C Programming - Control Instructions - Discussion

Discussion Forum : Control Instructions - Find Output of Program (Q.No. 6)
6.
What will be the output of the program, if a short int is 2 bytes wide?
#include<stdio.h>
int main()
{
    short int i = 0;
    for(i<=5 && i>=-1; ++i; i>0)
        printf("%u,", i);
    return 0;
}
1 ... 65535
Expression syntax error
No output
0, 1, 2, 3, 4, 5
Answer: Option
Explanation:

for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.

In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.

Loop condition always get evaluated to true. Also at this point it increases i by one.

An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)

Discussion:
49 comments Page 4 of 5.

Ritu said:   1 decade ago
I can't understand this for loop. Please explain the working method.

Bhuvana said:   1 decade ago
This question confuses the for loop structure. Can you explain me?.

Supriya said:   5 years ago
I didn't understand.

When i=6 then the condition will false right.
(4)

Praveen said:   8 years ago
This doesn't getting terminated even after the value exceeds 65535.

Zooglaw said:   1 decade ago
It's an endless loop, isn't it? ++i will never break the loop.

Sundar said:   1 decade ago
%u --> It is to display the data as unsigned integer.

Teja said:   1 decade ago
@shilpa great explanation. Now I cleared my doubt.

Santosh said:   1 decade ago
++i will break the loop when the value of i is 0.

Jit said:   1 decade ago
@shilpa---you made me out of of the loop!!

Debalipta said:   1 decade ago
Thank you silpa for clearing my doubt.


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