C/C++ (C++) How would I display a value rounded to the nearest integer?

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The discussion focuses on how to round a floating-point number to the nearest integer in C++. The initial code provided by the user does not achieve this because it uses an integer variable. To correctly round a number, it is suggested to use a floating-point variable instead. Several methods for rounding are discussed, including using `std::floor(num + 0.5)` for rounding to the nearest integer, `static_cast<int>(num)` for truncation, and `std::round(num)` for rounding in C++11 and later. The importance of avoiding direct comparisons with floating-point numbers is also mentioned. Overall, the conversation emphasizes the need for proper data types and methods to achieve accurate rounding in C++.
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Like for example

Prompting message reads.

"Please enter a positive value"

Example:

Please enter a positive value: 234.7
Rounded to the nearest integer the number is: 235


Please enter a positive value: 10.3
Rounded to the nearest integer the number is: 10


What I have

int num;

cout<< "Please enter a positive value";

cin>> num;

cout<< num << endl;




The problem is that doesn't round the number to the nearest interger so what would I do to round it?
 
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soul5 said:
Like for example

Prompting message reads.

"Please enter a positive value"

Example:

Please enter a positive value: 234.7
Rounded to the nearest integer the number is: 235


Please enter a positive value: 10.3
Rounded to the nearest integer the number is: 10


What I have

int num;

cout<< "Please enter a positive value";

cin>> num;

cout<< num << endl;




The problem is that doesn't round the number to the nearest interger so what would I do to round it?

(int) (3.5+0.5)
 
rootX said:
(int) (3.5+0.5)

Dude that's not it.
 
Last edited:
rootX gave a correct example of the calculation needed. You need to input as a float, not an int, then convert to an int by rounding as in rootX's suggestion.
 
It does help if num is a floating point variable, rather than an int.

Round to zero (truncation)
Code:
int rounded_num = static_cast<int>(num);

Round to +infinity
Code:
int rounded_num = std::floor(num + 0.5);

Round away from zero
Code:
int rounded_num = (num < 0.0)
    ? ((std::floor(num) == num - 0.5) ? std::floor(num) : std::floor(num + 0.5))
    : std::floor(num + 0.5);
Really shouldn't do direct == with floating points, but that's another subject.

Or in C++0x:
Code:
int rounded_num = std::round(num);

Round to even
Um, have fun...
 
KTC said:
It does help if num is a floating point variable, rather than an int.

Round to zero (truncation)
Code:
int rounded_num = static_cast<int>(num);

Round to +infinity
Code:
int rounded_num = std::floor(num + 0.5);

Round away from zero
Code:
int rounded_num = (num < 0.0)
    ? ((std::floor(num) == num - 0.5) ? std::floor(num) : std::floor(num + 0.5))
    : std::floor(num + 0.5);
Really shouldn't do direct == with floating points, but that's another subject.

Or in C++0x:
Code:
int rounded_num = std::round(num);

Round to even
Um, have fun...
very insteresting, but I think rootx's answer is enough and simple.
 
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