How Many Bytes Do Pointers Take Up? - Exchangetuts

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I am little bit confused about pointers and how many bytes they take up. In my textbook it first says that pointers on 16 bit systems take up 2 bytes, 32 bit systems 4 bytes, 64 bit system 8 bytes and so on. Then 10 lines after, it says that pointers take up that many bytes, that are needed to hold the addresses. Here are my questions :

1. So does this mean that if we are lets say on 64 bit system, address will need at most 8 bytes?
2. If we are on 16 bit system and pointers take 2 bytes, and address needs more the 2 bytes to be placed then what happens?

832 asked Dec 24 '13 16:12

GovernmentFX

People also ask

Is a pointer always 4 bytes?

The size of a pointer in C/C++ is not fixed. It depends upon different issues like Operating system, CPU architecture etc. Usually it depends upon the word size of underlying processor for example for a 32 bit computer the pointer size can be 4 bytes for a 64 bit computer the pointer size can be 8 bytes.

Is a pointer always 8 bytes in C?

As we already know, the size of the pointer in C is dependent only on the word size of a particular system. So, the size of a pointer to a pointer should have the usual values, that is, 2 bytes for a 16-bit machine, 4 bytes for a 32-bit machine, and 8 bytes for a 64-bit machine.

Why do pointers use 4 bytes?

Size of a pointer is fixed for a compiler. All pointer types take same number of bytes for a compiler. That is why we get 4 for both ptri and ptrc.

1 Answers

There is no fixed answer; it depends entirely on the architecture, the compiler implementation, and even the type of the pointer itself. Pointers to different types are not guaranteed to have the same size and/or representation.

For example, assume a word-addressed architecture, where the smallest addressable unit of storage is 16 bits wide (or wider). Each word can hold multiple char values; all other types take up a full word or more. On such an architecture, a char * and void * would need some extra bits to offset into the word compared to other pointer types.

Note also that a pointer type may be wider than the number of bits actually required to store an address. The original Macintosh ran on a Motorola 68000 CPU, which had a 32-bit word size, but only 24 bits on the address bus. Pointer types were 32 bits wide, leaving the upper 8 bits unused. Enterprising MacOS programmers took advantage of that to store some data to the uppermost byte of a pointer type, making the most of that precious 128 KB of RAM. Of course, Motorola eventually released a CPU with 32 address lines (the 68020), meaning all that code had to be rewritten.

On modern, commodity desktop and server hardware (read: x86), it's reasonably safe to assume that all pointer types are the same size as the native word size (32- or 64-bit), and that all pointer types have the same size and representation. Just be aware that this doesn't have to be true.

147 answered Sep 22 '22 08:09

John Bode

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