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polet [3.4K]
4 years ago
13

Ar count = 10;

Computers and Technology
1 answer:
Alinara [238K]4 years ago
6 0

Answer:

The value of count after the code is executed is 0

Explanation:

According to the code initially the value of count is 10

while (count>0)

{

count=count-2

}

means 10>0 condition is true

count=10-2=8

Again 8>0 condition of while loop is again  true

now count becomes=6

This process repeated untill the condition of while loop is true

and at last count becomes 0 the condition is false (0>0) so count becomes 0 after the code segment is executed

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oksian1 [2.3K]

Answer:

Following are the answer to this question:

Explanation:

A)

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\texttt{Number of address bits    \ \ \ \ \ \ \ \ \ \ \ \ \ \            Number of addresses}\\\\ \ \ \ \ 30  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                                 2^{30}= 1073741824

B) \texttt{I \ \ \ Opcode  \ \ \ Register address \ \ \  Immediate operand \ \ \    Memory address}\\\\\\\textt{1 \ bit  \  \ \ \ \ \ 128  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \    2 \ bits   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   24 \ bits  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \           30 \ bits}\\\\\\            = 2^7 \\\\ = 7 \ bits

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= 64-(1+7+24+30)\\\\=64 -62\ \  bits\\\\= 2\ \ bits\\

C)

Immediate value size while merging the additional benefit with the address field:

= 2^{24} + 2^{30}\\\\= 2^{54}\\\\\texttt{Range before combining(-,+) 24 bits   \ \ \ \ \ \       Range after combining( -,+)54bits}\\\\                         \textt{-2^{12} from + (2^{12}-1)   \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \                                    -2^{27} from + (2^{27}- 1)}

= \frac{24}{2} = 12\\\\= \frac{54}{2} = 27

The range is accomplished by dividing the bits by 2 into the two sides of the o and the number is one short to 0.

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