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yarga [219]
2 years ago
6

In a certain computer program, two positive integers are added together, resulting in an overflow error. Which of the following

best explains why the error occurs?
A) The program attempted to perform an operation that is considered an undecidable problem.
B) The precision of the result is limited due to the constraints of using a floating-point representation.
C) The program can only use a fixed number of bits to represent integers; the computed sum is greater than the maximum representable value.
D) The program cannot represent integers; the integers are converted into decimal approximations, leading to rounding errors.
Computers and Technology
1 answer:
Airida [17]2 years ago
8 0

The option that best explains why the error occurs is that The program can only use a fixed number of bits to represent integers; the computed sum is greater than the maximum representable value.

<h3>Can programs represent integers?</h3>

An integer value is known to be often listed out in the source code of a program in a way called a sequence of digits that is said to be optionally prefixed with + or −. Note that some programming languages do use other notations, like hexadecimal.

Computers are known to use a a fixed number of bits to show an integer. The most -used bit-lengths for integers are known to be 8-bit, 16-bit, 32-bit or 64-bit.

Learn more about errors from

brainly.com/question/11472659

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1. How many bits would you need to address a 2M × 32 memory if:
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Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

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w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

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