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nordsb [41]
3 years ago
13

What happens when variables in programming are not properly initialized? answers?

Computers and Technology
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0
Errors in compilation happen, or it runs but then doesn't recognize the input. Also you can't transform into other types of variables.
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Search engines - there was no single way to ask a question
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3 years ago
write a c++ program that reads from the user a positive integer (in a decimal representation), and prints its binary (base 2) re
Vesna [10]

Answer:

The program to this question can be described as follows:

Program:

#include <iostream> //defining header file

using namespace std;

int main() //defining main method

{

int x1,rem,i=1; //defining integer variable

long Num=0;// defining long variable

cout << "Input a decimal number: "; // print message

cin >> x1; //input value by user

while (x1!=0) //loop for calculate binary number

{

//calculating binary value    

rem= x1%2;

x1=x1/2;

Num =Num +rem*i;//holding calculate value

i=i*10;

}

cout <<Num;//print value

return 0;

}

Output:

Input a decimal number: 76

1001100

Explanation:

In the above code, four variable "x1, rem, i, and Num" is declared, in which three "x1, rem, and i" is an integer variable, and one "Num" is a long variable.

  • In the x1 variable, we take input by the user and use the while loop to calculate its binary number.
  • In the loop, first, we check x1 variable value is not equal to 0, inside we calculate it binary number that store in long "Num" variable, after calculating its binary number the print method "cout" is used to prints its value.  
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3 years ago
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aliina [53]

Answer:

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Explanation:

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3 years ago
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