Wrong answer of average function?

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...

herohuyongtao · Accepted Answer · 2014-10-24T15:11:40.430

4

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

edited Oct 24 '14 at 15:11

answered Oct 24 '14 at 14:41

herohuyongtao

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@user3889963 As programmers, we are all blind from time to time. :P – herohuyongtao Oct 24 '14 at 14:45
Instead of a cast, use a float literal: `avg1 = (num1 + num2 + num3) / 3.0f;` – John Bode Oct 24 '14 at 15:09

luk32 · Answer 2 · 2014-10-24T14:52:45.207

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.

gsamaras · Answer 3 · 2014-10-24T14:47:59.710

3

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.

edited Oct 24 '14 at 14:47

answered Oct 24 '14 at 14:42

gsamaras

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@chux Yes, `3.0` is a `double`. @G.Samaras If you want a `float` you can use `3.0f`. – Terseus Oct 24 '14 at 14:50

ani627 · Answer 4 · 2014-10-24T14:51:46.380

2

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.

edited Oct 24 '14 at 14:51

answered Oct 24 '14 at 14:41

ani627

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score 2 · Answer 5 · answered Oct 24 '14 at 14:42

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.

score 0 · Answer 6 · answered Oct 24 '14 at 14:59

0

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.

answered Oct 24 '14 at 14:59

Manjunath N

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Interesting suggestion with a subtle effect: This does a `(float) ((double) sum/(double))`. Had code been `return ((num1+num2+num3)/3.0f);` --> `(float) sum/(float)`. The first is typically more accurate while the later typically more efficient. IAC, it does depend on `FLT_EVAL_METHOD` and makes insignificant difference with OP's sample integers. – chux - Reinstate Monica Oct 24 '14 at 15:13

score 0 · Answer 7 · answered Oct 24 '14 at 15:04

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;
}

Wrong answer of average function?

7 Answers7