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kozerog [31]
1 year ago
6

O How basic blocks are identified and how the blocks are used to construct control flow graphs

Computers and Technology
1 answer:
OLEGan [10]1 year ago
5 0

Basic blocks are identified  because they are known to be a straight line that is known also as a code sequence that tends to have no branches in regards to its in and out branches and its exception is only to the entry and at the end.

Note that  Basic Block is said to be a composition  of statements that is known to be one that often always executes one after other, and this is often done in a sequence.

<h3>How do you create a flow graph from the basic blocks?</h3>

Flow graph  is gotten by:

  • Lets Block B1 be the initial node and also Block B2 will tend to follows B1, so from B2 to B1 there is seen a kind of an edge.

Note that the first task is for a person to partition a sequence of three-address code and this is done into basic blocks.

Hence, Basic blocks are identified  because they are known to be a straight line that is known also as a code sequence that tends to have no branches in regards to its in and out branches and its exception is only to the entry and at the end.

Learn more about basic blocks from

brainly.com/question/132319

#SPJ1

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A company currently uses Microsoft Active Directory as its identity provider. The company recently purchased Oracle Cloud Infrastructure (OCI) to leverage the cloud platform for its test and development operations. As the administrator, you are now tasked with giving access only to developers so that they can start creating resources in their OCI accounts. The step to achieve this is :

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Explanation:

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4 0
2 years ago
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kogti [31]

Explanation:

i hope this will help you

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2 years ago
Which of the following is a step in paraphrasing? Skim the original information to get a sense of what it is about. Select a few
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Consider the following high-level recursive procedure: long long int flong long int n, long long int k long long int b b k+2; if
Stolb23 [73]

Answer:

f(long long, long long):

push rbp

mov rbp, rsp

push rbx

sub rsp, 40

mov QWORD PTR [rbp-40], rdi

mov QWORD PTR [rbp-48], rsi

mov rax, QWORD PTR [rbp-48]

add rax, 2

mov QWORD PTR [rbp-24], rax

cmp QWORD PTR [rbp-40], 0

jne .L2

mov QWORD PTR [rbp-24], 8

jmp .L3

.L2:

mov rax, QWORD PTR [rbp-40]

lea rdx, [0+rax*4]

mov rax, QWORD PTR [rbp-24]

lea rbx, [rdx+rax]

mov rax, QWORD PTR [rbp-48]

lea rdx, [rax+1]

mov rax, QWORD PTR [rbp-40]

sub rax, 1

mov rsi, rdx

mov rdi, rax

call f(long long, long long)

add rax, rbx

mov QWORD PTR [rbp-24], rax

.L3:

mov rdx, QWORD PTR [rbp-24]

mov rax, QWORD PTR [rbp-48]

add rax, rdx

add rsp, 40

pop rbx

pop rbp

ret

Explanation:

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