Matlab vs C++ Double Precision

Question

I am porting some code from Matlab to C++.

In Matlab

format long
D = 0.689655172413793 (this is 1.0 / 1.45)
E = 2600 / D
// I get E          


        

Answer 1

If I understand what you are trying to achieve, using ceil function might help:

ans = ceil(ans); /* smallest integral value that is not less than ans. */
// now ans in C++ is also be 3970.

Here is usage reference.

Answer 2

You got confused by the different ways C++ and MATLAB are printing double values. MATLAB's format long only prints 15 significant digits while C++ prints 17 significant digits. Internally both use the same numbers: IEEE 754 64 bit floating point numbers. To reproduce the C++-behaviour in MATLAB, I defined a anonymous function disp17 which prints numbers with 17 significant digits:

>> disp17=@(x)(disp(num2str(x,17)))

disp17 = 

    @(x)(disp(num2str(x,17)))

>> 1.0 / 1.45

ans =

   0.689655172413793

>> disp17(1.0 / 1.45)
0.68965517241379315

You see the result in MATLAB and C++ is the same, they just print a different number of digits. If you now continue in both programming languages with the same constant, you get the same result.

>> D = 0.68965517241379315 %17 digits, enough to represent a double.

D =

   0.689655172413793

>> ans = 2600 / D %Result looks wrong

ans =

     3.770000000000000e+03

>> disp17(2600 / D) %But displaying 17 digits it is the same.
3769.9999999999995

The background for printing 17 or 15 digits:

• A double requires 17 significant digits to be stored without precision loss, which is what C prints.
• For up to 15 digits any number can be converted from string to double to string and results back in the original number, which is what MATLAB does.