c++ - Why does printf() promote a float to a double?

Question

From a previous question:

If you attempt to pass a float to printf, it'll be promoted to double before printf receives it

printf() is a variadic function right? So does a variadic function promote a float argument to a double before passing it?

Answer 1

Yes, float arguments to variadic function are promoted to double.

The draft C99 standard section 6.5.2.2 Function calls says:

[...]and arguments that have type float are promoted to double. These are called the default argument promotions.[...]

from the draft C++ standard section 5.2.2 Function call:

[...]a floating point type that is subject to the floating point promotion (4.6), the value of the argument is converted to the promoted type before the call. [...]

and section 4.6:

A prvalue of type float can be converted to a prvalue of type double. The value is unchanged

cppreference covers the default conversions for variadic function in C++ well:

• std::nullptr_t is converted to void*
• float arguments are converted to double as in floating-point promotion
• bool, char, short, and unscoped enumerations are converted to int or wider integer types as in integer promotion

We can see in C and presumably in C++ this conversion was kept around for compatibility with K&R C, from Rationale for International Standard—Programming Languages—C (emphasis mine):

For compatibility with past practice, all argument promotions occur as described in K&R in the absence of a prototype declaration, including the not always desirable promotion of float to double.