include <iostream>
using namespace std;
int main() {
int n; // number to convert to binary
while (cin >> n) {
if (n > 0) {
cout << n << " (decimal) = ";
while (n > 0) {
cout << n%2;
n = n/2;
}
cout << " (binary) in reverse order" << endl;
} else {
cout << "Please enter a number greater than zero." << endl;
}
}
return 0;
}//end main
Assuming the number is represented by a decimal integer, initialise a counter to zero, then repeatedly divide the number by 10 and until the number is zero. After each division, examine the remainder. Each time the remainder is zero, increment the counter. If the number is represented by a decimal float, repeatedly multiply by 10 until the value is an integer, then perform the previous algorithm.
write a c++ program to convert binary number to decimal number by using while statement
public class Dataconversion { public static void main(String[] args) { System.out.println("Data types conversion example!"); int in = 44; System.out.println("Integer: " + in); //integer to binary String by = Integer.toBinaryString(in); System.out.println("Byte: " + by); //integer to hexadecimal String hex = Integer.toHexString(in); System.out.println("Hexa decimal: " + hex); //integer to octal String oct = Integer.toOctalString(in); System.out.println("Octal: " + oct); } }
x -=y;
This is not a question.
prompt x floor(x + .5) -> x disp x
5 is an integer; it is a whole number without a decimal or fraction.
1550 is an integer and cannot be expressed as a mixed number. As a decimal, it is 1550 - exactly as in the question.
Assuming the number is represented by a decimal integer, initialise a counter to zero, then repeatedly divide the number by 10 and until the number is zero. After each division, examine the remainder. Each time the remainder is zero, increment the counter. If the number is represented by a decimal float, repeatedly multiply by 10 until the value is an integer, then perform the previous algorithm.
write a c++ program to convert binary number to decimal number by using while statement
public class Dataconversion { public static void main(String[] args) { System.out.println("Data types conversion example!"); int in = 44; System.out.println("Integer: " + in); //integer to binary String by = Integer.toBinaryString(in); System.out.println("Byte: " + by); //integer to hexadecimal String hex = Integer.toHexString(in); System.out.println("Hexa decimal: " + hex); //integer to octal String oct = Integer.toOctalString(in); System.out.println("Octal: " + oct); } }
Reference:cprogramming-bd.com/c_page2.aspx# reverse number
You first write the integer part, then write the decimal point and then the fractional part in decimal form.
pongada punda vayanungala ..................
The reason you're having so much difficulty writing it is that there's no such thing as an "interger". An "integer" is a whole number. Any time you write a number that doesn't have any fraction or decimal in it, you have written an integer.
Since 19 is an integer, it does not require a decimal point.
x -=y;