没有上限或下限的C ++舍入数字

Question

I'm attempting the following question from DS Malik's C++ Programming: From Problem Analysis to Program Design

Chapter 2, Question 7

Write a program that prompts the user to input a decimal number and outputs the number rounded to the nearest integer.

My program so far

 #include <iostream>

 using namespace std:cout;
 using namespace std:cin;

 int main()
  {
    double decimalNum;
    cout << "Please enter a decimal number: ";
    cin >> decimalNum;
    return 0;
  }

Is it possible to round the number to an integer without using floor or ceiling? Please note that I just started C++ programming a few days ago so nothing too overly complex please.

Answer 1

As pointed out in the comments, add 0.5 and typecast it to as int .

The reason behind adding 0.5 is to ensure that numbers are rounded off as opposed to the decimal part being truncated. So, 6.6 will be truncated to 6 , but rounded off to 7 .

Typecasting is the procedure of changing types. Some types are compatible with each other, but are different. For example, int , float , double , long all represent numbers. So when you write an int + float , the compiler first converts the int to a float automatically, and then does the operation. This is known as typecasting. You can do it explicitly, or the compiler does it many times.

So, to wrap it up, adding 0.5 will help you round off correctly, and typecasting will change the type.

Answer 2

You can try the following code -

#include <iostream>

using namespace std;

int main()
{
    double decimalNum;
    cout << "Please enter a decimal number: ";
    cin >> decimalNum;
    int roundNum = decimalNum + 0.5;
    cout << "The nearest rounded number : " << roundNum << endl;
    return 0;
}

Answer 3

Adding 0.5 then calling modf may be your best bet - the integral part is returned as a double so you still get correct results for numbers outside the range of even a 64-bit integral type. There are a lot of other options also available in the cmath header .