$ followed by a char in command line argument in C

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 have to "inhibit" you argument like this:

./a.out \$23

Some characters are interpreted by the shell. These characters include the following:

• \ Which inhibits (escapes) the character just behind it (usefull for space, tabs or in your case)
• * Which represents any single character or character strings
• $ Which represents a variable (in your case, the shell understands the variable $23, not the string "$23")
• || or | Which allows a resolution in your command or to pipe your command
• && or & Which allows combination of commands or which allows the use of job control
• " Which allows the shell to delimit a character string
• ' Which allows the shell to not interpret a character string with special characters
• ; Which delimits commands
• ` Which interprets the command enclosed by two of these and returns the command's output

Answer 2

The shell treats $23 as the positional parameter $2 followed by the literal character 3. To pass the string "$23", do either

./a.out \$23

or

./a.out '$23'

To pass the shell's 23rd positional parameter (unlikely, but possible), you would write

./a.out ${23}

Answer 3

Presumably the $2 is being treated as a shell variable. Try escaping the dollar sign:

./a.out \$23

Answer 4

You need to escape the $ character.

Try this:

./a.out \$23