Since it's a *=, I'd expect the operation to occur in double, quantize accordingly, and handle overflow as double-to-int would, rather than int multiplication w

Since it's a *=, I'd expect the operation to occur in double, quantize accordingly, and handle overflow as double-to-int would, rather than int multiplication would. For 10.0 specifically, double is more than large enough to, as far as I know, result in the same output. For arbitrary double literals, I wouldn't expect this to be the case. Note: overflow cast from double to int is undefined, last I looked, so one could argue that the numerical regions for which quantization would occur are outside of the practical range of this chunk of code.