Question

What are the rules governing precedence in C++11 type inference for float/double types, for example, when inferring from an expression containing multiple types, like so:

auto var = float(1) * double(1);

Answer 1

The result will be a double. This is called floating point promotion.

From the standard, ISO 14882:2011, 4.6 Floating point promotion:

1 A prvalue of type float can be converted to a prvalue of type double. The value is unchanged.
2 This conversion is called floating point promotion.

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As noted by @sftrabbit, in 5. Expressions, paragraph 9 in the new standard:

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result.

This pattern is called the usual arithmetic conversions, which are defined as follows:

— If either operand is of scoped enumeration type (7.2), no conversions are performed; if the other operand does not have the same type, the expression is ill-formed.
— If either operand is of type long double, the other shall be converted to long double.
— Otherwise, if either operand is double, the other shall be converted to double.
— Otherwise, if either operand is float, the other shall be converted to float.
— Otherwise, the integral promotions (4.5) shall be performed on both operands.