Question: Question # 2 [ 1 5 pts ] If invoking the program as . / memory 1 2 0 , 3 0 0 , 8

Question #2[15 pts]
If invoking the program as
./memory 120,300,80,30
were to produce this output:
0|1024|1152|2176|2688|3712|3744|4768|4832|
--------------
The 120-byte chunk was allocated at address 2176
The 300-byte chunk was allocated at address 2296
The 80-byte chunk was allocated at address 1024
The 30-byte chunk was allocated at address 3712
would you say that the heap allocator uses
first fit
2176
best fit
2296
worst fit
1024
none of the above?
3712
Explain why.

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