java – 理解try catch最后返回它返回的值和值

public static void main(String[] args) {
    System.out.println(returnString());
}
private static String returnString(){
    try {
        System.out.println("Executing try");
        return "Return try value";
    } catch (Exception e){
        System.out.println("Executing Catch");
        return "Return catch value";
    } finally {
        System.out.println("Executing finally");
        return "Return finally value";
    }
}

这个输出是

Executing try
Executing finally
Return finally value

如果我改变我的finally块而不返回任何类似的东西

public static void main(String[] args) {
    System.out.println(returnString());
}
private static String returnString(){
    try {
        System.out.println("Executing try");
        return "Return try value";
    } catch (Exception e){
        System.out.println("Executing Catch");
        return "Return catch value";
    } finally {
        System.out.println("Executing finally");
    }
}

然后输出是

Executing try
Executing finally
Return try value

现在我明白了,除非我们调用system.exit(0),否则最终总会被执行;调用或JVM崩溃.
我无法理解的是为什么返回值已经改变了？我仍然希望它返回try块的值.任何人都可以解释为什么最终值被考虑而不是try块的返回值？

请不要回答,因为即使在try块中有返回,最终也会执行…或者最后只有在有system.exit(0)时才执行;调用或JVM崩溃.据我所知.

编辑:(根据Dirk的Dirk评论)

public static void main(String[] args) {
    System.out.println(returnString());
}
private static String returnString(){
    try {
        System.out.println("Executing try");
        return printString("Return try value");
    } catch (Exception e){
        System.out.println("Executing Catch");
        return printString("Return catch value");
    } finally {
        System.out.println("Executing finally");
        return printString("Return finally value");
    }
}

private static String printString(String str){
    System.out.println(str);
    return str;
}

输出：

Executing try
Return try value
Executing finally
Return finally value
Return finally value