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iren [92.7K]
3 years ago
11

Which BIOS interrupt provides serial port services andlist down names of 9 pins of DB-9connector.

Computers and Technology
1 answer:
mote1985 [20]3 years ago
8 0

Answer:

To help interrupt serial driven communications, the PC utilizes two interrupts, IRQ 3 and IRQ 4.

INT 0Bh and INT 0Ch are the BIOS interrupt

It patches the COM1,COM2 when you call the Com Init Intr routine: interrupt vectors (int 0Ch int 0Bh), allowing IRQ 4 and IRQ3 in the 8259A PIC.

IRQ 3  0Bh  Serial port 2

IRQ 4  0Ch  Serial port 1

<u>9 pins of DB-9 connector -</u>

1       RTS, Request to Send

2       CTS, Clear to send

3       GND,Ground

4       RXD,Receive Data

5      TXD, Transmit Data

6       DTR, Data Terminal Ready

7       DSR, Data Set Ready

8       RI, Ring Indicator

9       DCD,Data carrier detect

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Answer:

Following are the program in python language the name of the program is factors.py

num= int(input("Please enter a positive integer: "))#Read the number by user

print("The factors of ",num,"are:")

for k in range(2,num): #iterating over the loop

   if(num%k==0): #checking the condition

       print(k)#display the factor

Output:

Please enter a positive integer: 12

The factors of 12 are:

2

3

4

6      

Explanation:

Following are the description of the program

  • Read the number by user in the "num" variable
  • Iterating the for loop from k=2 to less then "num".
  • In the for loop checking the factor of "num"  variable by using % operator.
  • Finally display the factor by using print function
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What would the range(3, 9) function generate?
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g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err
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Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

   

   // variables declared

   int n;

   int sum=0;

   float avg;

   

   do

   {

       // user input taken for number    

       cout<< "Enter a number between 1 and 100 (inclusive): ";

       cin>>n;

       

       if(n<1 || n>100)

           cout<<" Number is out of range. Enter valid number."<<endl;

       

   }while(n<1 || n>100);

   

   cout<<" "<<endl;

   

   // printing even numbers between num and 50  

   for(int num=1; num<=n; num++)

   {

       sum = sum + num;

   }

   

   avg = sum/n;

   

   // displaying sum and average

   cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

   cout<<"Average of numbers between 1 and "<<n<<" is ";

   printf("%.2f", avg);

   

       return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

 

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

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