+ 1
Explain The Answer
5 Respuestas
+ 3
Your Code:
#include <stdio.h>
int fun(int a,int b){
printf("%d %d\n",a,b);
if(b==0)
return 0;
if(b%2==0)
return fun(++a + a++,b/2);
return fun(++a + a++,b/2)+a;
}
int main() {
int x=3,y=2;
int z=fun(y,x);
printf("%d",z);
return 0;
}
1. You have a function named fun which takes in 2 integer (a and b) parameters.
2. In the func, you first print the 2 integers.
3. If b is 0 then you return the function with nothing.
4. If b is an even number then, then you run the function fun again (which is known as recurrsion) by passing 1st parameter as (a+1)+a and 2nd parameter as b/2.
5. As you already returned from the function the next return will not be considered.
6. In the driver function or the main function, you are passing 3 and 2 to the function fun and getting the value in z.
7. Here, you are printing z.
In this, program, whatever value you put in x and y in the main, you will get 1 in z.
So, therefore it will print 1, in whatever case.
+ 2
See the warning about sequence point. You are accessing and modifying data in a sequence, there is no definite or correct in this case. You may see different outputs with different compiler.
+ 2
This is not correct one Ipang .. In your point answer the question based on this compiler... Kindly refer operator precedence rule.. I checked 4 compilers all compilers give same answer..
+ 2
Irrespective of any other compiler, output here is
2 3
7 1
17 0
13
the output 2,3,7,1,17,0 are understandable (if not ,you can ask me) but I am figuring out how 13 is the final value returned by the function .
+ 1
Suresh Rajendiran
Operator precedence hardly play a role where sequence point violation occurs.
Notice that I wrote "you may see different outputs ...", I didn't say you *will*. I tested the code in C4Droid and it gave different output BTW.
And about sequence point, and the triggers, kindly refer this
https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points