c++ - #define to double - different value? -

There are two different ways that I am defining the same value. I want it to be present as a 64-bit (double exact) float point number (aka double).

  #define THISVALUE -0.148759f double myDouble = -0.148759;   
 If I do the following operation  
  double try = THISVALUE;   
 And when I look at the value while debugging or printing it, I can see that it's  tryIt  at  -0.14875899255275726  < / P>. 
 I think a floating point is not correct, but it's just a crazy difference that actually throws my math. Specifying double in the straightforward code block gives me the value of  -0.14875 9, 00000000000  in the debugger - it should be exactly the same way.  
 Any thoughts on what's going on?   
 
 
 Because  -0.14875 9f  a  double  No, this is a  name . Therefore it is almost certainly different precision which is doing the difference.  
 Either you should give the same result in these two variations:  
  #define THISVALUE -0.148759 double myDouble = - 0.148759; // Both doubled #define THISVALUE -0.148759f double myDouble = -0.148759f; // Both float IEEE 754 single precision values ​​(usually used in  float ) have only 32 bits available, so double precision value (which is 64 ) compared to limited precision values According to  
, the  thick  figures for range and accuracy are as follows:  
 
  singles , category  With a ± 10  a ± 38   7 digit precision.  
 For  double , the category  ± 10  15-digit accuracy ± 308  .   
 
 And, on one hand, nowadays there is more to use macros, either  or < / Em> Objects can be preceded by  inline  suggestions and good compilers, later with  const int  (or  const double  your case Can be done without losing any information between compilation steps (things like name and type information).