Wrong answer of average function? - c

Question

I'm practically new to C programming and I've been trying to get a simple average function right, but the fractional part of the answer keeps messing up...??
#include <stdio.h>
#include <float.h>
float cal(int num1,int num2,int num3);
int main(){
int a,b,c;
float avg;
a=10;
b=5;
c=11;
avg=cal(a,b,c);
printf("Average is : %E\n", avg);
return 0;
}
float cal(int num1,int num2,int num3){
float avg1;
avg1=(num1+num2+num3)/3;
return avg1;
}
The answer (avg) should be 8.66666666667, but instead I get 8.00000000...

Answer 1

You're doing integer division here. Cast it to float (at least one of them) or use float literals before division to force it to use float division.
For example, change
avg1=(num1+num2+num3)/3;
to
avg1=(num1+num2+num3)/(float)3; // 1. cast one to float
avg1=(num1+num2+num3)/3.0f; // 2. use float literals

Answer 2

Change this
avg1=(num1+num2+num3)/3;
to this
avg1=(num1+num2+num3)/(float)3;
That way you force a division by a float.
With your code you actually perform integer division, which means that the decimal digits get discarded. Then the result of the division is assigned to a float number, but the decimal digits are already gone. That's why you need to cast at least one operand of the division to a float, in order to get what you want.

Answer 3

It is because all operands are integer in here (num1+num2+num3)/3. So you get an integer division, that is later cast to a float (i.e. upon the assignment, but after the evaluation).
You need to make one of the division operands a float, so the rest will be converted. And the division will be a float division.
E.g:
(num1+num2+num3)/(float)3
(num1+num2+num3)/3.0f
((float)(num1+num2+num3))/3
Note that the additions are still integer additions, because of parentheses.
A nice read on the conversion rules is here.

Answer 4

Force the division to be perfomed in floating point
avg1=(num1+num2+num3)/3.0f;
What happens in your case is, you perform the an integer division and then convert it to float:
Resulting type of (num1+num2+num3)/3 is an integer, while the type of (num1+num2+num3)/3.0f is a float.
Integer division will give the result without the decimal point.

Answer 5

Change
avg1=(num1+num2+num3)/3;
to
avg1=(float)(num1+num2+num3)/3;
If you perform integer division then result will also be integer.
As avg1 is already declared as float, you can cast the result of the operation to get float value.

Answer 6

You can simplify your code further
float cal(int num1,int num2,int num3){
return ((num1+num2+num3)/3.0);
}
Just change the value 3 to 3.0 that's enough. because the up-casting has to be made manually the compiler only perform the down-casting.

Answer 7

In addition to the well pointed out need to use floating point division rather than integer division, typical float will not provide a precise number like 8.66666666667 but only to 6 or so digits. Further, conversion of typical int (32-bit) may result in truncation when converting to float.
For a more precise answer with 11 digits to the rights of ., use double instead of float
double cal(int num1,int num2,int num3){
double avg1;
avg1=(num1+num2+num3)/3.0; // 3 --> 3.0
return avg1;
}
int main(void){ // added void
int a,b,c;
double avg;
a=10;
b=5;
c=11;
avg=cal(a,b,c);
// printf("Average is : %E\n", avg);
printf("Average is : %.11E\n", avg); // Print to 11 digits after the dp.
return 0;
}