Why is the product of two positive integers a negative integer?

Question

This semester i took system proramming course. Why 50000*50000 will be negative? I try to understand logic of this. Here is the screenshot of the slide

slide image

Answer 1

32-bit signed integers are stored by using bits 0-30 as the number and bit 31 indicating the sign of the number.

This means that the maximum value that can be represented is 2,147,483,647 (all bits from 0-30 are set, bit 31 is 0 indicating a positive number).

The product of 50,000 and 50,000 is 25,000,000,000 is greater than this number and you have what is called an overflow. This means that data has "overflowed" from its expected bounds (the bottom 31 bits) into the sign bit).

You now have bit 31 set, indicating that this is a negative number. To figure out a negative number from its binary representation, you take the ones' complement (flip all the bits), add one and then throw a negative sign in front of it.

Be careful when you take the ones' complement that you limit yourself to a 32-bit range... you shouldn't be including bits higher than bit 31.

Check out signed number representations for more information.

Answer 2

Sample Program Pseudo Code

Print --> ("Size of int: " + (Integer.SIZE/8) + " bytes.");
int a=50000;
int b=50000;
Print --> (" Product of a and b " + a*b);   

Output :
Size of int: 4 bytes. 
Product of a and b:-1794967296

Analysis : 4 bytes= 4*8= 32bits.

Since signed int can hold negative values, one-bit is used for sign (- or +), so bits available for numeric range=31. Number range = -(2^31) , 0 and (2^31-1) [one positive number is sacrificed for 0]

-2147483648, 0 and 2147483647

Maximum possible positive int = 2147483647 (greater than 1600000000, so 40000*40000 is fine) Actual Product 50000*50000=2500000000 (greater than 2147483647)

In practice many portable C programs assume that signed integer overflow wraps around reliably using two's complement arithmetic. Yet the C standard says that program behavior is undefined on overflow, and in a few cases C programs do not work on some modern implementations because their overflows do not wrap around as their authors expected. http://www.gnu.org/software/autoconf/manual/autoconf-2.62/html_node/Integer-Overflow.html

Answer 3

This is because in most programming languages, the integer data type has a fixed size.

That means that each integer value have a defined MIN and MAX value.

For example in C# MAX INT is 2147483647 and MIN is -2147483648 In PHP 32 bits it's 2147483647 and -2147483648 In PHP 64 bits it's 9223372036854775807 and -9223372036854775808

What happen when you try to go over that value? Simply the computer will make what's called an integer overflow and the value will loop back to the min value.

In other words, in C# 2147483647 + 1 = -2147483648 (assuming you use an integer datatype, not long or float). That exactly what happen with 50000 * 50000, it just goes over max value and loop from the next value.

The exact min and max values are dependent on the language used, the platform the code is built, the platform the code is run on and the static type of the value.

Hope it clears everything out for you!