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 2 of 5.
Paridhi said:
1 decade ago
Thanks shilpa!
Sai said:
1 decade ago
If you compile the code in gcc, the output will be different.
After 32767, it is printing from 4294934528 to 4294967295.
After 32767, it is printing from 4294934528 to 4294967295.
Prakhar said:
1 decade ago
This option is wrong, actually there is no right option. The code will print first "1" to "32767" after it will increase as usual and print "-32768" to "-1" and now if increase in i++ the value will be "0" so the condition becomes false. Exit the loop.
Jose said:
1 decade ago
I've tested this piece of software on Linux with GCC and I agree with Prakhar.
Sai, you can see the output Prakhar is suggesting if you change the output format option from %u to %d. However you're also right about those long numbers representation. I think it happens because the compiler uses a 4-bit equivalent representation for the negative values when you use the %u option and I overflows, but the number of iterations are exactly the same, and when it reaches 0, as Prakhar said, the loop finishes.
Sai, you can see the output Prakhar is suggesting if you change the output format option from %u to %d. However you're also right about those long numbers representation. I think it happens because the compiler uses a 4-bit equivalent representation for the negative values when you use the %u option and I overflows, but the number of iterations are exactly the same, and when it reaches 0, as Prakhar said, the loop finishes.
Sarah said:
1 decade ago
Does this answer mean the i increases from 1 to 32767(0x7fff) and increases again to become -32768(0x1000), and again it becomes -32767(0x1001). By this incremental rule, the last two value is -1(0xffff) and 0(overflow?)
And because it printf as an unsigned int, it shows 1..65535?
I'm confused, can anyone please help me?
And because it printf as an unsigned int, it shows 1..65535?
I'm confused, can anyone please help me?
Anil said:
1 decade ago
%u returns the address of specified variable here that variable is i.
Akshay said:
10 years ago
Can anyone explain me why output starts from 1, not from 2 as initialization expression evaluates to 1 and after it, ++i increment i to 2?
Ankit kumar said:
9 years ago
Here is a syntax error, as the place of loop condition and increment have been interchanged.
Rejaul said:
9 years ago
Here i<=5 && i >=-1 is not a condition.it's working as a initialise statement.
So ,in the given statements i <= 5 and i >=-1 is true as i = 0, so they both return true,which is 1 and (1 && 1) is 1.
The condition is i ++ .the condition is always true and in each every iteration it will get increased by 1 and the last increment statements has no effect.because the format specifier is unsigned,it will print all possible values.
So ,in the given statements i <= 5 and i >=-1 is true as i = 0, so they both return true,which is 1 and (1 && 1) is 1.
The condition is i ++ .the condition is always true and in each every iteration it will get increased by 1 and the last increment statements has no effect.because the format specifier is unsigned,it will print all possible values.
Senthil said:
9 years ago
Thank you @Avishek Ghosh.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers