Programming closed

Question

Consider the following program segment

int main()
{
char *str = "GATECS";
printf("%d", madeeasy(str));
return 0;
}
int madeeasy(int *p1)
{
int *p2 = p1;
while(*++p1);
return p1-p2;
}

The out of above program will be ________. Assume that the object of data type int occupies 2 bytes.

Comments

Initially p1 points to GATECS

int *p2 = p1 means p2 is pointing also pointing to GATECS

while(*++p1); meaning at loop breaks  when p1 is pointing to S.

Now p1 is pointed to '\0'.

Each integer takes 2 bytes.So calculating number of jumps , ans of p1-p2  is 3 as p1 is at higher address.

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meaning of while (∗++p1)????

 increment value of p1....till we reach last charcter (S out of GATECS)
OR
 increment value of p1....till we reach NULL CHARACER(/0)

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Jaspreet as you can see in the question That the address of  GATECS is passed in str.. Nd nowhere GATECS is stored in array.. Therefore the pointer is incremented till it gets the value which will break the condition of while.. Now while is false only when we get null.. Therefore it will be incremented till we reach null character

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This is a bad question because of reason explained by arjun sir in this comment

You should not use an integer pointer to point to a character array or else behaviour is undefined in C. This is because C language supports memory alignment for integers where any integer can be assumed to start from an address which is a multiple of sizeof int.

Duplicate: https://gateoverflow.in/164145/programming

Answer 1

ans will  be 6 because we are  doing pointer subtraction

Answer 2

Answer is 6. 

This function basically counts number of characters in a string.

The function madeeasy(int *p1) basically counts number of characters in input string..

Inside madeeasy(int *p1) pointer p2 is initialized as p1. 

The statement while(*++p1) increments p1 till ‘ \ 0’ is reached! p1 is incremented by 6..

Finally the difference between p1 and p2 is returned which is 6.  

Comments

I am getting the same answer, i.e, 6. But Made Easy has given the answer as 3. :-(

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Ohhk. i got it! Because we are pointing integer pointer to a string.

So value returned will be p1-p2/sizeof(int).

and that's why they have mentioned size of int specifically.

OR

when we're incrementing p1, it will increment by [p1+(1*sizeof(int))]

and hence in 3 increments only we can reach to '\0'.

because each time integer pointer will increment by 2 characters.

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thank you. this comment is more helpful than the answer. :-)

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You should not use an integer pointer to point to a character array or else behaviour is undefined in C. This is because C language supports memory alignment for integers where any integer can be assumed to start from an address which is a multiple of sizeof int.

Answer 3

answer will be 3........as difference will be 6.....we will divide it by 2  as size of int pointer will be 2 bytes....thus 3....

Comments

@himunshu sir what is meaning of while (∗++p1)????

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increament the pointer till you get null.....so pointer will be increamented 6 times.....difference between these 2 pointers will be 6 bytes ...since it is int pointer and size of int pointer is 2 bytes....hence ans will be divided by 2......thus we will get 3...had it been char pointer ans would have been 6...since the size of char pointer is 1.....

sorry for the late reply