How to see pointer address in java

44964869

It’s impossible to compare with main to be true, and it should always be false.

阿迪雅

== compares the address of the object. do not you know?
-----------------------------------
SB, please read the code clearly and talk about it:

if (str== ???) //Here I want to make a == comparison with String str in the main function

舍不得

First of all, you should be clear that String in Java is immutable. That is to say, when you assign "aaa" to str in main, and then assign "bbb" to str in changeStr, the virtual machine implicitly creates a String and connects your str reference to the new The String. The original "aaa" has been discarded, and the JVM is responsible for recycling and destroying it.

Then, if you want to compare the addresses of two objects, just use == and that's it. a==b, true if the addresses are equal, it's that simple. As for how to look at the pointer address, I don't know if toString() comes out, but it is completely unnecessary in Java. Abandon the pointer, or you might as well use C++.

珈蓝易拉罐

The address is no longer visible.
To determine whether two reference type variables are referencing the same object, use "==".

珈蓝易拉罐

To learn Java, you must learn to think about problems in the Java language. If you just want to use Java as a tool for expressing C++ ideas, you may be very awkward and make yourself very confused.

atepie0011

== compares the address of the object. do not you know?
-----------------------------------
SB, please read the code clearly and talk about it:

if (str== ???) //Here I want to compare with String str in the main function ==
================================================= ===========
Host, you don’t understand basic programming. You pass the compared object as a parameter, how do you compare? Even in C, the same is true, unless you use global variables, why don't you get the hang of it? Just turn it a little bit.
public class StringTest {
public static void changeStr(String str){
String str2=str;
String str="bbb";
if (str==str2){
    System.out.println("==");
}
}
public static void main(String[] args) {
String str="aaa";
changeStr(str);
System.out.println(str);
}
}
Also, the concept of references in Java and pointers in C language are equivalent. It seems that you are just learning java. Many things need to be used to it. Although the reference in java is equivalent to the C language pointer, it is an incalculable pointer and can only be compared for equality and inequality, and cannot be added or subtracted. In addition, the referenced physical memory address cannot be obtained, although it has an actual image in the memory.

thiefgroup

>>SB, please read the code clearly before speaking:
>>if (str== ???) //Here I want to make a == comparison with String str in the main function

Please use your brain to think clearly before speaking.
Don't always reveal your ignorance of two languages ​​in SB way.

阿迪雅

In addition, the referenced physical memory address cannot be obtained, although it has an actual image in the memory.
---------------
That’s enough. I don’t need his physical memory. In fact, C is also the virtual memory mapping of windows.
I just want to see his mapped address in memory
public static void changeStr(String str){
String str2=str;
String str="bbb";
if (str==str2){

I've thought about this a long time ago, but I can't always add a temporary one like this. It's awkward.

珈蓝易拉罐

If you want to say this, you have to say "C language does have its convenience."
Under normal circumstances, the hashCode method of the Object class in Java can guarantee that different objects return different values, so it can be used as an ID-like thing, but that is only the practice of general language implementers, not the Java language itself Hard requirements.
Moreover, Object.hashCode can be overloaded, so even if that is done in Object, it is still not necessarily in derived classes.

珈蓝易拉罐

In addition, when debugging a Java program in Eclipse, you can observe the object ID from the Variables window.
(I use Sun's own JDK, and I don’t know if other vendors’ implementations will actually have an ID for each object)