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dexar
29 days ago
11

Consider the following skeletal C program: void fun1(void); /* prototype */ void fun2(void); /* prototype */ void fun3(void); /*

prototype */ void main() { int a, b, c; . . . } void fun1(void) { int b, c, d; . . . } void fun2(void) { int c, d, e; . . . } void fun3(void) { int d, e, f; . . . } 256 Chapter 5 Names, Bindings, and Scopes Given the following calling sequences and assuming that dynamic scoping is used, what variables are visible during execution of the last function called? Include with each visible variable the name of the function in which it was defined. a. main calls fun1; fun1 calls fun2; fun2 calls fun3. b. main calls fun1; fun1 calls fun3. c. main calls fun2; fun2 calls fun3; fun3 calls fun1. d. main calls fun3; fun3 calls fun1. e. main calls fun1; fun1 calls fun3; fun3 calls fun2. f. main calls fun3; fun3 calls fun2; fun2 calls fun1
Computers and Technology
1 answer:
ivann1987 [930]29 days ago
8 0

Answer:

Refer to the explanation section

Explanation:

a) The main function invokes fun1; fun1 then calls fun2, followed by fun2 calling fun3.

fun3()                                        d, e, f

fun2()                                        c, d, e

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK INDICATING VARIABLES OF EACH FUNCTION

   The preceding call stack diagram clearly indicates that the last function called is fun3().

   Within fun3(), the local variables "d, e, f" are available.

   The variable "c" from fun2() can also be accessed.

   The variable "b" belonging to fun1() remains visible.

   Finally, the variable "a" from main() is also visible.

b) Here, main invokes fun1, which then calls fun3.

fun3()                                        d, e, f

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING VARIABLES FOR EACH FUNCTION

   Again, the previous call stack clearly shows that fun3() is the last function executed.

   The local variables in fun3() include "d, e, f".

   The accessible variables from fun1() are "b, c".

   Variable "a" from main() is also retrievable.

c) The main function calls fun2, which calls fun3, and fun3 in turn calls fun1.

fun1()                                        b, c, d

fun3()                                        d, e, f

fun2()                                        c, d, e

main()                                        a, b,c

CALL STACK ILLUSTRATING THE VARIABLES IN EACH FUNCTION

   The last function call evident from the diagram is to fun1().

   In fun1(), the visible local variables are "b, c, d".

   The visible variables "e, f" are from fun3().

   The variable available from main() is "a".

d) The main function invokes fun1, which calls fun3, and lastly, fun3 invokes fun2.

fun2()                                        c, d, e.

fun3()                                        d, e, f

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING VARIABLES WITHIN EACH FUNCTION

   From the call stack, it's evident that the final function call is made to fun2().

   In fun2(), local variables "c, d, e" can be accessed.

   Variables "f" from fun3() and "b" from fun1() remain visible.

   Finally, variable "a" from main() is also visible.

The ultimately called function will include all of its own local variables and the variables from all preceding function calls up to the main function.

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Answer:

EMI and RFI

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EMI, or Electromagnetic Interference, can also be referred to as Radio-frequency Interference (RFI) within the radio frequency domain.

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Determine the number of bytes necessary to store an uncompressed binary image of size 4000 × 3000 pixels
maria [879]

Response: 1,500,000 bytes.

Clarification:

If we assume the image dimensions are 4000 pixels in width and 3000 pixels in height, the total uncompressed image will consist of 4000*3000= 12,000,000 pixels.

In the case of a binary image, each pixel can have only two values, which necessitates one bit for each pixel.

This indicates that we need to accommodate 12,000,000 bits.

Given that 1 byte equals 8 bits.

So, to store an uncompressed binary image sized 4000 x 3000 pixels, 12,000,000/8 bytes is required ⇒ 1,500,000 bytes.

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Someone else can drive your car if _____.
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Answer:

C. you possess insurance documentation

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Read 2 more answers
Write a function called simulate_several_key_strikes. It should take one argument: an integer specifying the number of key strik
oksian1 [797]

Answer:

The code for the solution is implemented using Python 3.

  1. import random
  2. import string
  3. def simulate_several_key_strikes(l):
  4.    char_set = string.ascii_lowercase
  5.    return ''.join(random.choice(char_set) for i in range(l))
  6. print (simulate_several_key_strikes(10))

Explanation:

This program is designed to create random characters, and the quantity of characters produced is determined by the user's input. Therefore, it is necessary to import the random library (Line 1). Additionally, we incorporate the string module to access its associated method for generating English letters (Line 2).

Then, we define the function simulate_several_key_strikes that accepts one parameter, l. Inside this function, the ascii_lowercase method is utilized to establish the character set of lowercase letters, which is stored in the variable char_set (Line 5). The random.choice method randomly selects a character from char_set, concatenating it with an empty string (Line 6). A for-loop is utilized to produce a total of l characters, generating the final output.

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1 month ago
Write an expression to print each price in stock_prices. Sample output with inputs: 34.62 76.30 85.05
8_murik_8 [892]

Answer:

Below is the Python code:

stock_prices = input().split() #this takes input and separates it into a list

for price in stock_prices: #this loops through each stock price

   print("$",price) #this outputs each stock price prefixed by a dollar sign

Explanation:

The logic behind the program is clearly outlined in the attached comments. To illustrate the program's workings, let's consider an example:

Imagine the user inputs the stock_prices values of

34.62 76.30 85.05

The input() method captures user input, while the split() method divides the input string, providing a list of strings as:

['34.62', '76.30', '85.05']

Following that, the loop statement for price in stock_prices: iterates through each item in the list, and print("$",price) displays each value from the list on the output screen with a dollar sign, as follows:

$ 34.62                                                                                                            

$ 76.30                                                                                                            

$ 85.05    

4 0
1 month ago
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