int to long assignment

问题:

I have been trying this int and long conversion where I have tried assigning an int variable to a long variable. The code is as follows:

public static void main(String []args){
    int i = 1024;
    long j = i;
    long k = i*i*i*i;
    System.out.println("j= " +j+ ", k= " +k);
}

So, while printing j, it simply gave an output of 1024. But while printing k, it showed an overflow (k=0). I sorted out this problem by using this technique, where I have explicitly casted each i to long. i.e.

public static void main(String []args){
    int i = 1024;
    long j = i;
    long k = ((long)i)*((long)i)*((long)i)*((long)i);
    System.out.println("j= " +j+ ", k= " +k);
}

Now, it showed correct values of both j and k. So, I wanted to know why is this needed to cast int in the second case but not in the first one. k, being a long can retain this value after this heavy assignment. But, why it is not been correctly assigned?

回答1:

The reason is that your line

long k = i*i*i*i;

can be thought of as

long k = ((i*i)*i)*i;

or...

int k1 = i*i;
int k2 = k1*i;
int k3 = k2*i;
long k = k3;

So when any of kn overflows, you get the error.

When you do your casts you're obviously circumventing this problem by always multiplying longs together.

Of course the simplest modification to your initial program is to define i as a long straight away, instead of an int.

long i = 1024L;

回答2:

This problem is based on the fact that different types use different amount of memory. int and long are both Integers, the difference is that int is 4 bytes, while long is 8 bytes.

Lets modify your code a bit:

public class Test {


  public static void main(String []args){
        int i = 1024;
        long j = i;
        long k = i*i*i;

        System.out.println("My multipliers are of type int:");
        System.out.println("j= " +j+ ", k= " +k);           
        System.out.println("Is k less than Integer.MAX_VALUE: "  + (k < Integer.MAX_VALUE? "YES" : "NO"));
        k = i*i*i*i;
        System.out.println("j= " +j+ ", k= " +k);

        //now multiplying with j instead of i
        System.out.println("\nNow Im working with a long type:");
        k = j*j*j;
        System.out.println("j= " +j+ ", k= " +k);           
        System.out.println("Is k less than Integer.MAX_VALUE: "  + (k < Integer.MAX_VALUE? "YES" : "NO"));
        k = j*j*j*j;
        System.out.println("j= " +j+ ", k= " +k);
        System.out.println("Is k less than Integer.MAX_VALUE: "  + (k < Integer.MAX_VALUE? "YES" : "NO"));
        k = j*j*j*j;

     }

}

The code above shows that when you have multiplied i 3 times the result is a value that is smaller than Integer.MAX_VALUE (which is 2147483647), and when youre multiplying it 4 times the result is 0 since there is no enough place in the poor 4 bytes of the multipliers. :) But you can see that when the multipliers (right side, j) are of type long, the correct value is printed, and the last printing shows that the statement k < Integer.MAX_VALUE is false, which is the reason for your zero. :)

回答3:

i*i*i*i is an all int multiplication so the resultant is integer only (Which is overflowing in your case). for a long resultant you only need to convert one of them to long so this is sufficient

long k = ((long)i)*i*i*i;

回答4:

In first the value of i is an int and the integer multipllication happening and i*i*i*i result is integer and hence the integer overflow.

Where as in second case you are explicitly telling/making the result as long.

回答5:

1099511627776 which is the result of multiplication is >Integer.MAX_VALUE as it's Integer Multiplication so after i*i*i value would be 1073741824 but after multiplication with 1024 it becomes out of range which causes integer overflow and k will be 0.

• You can cast one of the i to long
• Multiply with 1L (1L*i*i*i*i) But NOT i*i*i*i*1L
• You can also assign i*i*i to long and than multiply with i like this long k =(k=i*i*i)*i;