Today, I came across quite strange problem. I needed to calculate string length of a number, so I came up with this solution
// say the number is 1000
(int)(log(1000)/log(10)) + 1
This is based on mathematical formula
log10x = lognx/logn10 (explained here)
But I found out, that in C,
(int)(log(1000)/log(10)) + 1
is NOT equal to
(int) log10(1000) + 1
but it should be.
I even tried the same thing in Java with this code
(int) (Math.log(1000) / Math.log(10)) + 1
(int) Math.log10(1000) + 1
but it behave the same wrong way.
The story continues. After executing this code
for (int i = 10; i < 10000000; i *= 10) {
System.out.println(((int) (Math.log10(i)) + 1) +
" " + ((int) (Math.log(i) / Math.log(10)) + 1));
}
I get
2 2
3 3
4 3 // here second method produces wrong result for 1000
5 5
6 6
7 6 // here again
So the bug seems to occur on every multiple of 1000.
I showed this to my C teacher, and he said that it might be caused by some type conversion error during log division, but he didn’t know why.
So my questions are
- Why isn’t
(int) (Math.log(1000) / Math.log(10)) + 1equal to(int) Math.log10(1000) + 1, while it should be, according to the math.
- Why is it wrong only for multiples of 1000?
edit: It is not rounding error, because
Math.floor(Math.log10(i)) + 1
Math.floor(Math.log(i) / Math.log(10)) + 1
produce same, wrong output
2 2
3 3
4 3
5 5
6 6
7 6
edit2: I have to round down, because I want to know the number of digits.
log10(999) + 1 = 3.9995654882259823
log10(1000) + 1 = 4.0
If I just round, I get same result (4), which is wrong for 999, because it has 3 digits.
You provided the code snippet
to illustrate your question. Just remove the casts to
intand run the loop again. You will receivewhich immediately answers your question. As tliff already argued, the casts cut off the decimals instead of rounding properly.
EDIT: You updated your question to use
floor(), but like castingfloor()will round down and therefore drop the decimals!