GATE CSE 2015 | Set 3 | Question: 7

Question

Consider the following C program segment.

# include <stdio.h>
int main()
{
    char s1[7] = "1234", *p;
    p = s1 + 2;
    *p = '0';
    printf("%s", s1);
}

What will be printed by the program?

• A. $12$
• B. $120400$
• C. $1204$
• D. $1034$

Comments

Excellent Answer 🙌🏽

 

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This is why , I love gate overflow , people don't just answer for the questions asked , they think beyond original answers & they express their own perspectives , so that we can gain more knowledge .

 Thank you so much for this great community.
Don't miss the answers of @tusharp and @Kshitij Sharma's answers in comments and @Ahwan's answer under the best answer.

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this has to be the best explanation

Answer 1

p = s1 + 2;

Type of s1 is char[7] and sizeof *s1 is sizeof (char) = 1. So, s1 + 2 will return address in s1 + 2 *sizeof(char) = address in s1 + 2. So,  $p$ now points to the third element in s1.

*p = '0';

The third element in s1 is made $0$. So, $1234$ becomes $1204$. C choice. 

Comments

https://gateoverflow.in/16618/doubt-regarding-string-constant

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Great Explanation. Helped a lot in cracking the confusion.

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I still didn't get it when we declared the array of characters as s1[7], we wasted the memory because we used only 4 characters and a null at last.

so they did it purposefully in the question?

Answer 2

The answer is C.
Here
*p = '0';    So answer is 1204
If *p = 0;   Here answer will be 12

0 means ASCII 0, which is the Null character.

'0'  means ASCII 48, which is character '0'
 

Comments

*p = '\0' will also print  12

this assumption  is wrong because there is nothing like character '\0' in and may lead to error as we assigning 2 char in one location.

Am I right  @Arjun Sir?

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No. \0 is a valid character as \ is an escape character and it's ASCII code is 0.

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Anything which can be stored in 1Byte is called as char. Computer understands this only. Although we are able to see that 0 is integer. But computer checks whether the thing can be stored in 1 byte or not..

Answer 3

Here s1 is an array, So s1 points to base address.

so , p=s1+2 will point to 3rd element of s1.

and *p='0'

value at p (Which is the third element of s1) , so 1234 becomes 1204.

Answer 4

s1[7]="1234"

p=s1+2

it is pointing to the address of "3" of "1234"

if *p=0 then the value of 3 replaced by 0

so the printf("%s", s1) prints 1204

Comments

if *p=0 then the output will be 12 only.

because 0 acts as Null here and printf will print until NULL character.

It should be *p=’0’.

Answer 5

1204 will be correct answer.

Answer 6

# include <stdio.h>
int main()
{
    char s1[7] = "1234", *p;
    p = s1 + 2;
    *p = 48;
    printf("%s", s1);
}

 

output=====1204

because ASCII value of 48 is 0

https://theasciicode.com.ar/ascii-printable-characters/number-one-ascii-code-49.html#:~:text=To%20get%20the%20letter%2C%20character,%221%22%20in%20ASCII%20table.