Digit limitation from decimal point in C++

Question

Hello,

I'm new in C++.I have double variable double a=0.1239857 I want to limit variable a from decimal point two digits.So a will be 0.12. I know C++ have functions that return largest or smallest integer that is greater or lower than a like ceil or floor.Is there a function that implement digit limitation of floating-point variable? Or How can i change precision of a variable?

Best regards...

Answer 1

What do you mean by you want to limit the variable ? The value or its formatting. For the value, you can use floor + division. Something like:

double a = 0.12123
double b;

b = floor(a * 100) / 100

Answer 2

This will result in two digits after the decimal place.

a = floor(a * 100.0) / 100.0;

Answer 3

Are you actually trying to round the number, or just change its displayed precision?

For the former (truncating the extra digits):

double scale = 0.01;  // i.e. round to nearest one-hundreth
value = (int)(value / scale) * scale;

or (rounding up/down as appropriate, per jheriko's answer)

double scale = 0.01;  // i.e. round to nearest one-hundreth
value = floor(value / scale + 0.5) * scale;

For the latter:

cout << setprecision(2) << value;

where the parameter to setprecision() is the maximum number of digits to show after the decimal point.

Answer 4

If you just want to output the value, you can do something like

printf("%.3f", a); // Output value with 3 digits after comma

If you want to convert the value itself, you can do:

a = (int)(a * 1000) / 1000.0f;

Note that both do no rounding, they just truncate the value.

Answer 5

Use a ios_base::precision for formatting i/o.

Answer 6

You can set the precision on a stream, e.g.

double d = 3.14579;
cout.precision(2);
cout << d << endl;

// Or use a manipulator

#include <iomanip>
cout << setprecision(2) << d << endl;

Note that when you send a double or float to a stream like this, it will automatically round for you (which can trip you up sometimes if you aren't aware of this).

Answer 7

An actual rounding solution would be x = floor(100*x + 0.5) / 100; assuming the value to be rounded is in a variable "x".

The x = floor(100*x) / 100; recommended by others here will actually truncate the number to 2dp instead.