Why does the 32-bit system support 4G memory maximum?? I figured it out for myself.

Today suddenly enlightened, figured out ...

Here is my abstract idea:

32-bit system This number refers to the number of bits that the hardware sends over at a time, one byte equals 8 bits, and one storage unit of memory is a byte, or 8 bits.

Can also think of this bit, that is, the memory address in the stack bits number, then the 32-bit means that the memory address of the stack is 32 11, that is:

You will find that the maximum memory address of the 32-bit system is: 2 of 32 parties-1, the minimum address is: 0-----Because computers are all starting from 0. 32 x 0 is also a memory address.

Then the maximum addressing space for a 32-bit system is: 0-(2 of 32 parties-1)

Abstract representation, since 0 is also an address then the 32-bit addressable space (how many addresses can be found) is: (2 of 32 square-1) + 1 (because 0 is also an address) = 2 of the 32-time address.

Speaking of which, let's figure out if 4G is equal to 2 of the 32-time square.

4G =?

Fully understood, the 32-bit system has a maximum addressable space of 4G. Oh, I share it.

Why does the 32-bit system support 4G memory maximum?? I figured it out for myself.