JAVA Data type Overflow analysis __java

Package p20140207;

public class T20140207 {public

	static void Main (string[] args)  
    {  
		int test =integer.max_value;
		System.out.println ("no.1=" +test);
		System.out.println ("no.2=" + (test+1));
		System.out.println ("no.3=" + (test+1l));
    }  
      

Guess what the result would be?

The results are as follows: no.1= 2147483647

No.2=-2147483648

no.3= 2147483648

Package p20140207;

public class T20140207 {public

	static void Main (string[] args)  
    {  
		int test =java.lang.integer.max_value;
		System.out.println ("no.1=" +test);
		System.out.println ("no.2=" + (test+1));
		System.out.println ("no.3=" + (test+1l));
		System.out.println ("no.4=" + (Math.pow (2));//int can represent 2 of 32 times = 4,294,967,296 digits
		System.out.println ("no.5=" + ( Math.pow (2, 32)/2-1))//maximum number of representations
    }  
	//0~2147483647 represent 0~2147483647
	//2147483648~ 4294967295 respectively represents -2147483648~-1
	//So when the 2147483647+1 becomes the No. 2147483648 number, it turns into -2147483648
      
}.

PS: When the operation is to use 2 of the complement of the calculation, the details are not specifically listed.