Alexandre Oliva <[EMAIL PROTECTED]> writes: > On Dec 20, 2007, Ian Lance Taylor <[EMAIL PROTECTED]> wrote: > > > It is technically feasible but problematic for other reasons. > > i = i * m + ((i / j) + k) / n; > > On a two register machine like the x86 i will change several times > > during that calculation. > > No. The register used to hold its initial value will. Keep in mind > the separation between user variables and implementation locations. > The user variable 'i' is only supposed to change when assignment > operation is performed, (even if only in a theoretical level), when > the final value of the RHS is available and stored in the location > then assigned to hold the value of variable 'i'.
OK, fair enough. > Now, it is possible that the previous value of 'i' becomes unavailable > while the expression is evaluated. Then, in order to represent this > correctly, we just have to note that 'i' is no longer available as > soon as all locations holding its original value are clobbered, and > that it's available again when its new location holds the assigned > value. Right, which will significantly increase debugging size as you add two more notes around many lines. Ian
