Given the multiplication 65535 x 32768, will the result cause overflow on a 32-bit system? Will the result need more than 32 bits to be stored in?

65535 x 32768 = 214 7450880

which is in the range of 32 bit unsigned integers, so there will be no overflow and we wont be in a need of more bits than 32 bit.

Explanation:

An overflow error indicates that software attempted to write down data beyond the bounds of memory.

Each program features a section of memory allocated for a stack. The stack is employed to store internal data for the program and is extremely fast and keep track of return addressing. A program may jump to a neighborhood that reads some data from the disk drive, then it returns from that routine to continue processing the info. The stack keeps track of the originating address, and therefore the program uses that for the return. Kind of like leaving breadcrumbs to seek out your way back. The stack features a limited amount of space for storing. If software attempts to access a neighborhood of the stack beyond its limits, an overflow error occurs.

• signed 32-bit integers support the range [-2147483648,2147483647]

• unsigned 32-bit integers support the range [0,4294967295]

And if you go outside this range, even temporarily, you would like to be very careful. Most cases handle the overflow technically and provide you "wraparound" or "modulo" behavior where carry bits outside the range just disappear and you're left with the low-order bits like the precise result.

In computer architecture of 32 bit, memory addresses, or other data units are those who are 32 bits (4 octets) wide. Also, 32-bit CPU and ALU architectures are those who have supported registers, address buses, or data buses of that size. 32-bit microcomputers are computers during which 32-bit microprocessors are the norm.