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just olya [345]
3 years ago
8

Which type of electronic community allows real time discussion among members

Computers and Technology
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

Instant messaging is the type of electronic community that allows real-time discussion among members.

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3 years ago
Get two positive integers from the user. Write code to multiply the two numbers together using only addition or subtraction. You
padilas [110]

Answer:

number1 = int(input("Enter the first number: "))

number2 = int(input("Enter the second number: "))

result = 0

for i in range(number1):

   result += number2

print(str(result))

Explanation:

Ask the user for two numbers

Initialize result as 0 to hold the multiplication of the numbers

Create a for loop that iterates "number1" times. In each iteration, add number2 to the result.

When the loop is done, print the result

8 0
3 years ago
Write a C program that reads two hexadecimal values from the keyboard and then stores the two values into two variables of type
sattari [20]

Solution :

#include  $$

#include $$

#include $$

//Converts $\text{hex string}$ to binary string.

$\text{char}$ * hexadecimal$\text{To}$Binary(char* hexdec)

{

 

long $\text{int i}$ = 0;

char *string = $(\text{char}^ *) \ \text{malloc}$(sizeof(char) * 9);

while (hexdec[i]) {

//Simply assign binary string for each hex char.

switch (hexdec[i]) {

$\text{case '0'}:$

strcat(string, "0000");

break;

$\text{case '1'}:$

strcat(string, "0001");

break;

$\text{case '2'}:$

strcat(string, "0010");

break;

$\text{case '3'}:$

strcat(string, "0011");

break;

$\text{case '4'}:$

strcat(string, "0100");

break;

$\text{case '5'}:$

strcat(string, "0101");

break;

$\text{case '6'}:$

strcat(string, "0110");

break;

$\text{case '7'}:$

strcat(string, "0111");

break;

$\text{case '8'}:$

strcat(string, "1000");

break;

$\text{case '9'}:$

strcat(string, "1001");

break;

case 'A':

case 'a':

strcat(string, "1010");

break;

case 'B':

case 'b':

strcat(string, "1011");

break;

case 'C':

case 'c':

strcat(string, "1100");

break;

case 'D':

case 'd':

strcat(string, "1101");

break;

case 'E':

case 'e':

strcat(string, "1110");

break;

case 'F':

case 'f':

strcat(string, "1111");

break;

default:

printf("\nInvalid hexadecimal digit %c",

hexdec[i]);

string="-1" ;

}

i++;

}

return string;

}

 

int main()

{ //Take 2 strings

char *str1 =hexadecimalToBinary("FA") ;

char *str2 =hexadecimalToBinary("12") ;

//Input 2 numbers p and n.

int p,n;

scanf("%d",&p);

scanf("%d",&n);

//keep j as length of str2

int j=strlen(str2),i;

//Now replace n digits after p of str1

for(i=0;i<n;i++){

str1[p+i]=str2[j-1-i];

}

//Now, i have used c library strtol

long ans = strtol(str1, NULL, 2);

//print result.

printf("%lx",ans);

return 0;

}

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The answer is ............................ False
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