Why Does Comparing Floats in C Sometimes Produce Unexpected Results?

Understanding Floating Point Comparison in C

When comparing floating point values in C , it is important to be aware of potential inaccuracies due to their limited precision. In the code snippet provided, the unexpected output "1 is right" is observed when comparing the floats a and b with constants 0.7 and 0.5, respectively.

Causes of Unexpected Output:

The issue arises because of the following reasons:

• Implicit Type Conversion: When comparing a float to a double (e.g., if (a < 0.7)) or a float to a literal constant that is a double (e.g., if (a < .7)), the float is implicitly converted to a double.
• Limited Precision of Floats: Floats have a lower precision than doubles, meaning they can represent fewer decimal places accurately. As a result, 0.7 as a float may not be exactly the same as 0.7 as a double.
• Exact Representation of Constants: Constants like 0.5 that are powers of 2 can be represented exactly as floats and doubles. Therefore, b < 0.5 remains false even when b is a slightly inaccurate float.

Resolution:

To obtain the expected output, you can either:

• Use Doubles: Change the variables a and b to doubles (double a = 0.7; double b = 0.5;).
• Use Float Literals: Modify the constants to float literals (if (a < 0.7f) and if (b < 0.5f)). This forces the comparison to use floats throughout.

Code with Expected Output:

int main()
{
    double a = 0.7;
    double b = 0.5;
    if (a < 0.7)
    {
        if (b < 0.5)
            printf("2 are right");
        else
            printf("1 is right");
    }
    else
        printf("0 are right");
}

In this corrected code, the comparison of a and b to their respective doubles ensures accurate precision, resulting in the expected output of "0 are right."

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