Question

It makes sense to define mathematical constants as double values but what happens when one requires float values instead of doubles? Does the compiler automatically interpret the doubles as floats at compile-time (so they are actually treated as they were const floats) or is this conversion made at runtime?

Answer 1

If by “defining”, you mean using #define, here’s what happens:

Say you have:

#define CONST1 1.5
#define CONST2 1.12312455431461363145134614  // Assume some number too
                                             // precise for float

Now if you have:

float x = CONST1;
float y = CONST2;

you don’t get any warning for x because the compiler automatically makes CONST1 a float. For y, you get a warning because CONST2 doesn’t fit in a float, but the compiler casts it to float anyway.

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If by “defining”, you mean using const variables, here’s what happens:

Say you have

const double CONST1=1.5;
const double CONST2=1.12312455431461363145134614; // Assume some number too
                                                  // precise for float

Now if you have:

float x = CONST1;
float y = CONST2;

there is no way for the compiler to know the values of CONST1 and CONST2(*) and therefore cannot interpret the values as float at compile them. You will be given two warnings about possible loss of data and the conversion will be done at runtime.

(*) Actually there is a way. Since the values are const, the optimizer may decide not to take a variable for them, but replace the values throughout the code. This could get complicated though, as you may pass the address to these variables around, so the optimizer may decide not to do that. That is, don’t count on it.

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Note that, this whole thing is true for any basic type conversions. If you have

#define CONST3 1

then you think CONST3 is int, but if you put it in a float, it would become float at compile-time, or if you put it in a char, it would become char at compiler-time.