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SCORPION-xisa [38]
3 years ago
13

A program contains the following function definition: def cube(num): return num * num * num Write a statement that passes the va

lue 4 to this function and assigns its return value to the variable result.
Computers and Technology
1 answer:
Andrej [43]3 years ago
7 0

Answer:

  1. def cube(num):
  2.    return num * num * num  
  3. result = cube(4)

Explanation:

Given the function <em>cube()</em> as in Line 1 - 2.

To pass the value 4 to this function, we just place 4 inside the parenthesis of the function (Line 4). The value of 4 will be multiplied by itself for two times and the final multiplied result (4 * 4 * 4 = 64) will be returned to variable result.

If we use <em>print() </em>function to display the result, we shall see a value of 64 printed on the terminal.

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A 64-bit word computer employs a 128KB cache. The address bus in this system is 32-bits. Determine the number of bits in each fi
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Answer: provided in the explanation segment

Explanation:

taking a step by step process for this, we will analyze this problem,

we have that the Word size = 8 bytes

Cache size = 128 KB = 128×2¹⁰ bytes = 2¹⁷ bytes

(a). we are asked to cal for fully associative mapping with line size of  4 words.

  • fully associative mapping with line size of  4 words = 4 × 8 = 32 bytes
  • offset = log₂³² = 5 bit
  • tag = 32 - (5+0) = 27 bit
  • index bit = 0

(b). Direct mapping with the line size of 8 words:

  • here  the line size = 8×8 = 64 bytes
  • no of lines = 2¹⁷/2⁶ = 2¹¹ lines
  • offset = log ₂(2⁶) = 6 bit
  • index = log ₂(2¹¹) = 11 bit
  • tag = 32 - (5 +11) = 15 bit

(c). 2-way set associated mapping with the line size of 1 word:

  • no of lines = 2¹⁷/2³ = 2¹⁴ lines
  • offset = log₂(2³) = 3 bit
  • no of sets = 2¹⁴/2 = 2¹³ sets
  • index = log₂(2¹³) = 13 bit
  • tag = 32 - (3+13) = 16 bit

(d). 8-way set associated mapping with the line size of 2 words:

8-way set associated mapping with the line size of 2 = 2*8 = 16  bytes

  • no of lines = 2¹⁷/2⁴ = 2¹³ lines
  • no of sets = 2¹³/2³ = 2¹⁰ sets
  • offset = log₂¹⁶ = 4 bit
  • index = log ₂(2¹⁰) = 10 bit
  • tag = 32 - (4+10) = 18 bit

cheers i hope this helps!!!!

5 0
4 years ago
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