the output will be #include<stdio.h> int main () { int i; i = 9/2;

the output will be

#include<stdio.h>
int main ()
{
int i;
i = 9/2;
printf("%i",i);
return 0;
}

Question Posted / karunakar1916

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the output will be #include<stdio.h> int main () { int i; i = 9/2; printf("%i&quo..

Answer / manomit mitra

Is This Answer Correct ?

48 Yes

8 No

the output will be #include<stdio.h> int main () { int i; i = 9/2; printf("%i&quo..

Answer / sumesh

the output of this program will be 4.here i is an integer type not float.

Is This Answer Correct ?

17 Yes

3 No

the output will be #include<stdio.h> int main () { int i; i = 9/2; printf("%i&quo..

Answer / jac

The output will be 4. The division is being stored in a variable of type int - division operations upon it will comport to the rules of integer division, which says, "Divide but drop the remainder". Moreover, the formatting flag "%i" prints an integer value all the time, even if you try and pass it a floating-point type (which will probably give you garbage - without a cast, that is).

http://www.cs.cf.ac.uk/Dave/C/node4.html - search "integer division"
http://linux.die.net/man/3/printf - look for the "i" formatting string

Is This Answer Correct ?

11 Yes

1 No

the output will be #include<stdio.h> int main () { int i; i = 9/2; printf("%i&quo..

Answer / nupur

it will return 0

Is This Answer Correct ?

23 Yes

27 No

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