Print whole non-decimal part of a double

Question

Let's say I have a double d = 123456789 (I know that the double is always representing an integer). I am trying to print it using C++ so that exactly 123456789 is printed. Unfortunately, this is not the default behavior:

double d = 123456789;
cout << d << "\n";

output: 1.23457e+08

I discovered the following trick/cheat:

double d = 123456789;
cout << (long) d << "\n";

output: 123456789

Is this the way to do it, or is there a cleaner way using some output manipulators?

Does this answer your question? [How to make C++ cout not use scientific notation](https://stackoverflow.com/questions/5212018/how-to-make-c-cout-not-use-scientific-notation) — lukehod, Jan 18 '20 at 16:11
TL;DR potential dupe: `std::cout << std::fixed << std::setprecision(0) << d;`. — Evg, Jan 18 '20 at 16:21

Zheng Qu · Accepted Answer · 2020-01-18T16:39:17.020

3

std::setprecision can be used here.

#include <iomanip>   // std::setprecision
#include <iostream>  // std::cout

int main() {
  double d = 123456789;
  std::cout << std::setprecision(5) << d << "\n";
  // 1.2346e+08
  std::cout << std::setprecision(15) << d << "\n";
  // 123456789
}

The maximum precision can be set with

std::setprecision(std::numeric_limits<long double>::digits10 + 1)

edited Jan 18 '20 at 16:39

answered Jan 18 '20 at 16:07

Zheng Qu

781
6
22

When `d > 1e15` or so, output still uses `e+`. `double` typically has a range up to `1e308`. – chux - Reinstate Monica Jan 18 '20 at 16:20

OLP · Answer 2 · 2020-01-18T16:22:21.067

Edit: I'll leave this here for the info about casting, but the right way is to use setprecision as suggested in the other answers.

Casting to long is fine as long as your values can fit in a long. You can upgrade to long long or unsigned long long if you are dealing with values larger than can be represented by long. (use the signed version to support negative values).

Keep in mind that double/float can represent very large values so eventually you might encounter values that don't fit even in unsigned long long.

You can check for overflow with something like

if (d <= (double)std::numeric_limits<long long>::max()
     && d >= (double)std::numeric_limits<long long>::min())
{
    cout << (unsigned long long) d;
}
else
{
    cout << "value d is very large: " << d;
}

score 1 · Answer 3 · answered Jan 18 '20 at 16:14

1

You can use manipulator and set the output format as shown below

double d = 123456789;
std::cout << std::setprecision(9) << std::noshowpoint;
std::cout << d << std::endl;

answered Jan 18 '20 at 16:14

Nagappa

194
2
13

Print whole non-decimal part of a double

3 Answers3