C Programming - Control Instructions - Discussion

Discussion Forum : Control Instructions - Find Output of Program (Q.No. 8)
8.
What will be the output of the program?
#include<stdio.h>
int main()
{
    unsigned int i = 65536; /* Assume 2 byte integer*/
    while(i != 0)
        printf("%d",++i);
    printf("\n");
    return 0;
}
Infinite loop
0 1 2 ... 65535
0 1 2 ... 32767 - 32766 -32765 -1 0
No output
Answer: Option
Explanation:

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535.

Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.

Hence there is no output.

Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.

Discussion:
16 comments Page 2 of 2.

RAHUL said:   1 decade ago
When I run this program answer is A.

Kiran Kumar Reddy P said:   1 decade ago
When I simulated using gcc in linux. It results into infinite loop printing all the values.

Dheeraj bajpai said:   1 decade ago
In my point of view answer is<A> with Borland C++ compiler.

Priyanka.s said:   1 decade ago
What will be the output if that while loop became true ?

Sundar said:   1 decade ago
@Sandeep

Don't forget that the size of int should be 2 bytes. If you run the above program in 32 bit platform, for eg. GCC under Linux, it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.

To test the above program, just use Turbo C, you will understand.

I have tested it in both Turbo C and GCC.

Sandeep said:   2 decades ago
I think the answer to this question should be A; because when i run it on code block IDE it went in an infinite loop;

In case i am wrong plz do correct me!


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