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java优化怎么样？(How does java optimisation? Example code)

 考虑这个小班：  
public class Example {
  public int plus(int a){
    int b;
    b = 1;
    return b+a;
  }
}
 
 javap说：  
public int plus(int);
  descriptor: (I)I
  flags: ACC_PUBLIC
  Code:
    stack=2, locals=3, args_size=2
     0: iconst_1
     1: istore_2
     2: iload_2
     3: iload_1
     4: iadd
     5: ireturn
  LineNumberTable:
    line 4: 0
    line 5: 2
 
 我不明白为什么我们需要第1和第2行？  
 我听说Java Runtime环境会计算函数调用并在超过一定数量的调用时编译函数。  
 这是否意味着Java编译器将所有优化都留给了字节码编译器？ 例如Java程序开始变慢并且随着时间变得更快？  
 请解释！ 

Consider this small class: 
public class Example {
  public int plus(int a){
    int b;
    b = 1;
    return b+a;
  }
}
 
javap says: 
public int plus(int);
  descriptor: (I)I
  flags: ACC_PUBLIC
  Code:
    stack=2, locals=3, args_size=2
     0: iconst_1
     1: istore_2
     2: iload_2
     3: iload_1
     4: iadd
     5: ireturn
  LineNumberTable:
    line 4: 0
    line 5: 2
 
I don't understand why we need line 1&2? 
I heard that the Java Runtime environment counts function calls and compiles functions if they exceed a certain number of calls. 
Does that mean that the Java compiler leaves all optimisation to the byte code compiler? E.g. Java programs start slow and become faster by time? 
Please explain!

原文：https://stackoverflow.com/questions/43119920

更新时间：2023-10-28 21:10

最满意答案

 在第一个代码示例中，  
int (*Parr)[10] = &arr;
 
 Parr是指向10个int的数组的指针。 它只能指向这样一个对象。 例如，  
int (*Parr)[10];
int a[10];
Parr = &a; // OK
int b[42];
Parr = &b; // ERROR, b is of the wrong type
 
 在第二个代码示例中，  
int *Parr = arr;
 
 Parr是指向int的指针，初始化为指向arr的第一个元素。 但它可以指向任何int 。  
int* Parr;
int a[10];
Parr = &a; // OK, a decays to int*
int b[42];
Parr = &b; // OK, b decays to int*
int c = 42;
Parr = &c; // OK, c is an int*

In the first code sample,  
int (*Parr)[10] = &arr;
 
Parr is a pointer to an array of 10 ints. It can only point to such an object. For example, 
int (*Parr)[10];
int a[10];
Parr = &a; // OK
int b[42];
Parr = &b; // ERROR, b is of the wrong type
 
In the second code sample,  
int *Parr = arr;
 
Parr is a pointer to int, initialized to point to the first element of arr. But it can point to any int. 
int* Parr;
int a[10];
Parr = &a; // OK, a decays to int*
int b[42];
Parr = &b; // OK, b decays to int*
int c = 42;
Parr = &c; // OK, c is an int*

java优化怎么样？(How does java optimisation? Example code)

最满意答案

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