When do C and C++-compilers convert or promote a float to double, implicitly?

Question

For an embedded project I'd like to know when does a standard-compliant C-compiler (C99) and C++-compiler (C++11) will most likely implicitly promote a single-float variable/value to double-float.

I'm aware of two cases:

literals which are not suffixed with f. For example: 3.14
passing a float to function with variadic argument-list (...)

Are there any others? What about templates?

The answers of this question are very helpful for me as well - including it here for reference.

In all operations where the usual arithmetic conversions are used. — Vlad from Moscow, Mar 15 '16 at 09:13
You might want to read about [standard conversions in C++](http://en.cppreference.com/w/cpp/language/implicit_cast). And [for C as well](http://en.cppreference.com/w/c/language/conversion). — Some programmer dude, Mar 15 '16 at 09:14
@Patrick B. As for literals then there is no conversions. It is the definition of the lfloating literal. — Vlad from Moscow, Mar 15 '16 at 09:27
You can't pass a `double` to a template anyway. And the template instantiation will have a concrete function signature. — MSalters, Mar 15 '16 at 09:35

score 9 · Answer 1 · edited Mar 15 '16 at 10:01

In C:

A numeric literal with . and no suffix, e.g. 3.14, does not involve any promotion. It is double for its entire lifetime.

A float is promoted to double if the float is an argument to a function call, and the function being called has no prototype in scope, or the argument corresponds to the ellipsis (...) in the prototype in scope.

A float is converted to double in any of the following situations:

The float is an argument to a function call corresponding to a parameter of type double in a prototype in scope.
A binary operator has double and float as the two argument types. The operators this applies to are: * / + - < > <= >= == !=
The conditional operator has double and float as the second and third operand (in either order)
The float is cast to double
The float is assigned to a double (including compound assignment)

In C++, all of the above cases still apply, except for the cases about no prototype (since C++ requires all function calls to have a prototype in scope).

There is a new case: the standard conversion sequence which is too complicated to summarize briefly. But as an example, this C++ code contains an implicit conversion from float to double :

class T { public: T(double dummy) {} };
void foo(T); 
foo(3.14f); // Conversion sequence: float->double->T

I'm not sure if this is an exhaustive list for C++ though.

C++ has no "functions without prototypes" (i.e. In C++, `void foo()` is a function without arguments). Other than that, these cases all apply to C++ as well. — MSalters, Mar 15 '16 at 09:33

When do C and C++-compilers convert or promote a float to double, implicitly?

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