CODEWITHSUNDEEP
cfunctionvariablesinteger-division
user3889963
user3889963

Reputation: 477

Wrong answer of average function?

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...

Upvotes: 0

Views: 1500

Answers (7)

herohuyongtao
herohuyongtao

Reputation: 50677

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

Upvotes: 4

chux
chux

Reputation: 154075

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;
}

Upvotes: 0

Manjunath N
Manjunath N

Reputation: 1375

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.

Upvotes: 0

luk32
luk32

Reputation: 16080

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.

Upvotes: 3

ani627
ani627

Reputation: 6057

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.

Upvotes: 2

gsamaras
gsamaras

Reputation: 73394

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.

Upvotes: 3

2501
2501

Reputation: 25753

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.

Upvotes: 2

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