java - Help: Analyze the following code in detail, I am confused.
阿神
阿神 2017-05-17 09:59:16
0
3
698

public class T1 {

public static void rename(String aa){
    
    aa="AA";
    
}

public static void main(String args[ ]){
    
    String aa="BB";
    
    rename(aa);

    System.out.println(aa);
}

}
My understanding:
aa in the rename method is in the stack memory, and aa in the main method is in the heap memory. Executing the rename method will not modify aa in the main method. Worth it, right?

阿神
阿神

闭关修行中......

reply all(3)
迷茫

Call by value is the most commonly used evaluation strategy: the formal parameters of a function are copies of the actual parameters passed when called. Modifying the value of the formal parameter does not affect the actual parameter.

When called by reference, the formal parameters of the function receive implicit references to the actual parameters, rather than copies. This means that if the values ​​of function parameters are modified, the actual parameters will also be modified. At the same time both point to the same value.

Java Core Technology (I) mentioned that Java is all passed by value. First of all, for basic types, functions cannot modify its value. For reference types, functions cannot modify it to point to another object. So it's all passed by value.

Ty80

No, the aa inside the function points to the new address, and the external aa still points to the address of "BB", so the result is BB

伊谢尔伦

The function does not pass the AA in the MAIN function when passing value, but makes a copy of the value of AA and then passes it in

Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!