C Programming Tutorial
Description
It is possible to pass some values from the command line to your C programs when they are executed. These values are called command line arguments and many times they are important for your program especially when you want to control your program from outside instead of hard coding those values inside the code.
The command line arguments are handled using main() function arguments where argc refers to the number of arguments passed, and argv[] is a pointer array which points to each argument passed to the program. Following is a simple example which checks if there is any argument supplied from the command line and take action accordingly −
The arguments passed from command line are called command line arguments. These arguments are handled by main() function.
To support command line argument, you need to change the structure of main() function as given below.
int main(int argc, char *argv[] )
Here, argc counts the number of arguments. It counts the file name as the first argument.
The argv[] contains the total number of arguments. The first argument is the file name always.
Example
Let's see the example of command line arguments where we are passing one argument with file name.
#include <stdio.h>
void main(int argc, char *argv[] )
{
printf("Program name is: %s\n", argv[0]);
if(argc < 2){
printf("No argument passed through command line.\n");
}
else{
printf("First argument is: %s\n", argv[1]);
}
}
Run this program as follows in Linux:
./program hello
Run this program as follows in Windows from command line:
program.exe hello
Output:
Program name is: program
First argument is: hello
If you pass many arguments, it will print only one.
./program hello c how r u
Output:
Program name is: program
First argument is: hello
But if you pass many arguments within double quote, all arguments will be treated as a single argument only.
./program "hello c how r u"
Output:
Program name is: program
First argument is: hello c how r u
You can write your program to print all the arguments. In this program, we are printing only argv[1], that is why it is printing only one argument.
#include <stdio.h>
int main( int argc, char *argv[] ) {
if( argc == 2 ) {
printf("The argument supplied is %s\n", argv[1]);
}
else if( argc > 2 ) {
printf("Too many arguments supplied.\n");
}
else {
printf("One argument expected.\n");
}
}
When the above code is compiled and executed with single argument, it produces the following result.
$./a.out testing
The argument supplied is testing
When the above code is compiled and executed with a two arguments, it produces the following result.
$./a.out testing1 testing2
Too many arguments supplied.
When the above code is compiled and executed without passing any argument, it produces the following result.
$./a.out
One argument expected
It should be noted that argv[0] holds the name of the program itself and argv[1] is a pointer to the first command line argument supplied, and *argv[n] is the last argument. If no arguments are supplied, argc will be one, and if you pass one argument then argc is set at 2.
You pass all the command line arguments separated by a space, but if argument itself has a space then you can pass such arguments by putting them inside double quotes "" or single quotes ''. Let us re-write above example once again where we will print program name and we also pass a command line argument by putting inside double quotes −
#include <stdio.h>
int main( int argc, char *argv[] ) {
printf("Program name %s\n", argv[0]);
if( argc == 2 ) {
printf("The argument supplied is %s\n", argv[1]);
}
else if( argc > 2 ) {
printf("Too many arguments supplied.\n");
}
else {
printf("One argument expected.\n");
}
}
When the above code is compiled and executed with a single argument separated by space but inside double quotes, it produces the following result.
$./a.out "testing1 testing2"
Progranm name ./a.out
The argument supplied is testing1 testing2
