Why do these two float64s have different values?

Question

Consider these two cases:

fmt.Println(912 * 0.01)
fmt.Println(float64(912) * 0.01)

(Go Playground link)

The second one prints 9.120

Answer 1

As per spec:

Constant expressions are always evaluated exactly; intermediate values and the constants themselves may require precision significantly larger than supported by any predeclared type in the language.

Since

912 * 0.01

is a constant expression, it is evaluated exactly. Thus, writing fmt.Println(912 * 0.01) has the same effect as writing fmt.Println(9.12). When you pin 912 to float64, the other operand of the floating-point multiplication is implicitly pinned to float64, too. Thus, the expression float64(912) * 0.01 behaves like float64(912) * float64(0.01). 0.01 is not exactly representable in a float64, thus precision is lost in a different place than in the expression float64(912 * 0.01) which arises in the argument of fmt.Println() in your first example, explaining the different results.