Question

I just wrote a program in c language which uses command line arguments and i tried to print the first argument. when i execute program with following command

./a.out $23

and try to print the first argument using the below code

printf("%s", argv[1]);

the output is just

3

Am i missing something here, that command line arguments are treated differently if some special characters are present. can some one explain this behavior.

Answer 1

You need to escape the $ character.

Try this:

./a.out \$23