray.
Exercise
Try to see if you can solve the problem statement below.
Problem statement
Suppose we have coded the C program that is shown below. We compiled it and named its binary file display_args.
#include <stdio.h>
#include <unistd.h>
int main(int argc, char* argv[]) {
printf("%s", argv[0]);
return 0;
}
What will the code output be when we run the command ./display_args arg1 arg2?
Nothing
./display_args
arg1
arg2
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