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 
 

 >> 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）：

• a>


• 浮点数
  精度 b $ b
• 如何确定
  中的错误是否是我的答案
  的错误或错误？


• COLON运算符
  如何工作？

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.

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?