Problem converting double to float.

Hello,

I do not understand why do I get 20120312 in v2 value. Thanks!

 ``12`` `````` double v1 = 20120313.0; float v2 = (float) v1;``````
float is less precise so you get small errors.
It errors by one, it's not small. I still do not understand how this could happened - I can see the value of v1 = 20120313 in debugger, it's a whole number with no decimals. Why float conversion rounding it down?
floating point numbers cannot store all values exactly. A float has about 6-7 digits precision while double has 15-16 digits precision.
Some numbers can be stored exactly in float such as 0.5 but some numbers cannot. So if I do:

 ``12345`` `````` double v3 = 0.5; // Can be represented exactly. float v4 = (float) v3; double v5 = 0.4; // Cannot be represented exactly. float v6 = (float) v5;``````

The v3 and v4 values are both 0.5 exactly - that's what debugger shows.

On the other hand v5 is 3.99...7 and v6 is 4.0000006.

Back to v1 and v2, I see v1 in debugger is 20120313 exactly, no decimal numbers, so I assume it can be represented exactly - so it should be represented exactly both in double and float.

Am I missing something?
A floating point number consist of an exponent and a fraction part. 1.frac * 2^exp
1.19926411*2^24 == 20120313
The fractional part is actually stored in memory as binary 001100110000001011111001000001. This is 30 binary digits! double has a 52 bits fractional part so there is no problem.

The float on the other hand is smaller and has a 23 bits fractional part. We can't store a 30 bit number in only 23 bits so we have to make approximations. It turns out that 1.19926405*2^24 == 20120312 is the the closest we can get. The fractional part here is 001100110000001011111 (21 bits).
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Thank you for the effort doing all the calculations!
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