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Minchanka [31]
3 years ago
15

class student_record { public: int age; string name; double gpa; }; Implement a void function that has five formal parameters: a

dynamic array of student_records, the count (number of student_records in the array), an integer, a string, and a double. The function will delete a student record in the dynamic array with age, name and gpa fields that match the age, name and gpa fields that are given in the formal parameters. Remember, to check to see if the dynamic array is empty because you cannot delete from an empty array.
Computers and Technology
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

void delete_record(student_record *arr, int &size, int age, string name, double gpa) {

int index = -1;

if (arr != NULL && size > 0) {

for (int i = 0; i < size; ++i) {

if (arr[i].age == age && arr[i].name == name && arr[i].gpa == gpa) {

index = i;

break;

}

}

}

if (index != -1) {

for (int i = index; i < size - 1; ++i) {

arr[i] = arr[i + 1];

}

size--;

}

}

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Write a Python program that takes a file name as input and generates the following output: File size in bytes and KBs Time the f
Airida [17]

Answer:

See explaination

Explanation:

import os

import time

from stat import *

file = input("Enter file name: ")

details = os.stat(file)#stores statistics about a file in a directory

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print("File last accessed time: ", time.asctime(time.localtime(details[ST_ATIME])))#access time

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num_words = 0

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lines = f.readlines()

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8 0
3 years ago
Write an application that displays every perfect number from 1 through 1,000. A perfect number is one that equals the sum of all
matrenka [14]

Answer:

Written in Python

for num in range(1,1001):

    sum=0

    for j in range(1,num+1):

         if num%j==0:

              sum = sum + j

                   if num == sum:

                        print(str(num)+" is a perfect number")

Explanation:

This line gets the range from 1 to 1000

for num in range(1,1001):

This line initializes sum to 0 for each number 1 to 1000

    sum=0

This line gets the divisor for each number 1 to 1000

    for j in range(1,num+1):

This following if condition line checks for divisor of each number

<em>          if num%j==0: </em>

<em>               sum = sum + j </em>

The following if condition checks for perfect number

                   if num == sum:

                        print(str(num)+" is a perfect number")

5 0
3 years ago
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