C Programming - Bitwise Operators - Discussion

Discussion Forum : Bitwise Operators - Find Output of Program (Q.No. 6)
6.
What will be the output of the program?
#include<stdio.h>

int main()
{
    printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1);
    return 0;
}
4 1 8 1
4 >> 1 8 >> 1
2 >> 4 Garbage value >> Garbage value
2 4
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
36 comments Page 1 of 4.

Rather said:   6 years ago
Hello All.

For people who thinks why garbage values get printed , let us look at this ststement
printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1);

We have four %d's, means we should have four variables to print but here we have only two representing first two :ie 4>>1 & 8>>1, last two format specifiers have no variables listed to print , so in such cases, it will print them as garbage values.
(3)

Mayur said:   1 decade ago
4 = 0100 now shift to right 1 time. It will become 0010 = 2.

Then << is print on screen (becoz every special symbol or character print as well as written in double cotation " " )

Then it will print
8 = 1000 now shift to right 1 time. It will become 0100 = 4.
Then print Garbage value >> Garbage value
So,
Output will
2 >> 4 Garbage value >> Garbage value

Teja said:   1 decade ago
@Anusha,
Given code is
#include<stdio.h>

int main()
{
printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1);
return 0;
}

First we got the output from 4>>1,8>>1 is 2,4 directly.

But in printf() four %d's are given.so 2,4 assigns to first two %d's and the remaining two %d's will show the garbage values.

Got it?
(1)

Adam said:   9 years ago
#include<stdio.h>

int main()
{
printf("%d >> %d %d >> %d\n", 4 >> 1, 8 >> 1);
printf("%d >> %d >> \n", 4 >> 1, 8 >> 1);
printf("%d %d \n", 4 >> 1, 8 >> 1);
return 0;
}

Try this.

Sundar said:   1 decade ago
I got the following outputs:

In GCC (32-bit platform): 2 >> 4 134513662 >> 134513904

Turbo c (16 bit platorm): 2 >> 4 0 >> 344

Therefore, 2 >> 4 garbagevalue >> garbagevalue is the correct answer.

Gowda said:   1 decade ago
Here is my code.

#include<stdio.h>

int main ()
{
int i ;
for (i=0;i<3;++i)
printf ("%d\n", i) ;
return 0;
}

I am getting output as 0 1 2.

Can any one explain why am not getting from 1 as I used pre-increment operator.

Vishu said:   9 years ago
@Nizamuddin,
Array always represents starting from '0' position so there is two matrices represented as a a[0] & a[1].
so there is no a[2] so it shows a value as '0'.

If I made any mistake correct me. Thank you.

Jay said:   1 decade ago
First it solve 4>>1 and then solve 8>>1.
Answer is 2 and 4.

Then it print the value.
But here %d is 4 times in printf("").

So it print only two values and other two are garbage value.

Reenu said:   1 decade ago
@maya.

As the number gave after >> is 1..we have right shift one place..If that number is 2(>> 2)then we have to right shift 2 place..It depends on the number given after >>.

Sai said:   6 years ago
4>>1 means divide 4 by 2 power 1 and 8>>1 means divide 8 by 2 power 1 then you will get answer as 2 and 4 if it is less then less then multiply by 2.
(2)


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