关于浮点精度:为什么迭代次数不相等? [英] About floating point precision: why the iteration numbers are not equal?
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问题描述
i = 0;
x = 0.0;
h = 0.1;
,而x < 1.0
i = i + 1;
x = i * h;
disp([i,x]);
end
另外:
i = 0;
x = 0.0;
h = 0.1;
,而x < 1.0
i = i + 1;
x = x + h;
disp([i,x]);
end
我不明白为什么浮点加操作和
解决方案
比较以下输出:
>> fprintf('%0.20f \ n',0.1。*(1:10))
0.10000000000000001000
0.20000000000000001000
0.30000000000000004000
0.40000000000000002000
0.50000000000000000
0.60000000000000009000
0.70000000000000007000
0.80000000000000004000
0.90000000000000002000
1.00000000000000000000
>> fprintf('%0.20f\\\
',cumsum(repmat(0.1,1,10)))
0.10000000000000001000
0.20000000000000001000
0.30000000000000004000
0.40000000000000002000
0.50000000000000000 $
0.59999999999999998000
0.69999999999999996000
0.79999999999999993000
0.89999999999999991000
0.99999999999999989000
<比较使用MATLAB的COLON算子:
>> fprintf('%0.20f \ n',0.1:0.1:1)
0.10000000000000001000
0.20000000000000001000
0.30000000000000004000
0.40000000000000002000
0.50000000000000000000
0.59999999999999998000
0.69999999999999996000
0.80000000000000004000
0.90000000000000002000
1.00000000000000000000
如果您想要请参阅64位二进制表示法,使用:
>> format hex
>> [(0.1:0.1:1)'(0.1。*(1:10))'cumsum(repmat(0.1,10,1))]
3fb999999999999a 3fb999999999999a
3fc999999999999a 3fc999999999999a 3fc999999999999a
3fd3333333333334 3fd3333333333334 3fd3333333333334
3fd999999999999a 3fd999999999999a 3fd999999999999a
3fe0000000000000 3fe0000000000000 3fe0000000000000
3fe3333333333333 3fe3333333333334 3fe3333333333333
3fe6666666666666 3fe6666666666667 3fe6666666666666
3fe999999999999a 3fe999999999999a 3fe9999999999999
3feccccccccccccd 3feccccccccccccd 3feccccccccccccc
3ff0000000000000 3ff0000000000000 3fefffffffffffff
一些建议阅读(与MATLAB有关):
There are two similar matlab programs, one iterates 10 times while another iterates 11 times.
One:
i = 0;
x = 0.0;
h = 0.1;
while x < 1.0
i = i + 1;
x = i * h;
disp([i,x]);
end
Another:
i = 0;
x = 0.0;
h = 0.1;
while x < 1.0
i = i + 1;
x = x + h;
disp([i,x]);
end
I don't understand why there is difference between the floating point add operation and the multiple.
解决方案
Compare the output of the following:
>> fprintf('%0.20f\n', 0.1.*(1:10))
0.10000000000000001000
0.20000000000000001000
0.30000000000000004000
0.40000000000000002000
0.50000000000000000000
0.60000000000000009000
0.70000000000000007000
0.80000000000000004000
0.90000000000000002000
1.00000000000000000000
>> fprintf('%0.20f\n', cumsum(repmat(0.1,1,10)))
0.10000000000000001000
0.20000000000000001000
0.30000000000000004000
0.40000000000000002000
0.50000000000000000000
0.59999999999999998000
0.69999999999999996000
0.79999999999999993000
0.89999999999999991000
0.99999999999999989000
Also compare against using MATLAB's COLON operator:
>> fprintf('%0.20f\n', 0.1:0.1:1)
0.10000000000000001000
0.20000000000000001000
0.30000000000000004000
0.40000000000000002000
0.50000000000000000000
0.59999999999999998000
0.69999999999999996000
0.80000000000000004000
0.90000000000000002000
1.00000000000000000000
If you want to see the 64-bit binary representation, use:
>> format hex
>> [(0.1:0.1:1)' (0.1.*(1:10))' cumsum(repmat(0.1,10,1))]
3fb999999999999a 3fb999999999999a 3fb999999999999a
3fc999999999999a 3fc999999999999a 3fc999999999999a
3fd3333333333334 3fd3333333333334 3fd3333333333334
3fd999999999999a 3fd999999999999a 3fd999999999999a
3fe0000000000000 3fe0000000000000 3fe0000000000000
3fe3333333333333 3fe3333333333334 3fe3333333333333
3fe6666666666666 3fe6666666666667 3fe6666666666666
3fe999999999999a 3fe999999999999a 3fe9999999999999
3feccccccccccccd 3feccccccccccccd 3feccccccccccccc
3ff0000000000000 3ff0000000000000 3fefffffffffffff
Some suggested readings (MATLAB related):
- Cleve's Corner 1996 article
- A Glimpse into Floating-Point Accuracy
- How do I determine if the error in my answer is the result of round-off error or a bug?
- How does the COLON operator work?
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