两个正整数乘以long的结果为负值

Question

I have code like this:

int a = 629339;
int b = 4096;
long res = a*b;

The result is -1717194752 but if I add one manual cast to long long res = ((long)a)*b; or long res = (long) a*b; the result is correct 2577772544 Who can explain how does it works.

Answer 1

You have to break the assignment statement into its parts to understand what is doing on:

long res = a*b;

Step 1 is to get the values of a and b .

Step 2 is to evaluate a * b . Since a and b are both int s, this is an int multiplication. So we multiply 629339 by 629339 which would be 2577772544 . Unfortunately, 2577772544 is larger than the largest possible Java int value... so the multiplication operation silently overflows ... and we get -1717194752 instead.

Step 3 we assign the value of the RHS to the LHS. Since the RHS is int and the LHS is float , the JLS says we perform a primitive widening conversion... which simply turns -1717194752 into a long with the same value. The widened value is then assigned to res .

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To get the answer that you are expecting, we have to force multiplication to be performed using long arithmetic. For example:

long res = ((long) a) * b;

In this case, we have a multiplication of a long by an int , and this is handled by widening the int to a long and performing a long multiply. This no longer overflows (because 2577772544 is well below the largest long value), so when we finally assign the value to res , it is the number that you were expecting.

Answer 2

a*b is an integer, not a long.

Because it's only an integer, it has already wrapped around the 32-bit limit.
Casting this integer back to long will not magically recover that data.

Answer 3

long res = a*b;

a*b will be treated as integer unless you add 'l' at end (or) cast.

As per java tutorial

The int data type is a 32-bit signed two's complement integer. It has a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive). For integral values, this data type is generally the default choice unless there is a reason (like the above) to choose something else.