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;
}
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 3 of 5.
Adil said:
8 years ago
What about the syntax of for loop?
Is this for loop satisfied the syntax of for loop. As for(initialization ; termination condition ; increment / decrement.
Is this for loop satisfied the syntax of for loop. As for(initialization ; termination condition ; increment / decrement.
Stliya said:
8 years ago
@Shilpa.
You are wrong. If it is %d then infinite loop. It is % you so answer option A is correct. Loop terminates at 65535.
You are wrong. If it is %d then infinite loop. It is % you so answer option A is correct. Loop terminates at 65535.
Praveen said:
8 years ago
This doesn't getting terminated even after the value exceeds 65535.
Venkat said:
7 years ago
There format specifier taken as %u so its upto 65535.
If you take %d then its prints -32767 to 32767 so here loop will iterate 65535.
If you take %d then its prints -32767 to 32767 so here loop will iterate 65535.
Sandeep said:
6 years ago
What happens after i value is greater than 65535, is loop terminated or 0 will be printed resulting in an infinite loop.
Neha said:
1 decade ago
Limit is 65535 because 'u' means unsigned.
Range of short int is 2 byte = 16 bits = -32768 to 32767.
Due to 'u' (unsigned bit) its range become = 65535.
Because unsigned bit doesnt take -ve number as its name shows.
(32767*2)+1 or 32767+32768=65535.
Range of short int is 2 byte = 16 bits = -32768 to 32767.
Due to 'u' (unsigned bit) its range become = 65535.
Because unsigned bit doesnt take -ve number as its name shows.
(32767*2)+1 or 32767+32768=65535.
Zooglaw said:
1 decade ago
It's an endless loop, isn't it? ++i will never break the loop.
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?.
Nandy said:
1 decade ago
@Zooglw: here ++i statement act as conditional statement .it's nt a inc/dec statement.so it has no effect on the for loop
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