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The Cray X1 uses byte addressing with 64-bit addresses. It does not directly support memory accesses smaller than 64 bits, and such accesses must be emulated in software. The C compiler for the X1 was the first Cray compiler to support emulating 16-bit accesses. [1] The DEC Alpha uses byte addressing with 64-bit addresses. Early Alpha ...
For example, an 8-bit-byte-addressable machine with a 20-bit address bus (e.g. Intel 8086) can address 2 20 (1,048,576) memory locations, or one MiB of memory, while a 32-bit bus (e.g. Intel 80386) addresses 2 32 (4,294,967,296) locations, or a 4 GiB address space. In contrast, a 36-bit word-addressable machine with an 18-bit address bus ...
An eight-bit processor like the Intel 8008 addresses eight bits, but as this is the full width of the accumulator and other registers, this could be considered either byte-addressable or word-addressable. 32-bit x86 processors, which address memory in 8-bit units but have 32-bit general-purpose registers and can operate on 32-bit items with a ...
The effective 20-bit address space of real mode limits the addressable memory to 2 20 bytes, or 1,048,576 bytes (1 MB). This derived directly from the hardware design of the Intel 8086 (and, subsequently, the closely related 8088), which had exactly 20 address pins. (Both were packaged in 40-pin DIP packages; even with only 20 address lines ...
For example, the PDP-10 byte pointer contained the size of the byte in bits (allowing different-sized bytes to be accessed), the bit position of the byte within the word, and the word address of the data. Instructions could automatically adjust the pointer to the next byte on, for example, load and deposit (store) operations.
The Intel 80286 CPU used a 24-bit addressing scheme. Each memory location was byte-addressable. This results in a total addressable space of 2 24 × 1 byte = 16,777,216 bytes or 16 megabytes. The 286 and later could also function in real mode, which imposed the addressing limits of the 8086 processor. The 286 had support for virtual memory.
If you’re stuck on today’s Wordle answer, we’re here to help—but beware of spoilers for Wordle 1250 ahead. Let's start with a few hints.
The bitwise XOR (exclusive or) performs an exclusive disjunction, which is equivalent to adding two bits and discarding the carry. The result is zero only when we have two zeroes or two ones. [3] XOR can be used to toggle the bits between 1 and 0. Thus i = i ^ 1 when used in a loop toggles its values between 1 and 0. [4]