Java language is inaccurate for calculations? -

this question has answer here:

• is floating point math broken? 20 answers
• java float unexpectedly rounded 3 answers

i tried following for loop:

for(double = 0.0; i<=0.001; i+=0.0001)     system.out.println(i); 

and following output:

0.0
1.0e-4
2.0e-4
3.0000000000000003e-4
4.0e-4
5.0e-4
6.000000000000001e-4
7.000000000000001e-4
8.000000000000001e-4
9.000000000000002e-4

my questions are:

• how these .000000000000001s come?
• will these numbers come, or there problem in code?
• do these errors occur in java language, or other programming languages too?
• is double suitable for loops?

sadly, not numbers can represented in floating point:

for example, decimal number 0.1 not representable in binary floating-point of finite precision; exact binary representation have "1100" sequence continuing endlessly:

e = −4; s = 1100110011001100110011001100110011..., where, previously, s significand , e exponent.

when rounded 24 bits becomes

e = −4; s = 110011001100110011001101, 0.100000001490116119384765625 in decimal.