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The boss said: whoever wants to use double to define the amount of goods, just pack up and go
2022-07-04 01:49:00 【Tile grabbing engineering lion】
Let's look at the phenomenon first
Involving such as float perhaps double The processing of these two floating-point data , There will always be some Strange phenomenon , I wonder if you have noticed , Give me a few common chestnuts ：
Typical phenomena （ One ）： Conditional judgment exceeds expectations
System.out.println( 1f == 0.9999999f ); // Print ：false
System.out.println( 1f == 0.99999999f ); // Print ：true what ？
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Typical phenomena （ Two ）： Data conversion exceeded expectations
float f = 1.1f;
double d = (double) f;
System.out.println(f); // Print ：1.1
System.out.println(d); // Print ：1.100000023841858 what ？
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Typical phenomena （ 3、 ... and ）： The basic operation is better than expected
System.out.println( 0.2 + 0.7 );
// Print ：0.8999999999999999 what ？
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Typical phenomena （ Four ）： The data increased more than expected
float f1 = 8455263f;
for (int i = 0; i < 10; i++) {
System.out.println(f1);
f1++;
}
// Print ：8455263.0
// Print ：8455264.0
// Print ：8455265.0
// Print ：8455266.0
// Print ：8455267.0
// Print ：8455268.0
// Print ：8455269.0
// Print ：8455270.0
// Print ：8455271.0
// Print ：8455272.0
float f2 = 84552631f;
for (int i = 0; i < 10; i++) {
System.out.println(f2);
f2++;
}
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
// Print ：8.4552632E7 what ？ No +1 Did you? ？
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You see , These simple scenarios are difficult to meet our needs , So use floating point numbers （ Include double and float） There are many problems to deal with Obscure pit Waiting for us ！
No wonder the technical director made harsh remarks ： Who dares to deal with things like The amount of goods 、 Order transaction 、 as well as Monetary calculation Time use floating point data （double/float）, Just let us go ！
What's the reason ？
Let's take the first typical phenomenon as an example to analyze ：
System.out.println( 1f == 0.99999999f );
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Compare directly with code 1 and 0.99999999, Actually print out true！
What does that mean ？ This shows that computers are not at all I can't tell These two numbers . Why is that ？
We might as well think briefly ：
We know that these two floating-point numbers are only the specific values seen by human eyes , Is what we usually understand as a decimal number , But the bottom of the computer does not calculate according to the decimal system , Anyone who has learned the basic group counting principle knows , The bottom of the computer is ultimately based on things like 010100100100110011011 such 0、1 Binary to complete .
So in order to understand the actual situation , We should put these two decimal floating-point numbers Convert to binary space Take a look at .
Decimal floating point number to binary How to turn 、 How to calculate , I think this should belong to basic computer base conversion common sense , stay 《 The principle of computer organization 》 I must have learned in similar classes , I won't repeat it here , Give the result directly （ Convert it to IEEE 754 Single precision 32-bit, It's just float The precision corresponding to the type ）
1.0（ Decimal system ）
↓
00111111 10000000 00000000 00000000（ Binary system ）
↓
0x3F800000（ Hexadecimal ）
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0.99999999（ Decimal system ）
↓
00111111 10000000 00000000 00000000（ Binary system ）
↓
0x3F800000（ Hexadecimal ）
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Sure enough , The underlying binary representation of these two decimal floating-point numbers is the same , It's no wonder that == The judgment result of returns true！
however 1f == 0.9999999f The result returned is expected , Print false, We also convert them to binary mode to see what happens ：
1.0（ Decimal system ）
↓
00111111 10000000 00000000 00000000（ Binary system ）
↓
0x3F800000（ Hexadecimal ）
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0.9999999（ Decimal system ）
↓
00111111 01111111 11111111 11111110（ Binary system ）
↓
0x3F7FFFFE（ Hexadecimal ）
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Oh , Obviously , The binary representation of the two is really different , This is a logical result .
