Comparison of a float with a value in C

Predict the output of the following C program. 

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#include<stdio.h>
int main()
{
    float x = 0.1;
    if (x == 0.1)
        printf("IF");
    else if (x == 0.1f)
        printf("ELSE IF");
    else
        printf("ELSE");
}

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The output of above program is “ELSE IF” which means the expression “x == 0.1” returns false and expression “x == 0.1f” returns true. 

Let consider the following program to understand the reason behind the above output.  

C

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#include<stdio.h>
int main()
{
   float x = 0.1;
   printf("%d %d %d", sizeof(x), sizeof(0.1), sizeof(0.1f));
   return 0;
}

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The output of above program is "4 8 4" on a typical C compiler.
It actually prints size of float, size of double and size of float.

The values used in an expression are considered as double (double precision floating point format) unless a ‘f’ is specified at the end. So the expression “x==0.1” has a double on right side and float which are stored in a single precision floating point format on left side. In such situations, float is promoted to double (see this). The double precision format uses more bits for precision than single precision format. 

The binary equivalent of 0.110 can be written as (0.00011001100110011…)2 which goes up to infinity(See this article to know more about conversion). Since the precision of float is less than the double therefore after a certain point(23 in float and 52 in double) it would truncate the result. Hence, after promotion of float into double(at the time of comparison) compiler will pad the remaining bits with zeroes. Hence, we get the different result in which decimal equivalent of both would be different. For instance, 



In float 
=> (0.1)10 = (0.00011001100110011001100)2
In double after promotion of float ...(1)
=> (0.1)10 = (0.00011001100110011001100000000000000000...)2
                                      ^ padding zeroes here
In double without promotion ... (2)
=> (0.1)10 = (0.0001100110011001100110011001100110011001100110011001)2

Hence we can see the result of both equations are different.
Therefore 'if' statement can never be executed.

Note that the promotion of float to double can only cause mismatch when a value (like 0.1) uses more precision bits than the bits of single precision. For example, the following C program prints “IF”.

C

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#include<stdio.h>
int main()
{
    float x = 0.5;
    if (x == 0.5)
        printf("IF");
    else if (x == 0.5f)
        printf("ELSE IF");
    else
        printf("ELSE");
}

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Output: 

IF

Here binary equivalent of 0.510 is (0.100000…)2 

(No precision will be lost in both float and double type). Therefore if compiler pad the extra zeroes at the time of promotion then we would get the same result in the decimal equivalent of both left and right side in comparison(x == 0.5). 

You can refer Floating Point Representation – Basics for the representation of floating-point numbers.

This article is contributed by Abhay Rathi and improved by SHUBHAM BANSAL. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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