C Programming - C Preprocessor - Discussion
Discussion Forum : C Preprocessor - Find Output of Program (Q.No. 9)
9.
What will be the output of the program?
#include<stdio.h>
#define FUN(arg) do\
{\
if(arg)\
printf("IndiaBIX...", "\n");\
}while(--i)
int main()
{
int i=2;
FUN(i<3);
return 0;
}
Answer: Option
Explanation:
The macro FUN(arg) prints the statement "IndiaBIX..." untill the while condition is satisfied.
Step 1: int i=2; The variable i is declared as an integer type and initialized to 2.
Step 2: FUN(i<3); becomes,
do
{
if(2 < 3)
printf("IndiaBIX...", "\n");
}while(--2)
After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.
Hence the output of the program is "IndiaBIX... IndiaBIX..."
Discussion:
22 comments Page 1 of 3.
Sathiesha reddy said:
1 decade ago
The macro FUN(arg) prints the statement "IndiaBIX..." until the while condition is satisfied.
FUN(i<3); becomes true=1, hence condition satisfies thus calls the fun given in the code. Thus runs the below function twice.
do
{
if(2 < 3)
printf("IndiaBIX...", "\n");
}while(--2)
After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.
Hence the output of the program is "IndiaBIX... IndiaBIX..."
FUN(i<3); becomes true=1, hence condition satisfies thus calls the fun given in the code. Thus runs the below function twice.
do
{
if(2 < 3)
printf("IndiaBIX...", "\n");
}while(--2)
After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.
Hence the output of the program is "IndiaBIX... IndiaBIX..."
Chaitu said:
1 decade ago
Actually I have seen this some where in some other question saying..comma operator used here in printf() doesn't act as operator..similarly in scanf()..for()..the comma used here is different than..the above one(comma operator)..in this case we can see the effect of comma operator..
//effect of comma operator
main()
{
int i=5;
int j=i+(1,2,3,4,5);//1,2,3,4 r discarded & 5 is considered
printf("%d",j);
getch();
}
o/p: 10.
//effect of comma operator
main()
{
int i=5;
int j=i+(1,2,3,4,5);//1,2,3,4 r discarded & 5 is considered
printf("%d",j);
getch();
}
o/p: 10.
Abhijit said:
8 years ago
do
{
if(2 < 3)
printf("IndiaBIX...", "\n");
}while(--2)
There is a "\n". it should have taken printf to the next line. I am a little bit confuse, please someone explain it.
{
if(2 < 3)
printf("IndiaBIX...", "\n");
}while(--2)
There is a "\n". it should have taken printf to the next line. I am a little bit confuse, please someone explain it.
(3)
Shwetal said:
2 years ago
If there is printf("India","bix");
the output will be = India.
And there would be a warning -too many arguments;
The second " " is simply ignored.
the output will be = India.
And there would be a warning -too many arguments;
The second " " is simply ignored.
(1)
ARupinderjit said:
1 decade ago
Yar if you do printf("India","Bix"); then it'll print only first quoted string,that is,India.so is the answer.
Vince Bala said:
10 years ago
I don't get it.
When i becomes 0, the condition is still true (0<3).
So it will call the print function 3 times.
When i becomes 0, the condition is still true (0<3).
So it will call the print function 3 times.
Charan said:
1 decade ago
Actually the answer will be only IndiaBIX.
Because let us consider FUN(2<3).
Which means true = 1.
i.e FUN(1).
Because let us consider FUN(2<3).
Which means true = 1.
i.e FUN(1).
Purnima said:
8 years ago
In while(0) we will just check the value of I which is zero so loops terminate. We are not checking i<3.
Mayur said:
1 decade ago
But we has used, operator ;by rule of comma first is evaluate then discard it, so only "\n" is executed?
Nagalakshmi said:
9 years ago
Yea it is correct, here while(0) becomes an infinite loop. That's why the condition became fail at i=0.
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