New answers tagged multiplication
1
Your explanations are a bit incorrect. What line 10 does is to add the high 64-bit word of $x_l y_l$ (%rdx after line 9) to $2^{64} (x_h y_l + x_l y_h)$. The low 64-bit word has been computed in line 9 and is in %rax.
Now, all the overflows that have been ignored will affect bit 128 only, i.e. the obtained result can differ from the correct result only by a ...
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