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b<sub>'''1'''</sub> c<sub>'''1'''</sub>d<sub>'''1'''</sub> 3<sub>'''0'''</sub>
From this tree it can be seen that we need 2 registers to compute the left subtree of the '*', but only 1 register to compute the right subtree. Therefore we shall start to emit code for the left subtree first, because we might run into the situation that we only have 2 registers left to compute the whole expression. If we now computed the right subtree first (which needs only 1 register), we would then need a register to hold the result of the right subtree while computing the left subtree (which would still need 2 registers), therefore needing 3 registers concurrently. Computing the left subtree first needs 2 registers, but the result can be stored in 1, and since the right subtree needs only 1 register to compute, the evaluation of the
==Advanced Sethi-Ullman algorithm==
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