Why does comparing floats to double literals in C produce unexpected results?

Floating Point Comparison Puzzle

Consider the following C code:

int main()
{
    float a = 0.7;
    float 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");
}

You would intuitively expect the output to be "0 are right." However, the surprising result is "1 is right." Why does this occur?

The Pitfalls of Floating-Point Comparison

The key lies in the differences between floating-point and double-precision numbers in C. In the code, the variables a and b are declared as floats, which are 32-bit floating-point numbers. However, both comparisons (a < 0.7 and b < 0.5) involve doubles, as the literals 0.7 and 0.5 are treated as doubles.

During the comparison, the float variables get promoted to doubles, allowing for a higher range and precision. However, this conversion can introduce subtle artifacts due to the limited precision of floats. In this case, 0.7 as a float is not exactly equivalent to 0.7 as a double.

Specifically, 0.7 as a float is represented as 0x3f000000 in the IEEE 754 standard. When promoted to a double, this value is not an exact representation of 0.7. Instead, it becomes slightly greater, around 0x3f00000000000000 (approximately 0.7000000000000001).

The Cause of the Unexpected Result

As a result of this promotion, the condition a < 0.7 becomes true because the double representation of a is marginally less than 0.7. Subsequently, the second comparison b < 0.5 evaluates to false because b (represented exactly as a double) is equal to 0.5. Thus, the code prints "1 is right."

Solutions

To resolve this issue, you can either:

• Change the variables a and b to be declared as doubles, or
• Change the literals 0.7 and 0.5 to be specified as float

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