Passing float to a function in C++ appears to change precision

Question

This is a very noob question, but I am curious to know the reason behind this: -If I debug the following C++ code:

void floatreturn(float i){
      //nothing
}

int main(){
    float a = 23.976;
    floatreturn(a);
    return 0;
}

Monitoring the passed value of a, it appears to be 23.9759998 when entering floatreturn. As a result, any processing of the value in the function would require to manually tweak the precision. Is there a reason for this, and any way to avoid it?

What's the change? The decimal part is obviously not going to be 23.976, but I don't notice any change even when switching to double. ideone.com/7NgWb0 — silverfox
– silverfox, Commented Nov 24, 2021 at 4:35
23.976 is a double, not float. Use float a = 23.976f; instead. But you'll never get the exactly result in binary floating point — phuclv
– phuclv, Commented Nov 24, 2021 at 5:47
Try to represent 0.976 as a binary fraction, and you'll see the representation is infinite (the same reason that the fraction one-third in decimal has an infinite representation of 0.3 with the 3 repeating forever). So neither 0.976 nor 23.976 can be exactly represented in any floating point type (with a base-2 mantissa) and a nearest approximation is stored. On your platform, the nearest approximation represented by a float is apparently what you are seeing. (There is also a minor wrinkle that 23.976 has type double, and more precision may be lost on converting to float) — Peter
– Peter, Commented Nov 24, 2021 at 8:35

2 revs · Accepted Answer · 2021-11-24 16:43:03Z

1

The issue happened before floatreturn(a);.
It happened at float a = 23.976;
floatreturn(a); is irrelevant.

There are about 2^32 different values that float can encode exactly. 23.976 is not one of them. The nearest encodable float is about 23.9759998...

To avoid, use values that can exactly encode as a float or tolerate being close - about 1 part in 2²⁴

edited Nov 24, 2021 at 16:43

community wiki

2 revs
chux - Reinstate Monica

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4 Comments

mv_p Over a year ago

Thanks Mark for the link and chux for the concise answer.

mv_p Over a year ago

Just one more educational point for me: why does cout default to a precision of 3 decimals if we reprint the variable without setting precision for the output? I get that storing the float will always return the closest binary float possible, and that cout is rounding.

chux Over a year ago

@mv_p What was the exact output you saw?

mv_p Over a year ago

with a simple 'float x = 23.976; cout << x;' I see "23.976"

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