So why 0.99999999 The underlying binary representation of is ：00111111 10000000 00000000 00000000 Well ？
This is not a floating point number 1.0 Binary representation of ？
This is about the precision of floating point numbers .
The precision of floating point numbers ！
Did you learn 《 The principle of computer organization 》 Everyone in this class should know , The storage of floating-point numbers in the computer follows IEEE 754 Floating point count standard , It can be expressed by scientific counting as ：
Just give ： Symbol （S）、 Step code part （E）、 Mantissa part （M） The information of these three dimensions , The representation of a floating point number is completely determined , therefore float and double Floating point numbers are stored in the following two memory structures ：
、 The symbolic part （S）
0- just 1- negative
2、 Step code part （E）（ Index part ）：
- about float Type floating point number , Index part 8 position , Consider positive and negative , Therefore, the index range that can be expressed is -127 ~ 128
- about double Type floating point number , Index part 11 position , Consider positive and negative , Therefore, the index range that can be expressed is -1023 ~ 1024
3、 Mantissa part （M）：
The precision of a floating-point number is determined by the number of bits of the mantissa ：
- about float Type floating point number , Mantissa part 23 position , Conversion to decimal is 2^23=8388608, So the decimal precision is only 6 ~ 7 position ;
- about double Type floating point number , Mantissa part 52 position , Conversion to decimal is 2^52 = 4503599627370496, So the decimal precision is only 15 ~ 16 position
So for the above values 0.99999999f, It's obviously more than float Precision range of type floating-point data , Problems are inevitable .
How to solve the problem of accuracy
So if it involves The amount of goods 、 Transaction value 、 Monetary calculation What should we do in such a scene that requires high accuracy ？
Method 1 ： Use string or array to solve multi bit number problem
All the partners who have brushed the algorithm problems should know , Using strings or arrays to represent large numbers is a typical problem-solving idea .
For example, classic interview questions ： Write the addition of two arbitrary large numbers 、 Subtraction 、 Multiplication and so on .
At this time, we can use strings or arrays to represent such large numbers , Then the specific calculation process is simulated manually according to the rules of four operations , We also need to consider various factors such as ： carry 、 Borrow position 、 Symbol And so on , It's really complicated , This article does not elaborate .
Method 2 ：Java The large number of categories is a good thing
JDK We have already considered the calculation accuracy of floating-point numbers , Therefore, it provides a special for high-precision numerical calculation Class of large numbers To facilitate our use .
In the foreword 《 To be honest with you , I recently talked to Java The source code is on the bar 》 Said in ,Java A large number of classes are located in java.math It's a bag ：
You can see , frequently-used BigInteger and BigDecimal It is a sharp tool for dealing with high-precision numerical calculation .
BigDecimal num3 = new BigDecimal( Double.toString( 1.0f ) );
BigDecimal num4 = new BigDecimal( Double.toString( 0.99999999f ) );
System.out.println( num3 == num4 ); // Print false
BigDecimal num1 = new BigDecimal( Double.toString( 0.2 ) );
BigDecimal num2 = new BigDecimal( Double.toString( 0.7 ) );
// Add
System.out.println( num1.add( num2 ) ); // Print ：0.9
// reduce
System.out.println( num2.subtract( num1 ) ); // Print ：0.5
// ride
System.out.println( num1.multiply( num2 ) ); // Print ：0.14
// except
System.out.println( num2.divide( num1 ) ); // Print ：3.5
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Yes, of course , image BigInteger and BigDecimal The operation efficiency of this large number class is certainly not as efficient as that of the native type , The price is still relatively expensive , The selection needs to be evaluated according to the actual scene .
by the way , Let's talk about it here , I'm currently on the job Java Development , If you are learning now Java, understand Java, Eager to be a qualified Java Development Engineer , At the beginning of learning Java The process of the lack of basic introductory video tutorial , You can pay attention to me ：01. obtain . I have the latest Java Basic complete video tutorial .
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