CODEWITHSUNDEEP
cfloating-pointsigfpe
Suvi_Eu
Suvi_Eu

Reputation: 275

Operation with division resulting in error execution because of SIGFPE, Arithmetic exception

I have wrote an algorithm for MCU that executes the expression of two paralel Resistances before doing more calculations with the division result. When debugging the values had no sense; they were too much big and float var did not update their initial value.

So I went to https://www.onlinegdb.com/ and I tried a part of my algorithm there. When execute the code it launches a non normally exiting with the arithmetic exception at line: 

a = 1/(1/10 + 1/b) + 0.15;

First I was working with float but I have thought that exception could be caused by an overflow, hence I did the variable storage bigger using double, but the same exception appears. Then I tried doing the same but saying:

a = (1/10 + 1/b) + 0.15;

and the execution worked!

So, I have seen the '1' operator was the cause. But I don't understand exactly why and how to fix it (without using math.h).

The code is this:

#include <stdio.h>

float a = 0.0;
int b = 100;

int main()
{
    a = 1/(1/10 + 1/b) + 0.15;//Req 
    b = a; //get int part
    a *= 10;//fractionary part converted to dec
    if ((a - b*10)>5) b++;

    printf("float: %f , int: %i",a,b);

    return 0;
}

I was expecting to get (debugging):

which I think is not a big value to fit in a float or a double var. Instead of this I get the exception.

How to work with floats/doubles and int values to avoid exceptions when 1/(something smaller)?

Upvotes: 0

Views: 64

Answers (2)

unwind
unwind

Reputation: 399871

This:

 a = 1/(1/10 + 1/b) + 0.15;//Req

is doing integer division in the first part, since 1 and 10 are both of type int. You meant:

a = 1.f / (0.1f + 1.f / b) + 0.15f;

especially for embedded targets, it can be quite important to make sure your literals are floats and not doubles, as the support can be different.

Upvotes: 1

Some programmer dude
Some programmer dude

Reputation: 409196

Both 1/10 and 1/b are integer expressions, which both result in the integer 0.

So you have a plain division by zero error.

Solve it by using e.g. 1.0f as the dividend, like 1.0f / (0.1f + 1.0f / b) + 0.15f.

Upvotes: 3

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