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.

Apple said:   1 decade ago
What is %u for ?

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

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

Arun G Pandian said:   1 decade ago
for loop not understood.

Sundar said:   1 decade ago
I have tested the give program. It prints 1 ... to 65535.

But it takes some time to completely print all values.

Vino said:   1 decade ago
explain for loop....

Elizabeth said:   1 decade ago
limit 65535!! how?

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

Nemichand said:   1 decade ago
When the value of i is 6, then how loop work? because statement is
(i<=5 && i>=-1)

Here must be true both condition.

Teju said:   1 decade ago
Will you please explain me following program?

void fact(int n);
void main()
{
int x=5;
fact(x);
getch();
}
void fact(int n)
{
if(n>=0) fact(--n);
printf("%d",n);
}

// Output:
-1-101234

Please explain.


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