Char *ptr="solo learn" ;printf(ptr+5); what is its output?

Question

What my doubt is in printf, there is no format specifier. Will give error for not mentioning? Please explain.

Answer 1

There are problems with that "code"

1. Char is unknown type, it could be auto-caps feature, but still a problem.

2. printf() call is missing format specifier (you said so yourself).

3. ptr *+ 5 isn't a valid expression (edited).

(Edit)
The question had been edited

Answer 2

Ipang Sry the expression is ptr+5

Answer 3

Gajjala Venkata Sai Sri Harsha Hanumantha Rao
Oh I see ...
Well then you may see 'learn' as output, of course you'd need "%s" specifier.

(Edit)
The C-string itself (ptr) is the format specifier.

Answer 4

Ipang so the answer to question will be an error as there is no format specifier?

Answer 5

Gajjala Venkata Sai Sri Harsha Hanumantha Rao
Error primarily because no format specifier, the ptr *+ 5 (if that's how it was written in challenge). And the 'Char' typo also ...

Answer 6

Guys, in that case then, I'd assume the <ptr> is used as format specifier by printf(), it is possible. But I'd set <ptr> const char* instead to match the printf() prototype ...

Answer 7

1 It points to string
2 Index starts at 0
3 6 char is l (0..5) ,,🙏

Answer 8

Here , In the character pointer "ptr" , Base address of the given string is assigned. So pointer ptr stores the base address of "s" also. Now as given in the Printing statement 'printf(ptr+5)' , 5 is added to ptr , so pointer ptr will point 5 characters ahead now and the Character will be "l".
Hence , The output of the given code will be "l".

Answer 9

Aman kumar when i run the code, output is "learn" not "l"

Answer 10

codemonkey the doubt is there is no format specifier, will it give an error?

Answer 11

codemonkey OK thank you .i got the question in challenge and answer to it is learn. I got a doubt as there is no format specifier, it would give an error and asked a doubt.

Answer 12

Ipang in challenge written as ptr+5 only

Answer 13

codemonkey format specifier is not compulsory for strings in c

Answer 14

codemonkey do you know any reason how compiler interpreted internally

Answer 15

codemonkey i dont know it is not compulsory

Answer 16

codemonkey I Am asking as question only

Answer 17

codemonkey oh k thanks

Answer 18

codemonkey int *ptr=5, printf(ptr) will give compilation error? as pointer points to integer and not char data type.

Answer 19

codemonkey thank you very much

Answer 20

You've got type errors in this code. You have to do type casting. Also, you can't define var with uppercase.