All floats are doubles?

Question

I'm reading textbook that says this:

C++ treats all floating-point numbers you type in a program's source code (such as 7.33 and 0.0975) as double values by default.

I find this a bit odd and have never heard of it. Seems wasteful? Why get extra precision if you don't specify it? Why have two different types that mean the same thing? What about a long double?

Answer 1

This is a part of language specification. If you want a double, write:

auto a = 12.3;

If you want a float, write:

auto a = 12.3f;

If you want a long double, write:

auto a = 12.3L;

Source: MSDN

The whole topic is extensively described in C++ standard in chapter 2.14 Literals.

Answer 2

This is referring to floating-point literals only.

This is the same as saying that any integer number you write in code is always treated as a (signed) int. As soon as you assign this to a variable, you will get the type of the variable.

However, when using standalone literals in computation you will get the type of the literal for that computation, potentially triggering implicit type conversions:

float f = 3.141;    // f is of type float, even though the literal was double
auto d = f * 2.0;   // d will be of type double because of the literal 2.0
auto f2 = f * 2.0f; // f2 will be of type float again

The computation on the second line involves two different types: The type of the variable f is float. Even though it was constructed from a double literal, the type of the variable is what counts. The type of the literal 2.0 on the other hand is double and hence triggers an implicit conversion for the computation. The actual multiplication is therefore performed as a multiplication of two doubles.

If you want a standalone value to have a specific type, use the matching literal.