Digits in float and double variable

Question

Why is the number of digits stored by both float as well as double variable same, isn't double supposed to store more digits than float

#include <iostream>
using namespace std;
int main(){
double a = 3.141528579238;
float b = 3.141528579238; 
cout << a << " " << b;
return 0;
}

I'm getting this as my output

3.14153 3.14153

The number of digits stored is not the same as the number of digits displayed. — aschepler, Jun 26 '16 at 19:39
http://stackoverflow.com/questions/588004/is-floating-point-math-broken — πάντα ῥεῖ, Jun 26 '16 at 19:42
@OP: For you to make any observations about what is stored, you need to inspect the numbers in the debugger / memory, not use output statements to display the number. — PaulMcKenzie, Jun 26 '16 at 19:42
doubles and floats are storing _binary_ digits rather than decimal ones. And yes, double have more of those. — Mr Lister, Jun 26 '16 at 19:43
Also, any reason why you are using this particular number, so close to π? — Mr Lister, Jun 26 '16 at 19:46
No particular reason actually..just started with 3.1415 and just typed random numbers after that — von_oldmann, Jun 27 '16 at 03:53

score 1 · Accepted Answer · answered Jun 26 '16 at 19:46

The problem that you are facing is because first you should know that float has 4 bytes and double has 8 bytes and 3.141528579238 maybe can be stored in both and if you want to show more digits i suggest first including: #include <iomanip> then you can use the following to print any amount of digits after the "." : cout << std::setprecision(7) << myFloat; this will display 3.1415285.

score 0 · Answer 2 · answered Jun 26 '16 at 19:43

It isn't about digits it's about bits!

floats are 32bit and doubles are 64bit. A floating point number can be expressed like this: a*2^x where a < 1 and a => 0. a and x will be stored in either 32- or 64-bit memory.

The reason why your program is printing the same thing is because it just doesn't print your whole number.

Digits in float and double variable

2 Answers2