Output of the following program:
main()
{
char str[]="S\065AB";
printf("\n%d", sizeof(str));
}

Options
- 7
- 6
- 5
- error

CORRECT ANSWER : 5

Discussion Board
g

g

g 07-1-2017 03:47 AM

ccc

ccc

shubham bhuskade 07-1-2017 03:46 AM

Correct

\0 is escaped.hence size is 5!

Bipin Singh 09-18-2016 11:23 AM

I think the answer is 1.

AS \0 occurs,the string will end.so it takes the length before the string.so S is the only char and o/p is 1.

jakeera 07-24-2016 07:13 AM

Answer is wrong

It will output S5AB.

Sridhar 07-12-2015 02:41 PM

\065 is '5' :)

\065 is an octal code palced in our string, so the REAL string is "S5AB", the length is 4char + null terminator

Mihai 11-28-2014 10:30 AM

Explanation

5 is the correct answer. The two top comments are wrong. Barman should try running the code himself.

The null character (\0) at end of string IS counted. The embedded ascii control code '\06' (ACK) is discarded and therefore (the apparent 3 chars are) not counted. So the resulting output string is "S5AB", and the size of string is 5 (4 visible chars + null character).


gds 06-24-2014 03:17 AM

correct answer

/0 is not counted


ajinkya 04-24-2014 08:57 AM

answer is wrong

in C char[] or string is same thing and always NULL terminated
therefore char str[]="S\065AB"; has a size of 7+1=8
u can try running the code

Chinmoy Barman 01-22-2014 06:19 AM

ques.

can u please explain it?

sharmin 09-24-2013 04:35 AM

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