Method overload (overload)
- Although the functions are different, when the functions are "similar", the method overload mechanism can make similar methods as if they are using one method. [Java supports this mechanism but some languages do not, such as JavaScript]
- When the functions are similar, the method names can be the same.
1. When should I consider using method overloading?
- When the functions are similar, try to make the method names the same [but when the functions are different/not similar, make the method names as different as possible]
2. What conditions constitute method overloading after being met
- In the same class
- Same method name
- The parameter list is different:
1. The quantity is different:public static void m1(){} 与 public static void m1(int a){}
2. The order is different:public static void m2(int a,double b){} 与 public static void m2(double a,int b){}
3. The type is different:public static void m3(int x){} 与 public static void m3(double x){}
方法重复，编译错误。
public static void m4(int a,int b){
}
public static void m4(int b,int a){
}
3. Method overloading has nothing to do with what, and nothing
- Method overloading is related to method name + parameter list
- Method overloading has nothing to do with return value type
- Method overloading has nothing to do with modifier list
Method recursion
- What is recursion?
That is: the method itself calls itself.
a(){
a();
}
- Recursion consumes a lot of stack memory. Try not to use recursive algorithms when they are not in use.
- Such an error occurs when the following program is running [not an exception, but an error error]:
java.lang.StackOverflowError
Stack memory overflow error The
error cannot be recovered later, there is only one result. That is, the JVM stops working. - Recursion must have an end condition. Without an end condition, a stack memory overflow error will occur.
- Even if the recursion has an end condition, even if the end condition is correct, a stack memory overflow error may occur because the recursion is too deep.
note:
- Recursion can not be used as much as possible.
- But in some cases the realization of this function must rely on recursion.
example
递归计算前n个数的和：
public class RecursionTest01{
publci static void main(String[] args){
int n=4;
int reValue=sum(n);
System.out.println(reValue);
}
public static int sum(int n){
if(n==1){
return 1;
}
return n+sum(n-1);
}
}
递归计算n的阶乘：
public class RecursionTest02{
publci static void main(String[] args){
int n=5;
int reValue=method(n);
System.out.println(reValue);
}
public static int method(int n){
if(n==1){
return 1;
}
return n*method(n-1);
}
}