bab 10

29 Jun

1. What is the definition used in this chapter for “simple” subprograms?
subprograms cannot be nested and all local variables are static

2. Which of the caller or callee saves execution status information?
Saving the execution status of the caller could be done by either

4. What is the task of a linker?
a.find the files that contain the translated subprograms referenced in that program
and load them into memory
b.set the target addresses of all calls to those subprograms in the main program to
the entry addresses ofthose subprograms.

5. What are the two reasons why implementing subprograms with stackdynamic
local variables is more difficult than implementing simple
subprograms?
a.The compiler must generate code to cause the implicit allocation and deallocation
of local variables.
b.Recursion adds the possibility of multiple simultaneous activations of a
subprogram, which means that there can be more than one instance (incomplete
execution) of a subprogram at a given time, with at least one call from outside
the subprogram and one or more recursive calls. The number of activations is
limited only by the memory size of the machine. Each activation requires its
activation record instance.

problem set

gak ngerti pak.. nanti saya cari.. bahasa asing pak.. saya cuma bisa C sedikit

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