made easy

Question

int main

{

int * str ="2017";

printf("%d", madeEasy(str));

return 0;

}

int maseEasy(int * p1)

{

int * p2 = p1;

while(*++p1);

return (p1-p2);

}

assume object of data type int occupies 2 bytes then output of the program is.............??

Comments

2 Byte

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i think it will run infinitely...

it will terminate when int size memory location contains all 0's..

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No C code will be asked in GATE which has a compile error/warning. So, this is out of GATE scope. Those who have enough time can waste it.

Answer 1

My ans is 4 ,bcoz ptr differnce = (p1 -p2) / size of datatype = 8 /2 = 4

Comments

But can we use an integer pointer to point to a string literal? What will happen to the integer alignment requirements? As per C standard this causes undefnied behaviour.

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yes undefined behaviour.Is it due to n-byte aligned of int (where n=2 or 4) whereas char 1-B alignment.

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In C we can use a char pointer to point to any other type. No other type pointer (not including void*) can be used for pointing to a different type.

Answer 2

Ans is 4

due to address arithmatic (P1-p2)/sizeof(char)

Comments

can u explain/.................

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let the Base Address of array be 100

P2=100

take int size 2 byte then P1=108

p1-p2=(108-100)/2 divide by 2 is used because of addrress arithmetic

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i have doubt

when p1 is incremented and points to 102 at that time only *p1is 0 which will make cndition as false then how p1 =108?

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yes, you are right i didnt notice it

and my ans is work when int * is replaced with char *

this program produce Error cannot convert "const char * " to " int * " in intialization