Question

I have a simple program which is supposed to print a string. But I am not getting the expected output. Can anyone tell me what is wrong with the program ?

Here is my code:

main()
{
   char arr[] = "Test_string";
   printf("%20s"+1,arr);
   return 0;
}

output: 20s

Expected output is:Test_string

"Test_string" getting printed in 20 places as we are giving "%20s" as format specifier.

Answer 1

It is very simple if you carefully look at your printf call.

Here is the prototype of printf : int printf(const char *format, ...);.

printf expects a pointer to format string as the first argument. In your program you are passing a pointer to this string : "20s" and printf promptly prints what you are passing.

Let me explain why the pointer passed is pointing to "20s" and not "%20s".

Quoted strings in C are interpreted as character pointers.
Character arrays which, when passed to a function, decay into a pointer.

printf("%20s",arr); is equivalent to :

const char * ptr = "%20s";
printf(ptr,arr);

similarly printf("%20s"+1,arr); is equivalent to :

const char * ptr = "%20s";
printf(ptr+1,arr);

Because you are passing "%20s"+1, the actual pointer which is passed to printf is pointing to a string "20s".