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Assinale a alternativa que apresenta a quantidade mínima de registradores necessários para se fazer a alocação no código apresentado abaixo:
t1 := a - mem[...]
t2 := d / mem[...]
t3 := t2 + mem[...]
t4 := t1 + t3
A
6
B
5
C
4
D
3
E
2
Explanation:
This question is about register allocation in compiler optimization, specifically determining the minimum number of registers needed for the given code sequence. Let's analyze the code:
t1 := a - mem[...]
t2 := d / mem[...]
t3 := t2 + mem[...]
t4 := t1 + t3
t1 := a - mem[...]
t2 := d / mem[...]
t3 := t2 + mem[...]
t4 := t1 + t3
We need to track the live ranges of variables:
t1 is defined using a and mem[...]. After this line, t1 is live.t2 is defined using d and mem[...]. After this line, both t1 and t2 are live.t3 is defined using t2 and mem[...]. After this line, t1 and t3 are live (since t2 is no longer needed after being used to compute t3).t4 is defined using t1 and t3. After this line, only t4 is live.Register usage analysis:
t1, a, d, and two memory locations (but memory locations don't need registers if accessed directly)t1, t2, and a memory locationt1, t3The key insight is that we can reuse registers when variables are no longer live:
t1 (from line 1 to line 4)t2 (from line 2 to line 3)t3 (from line 3 to line 4)However, we can optimize further:
t2 is no longer needed, so its register can be reusedt1 and t3 simultaneously for line 4Minimum registers calculation: Using graph coloring/register allocation:
Live variable analysis:
Maximum simultaneously live variables = 2 (either {t1, t2} or {t1, t3})
However, we also need to consider that operations require registers for their operands. Looking at the operations:
t1 := a - mem[...] - needs registers for a and memory operandt2 := d / mem[...] - needs registers for d and memory operandt3 := t2 + mem[...] - needs registers for t2 and memory operandt4 := t1 + t3 - needs registers for t1 and t3Since memory operands can be accessed directly without needing registers (in many architectures), and a and d are source values that might be in registers or memory, the critical constraint is that we need to hold t1 and t3 simultaneously for the final addition.
Therefore, the minimum number of registers needed is 4 (Option C).
This allows us to:
t1t2 (which becomes t3)t4Why not fewer?
t1 and t3 simultaneously