C Programming ::

Question

#include<stdio.h> int main() { unsigned int i = 65535; /* Assume 2 byte integer*/ while(i++ != 0) printf("%d",++i); printf("\n"); return 0; }

[A]. infinite loop [B]. 0 1 2 ... 65535 [C]. 0 1 2 ... 32767 - 32766 -32765 -1 0

Comments

Option a)infinite loop

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yes. explain it.

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Because i is incremented by two in each loop.

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it will print like this,

1, 3, 5, 7........32797, -32797, -32765, -32763......-1, 1, 3, 5, 7, 9...........

you can see series is oscillating and thus it will give infinite loop..

Answer 1

Given ,

size of unsigned int = 2 bytes  =  16 bits

Hence magnitude of the variable 'i'  lies between  0  and   2¹⁶ - 1 which is 65535..

Hence after increment of 65535 , we will get 0 as the next value ..

But we have to see how the increment is going on here..

Here i  =  65535 initially..

So condition 

i++  !=   0

works fine as it is post increment operator..But after evaluating the while condition i is incremented to 0 as explained above.Now in the body of while loop , in printf() statement , i is incremented once again , thereby causing i = 1..

So next time when the condition of while() is checked , i = 1 which is not equal to 0..And effectively in previous step we had 2 increments from 65535 to 1..

Similarly in this iteration we will have increment from 1 to 3.Likewise it will be continued ..Eventually 65533 to 65535 and then the same story continues..:) 

Thus the above program leads to infinite loop..

Hence A) should be the correct answer..