在使用分段寻址模式的 16 位系统上，`size_t`、`uintptr_t`、`intptr_t` 和 `ptrdiff_t` 类型的实际大小是多少？

Question

Well, size_t , uintptr_t , intptr_t and ptrdiff_t type defined in C programming language really puzzles me.

I noticed that on 32-bit systems using flat memory mode , the real size of above four types is:

sizeof(size_t) = sizeof(uintptr_t) = sizeof(intptr_t) = sizeof(ptrdiff_t) = 4

And on 64-bit systems using flat memory mode , the real size of above four types is:

sizeof(size_t) = sizeof(uintptr_t) = sizeof(intptr_t) = sizeof(ptrdiff_t) = 8

But, I'm still curious about the real size of these four types on 16-bit systems using segmented addressing mode .

According to Windows 3.1x , Windows 3.1x is a series of 16-bit operating systems. But I can't find a suitable C language compiler to test and validate the real answer.

Answer 1

In real mode , size_t and ptrdiff_t are 16 bit types. The size of a pointer depends on whether it's a near, far, or huge pointer. Near pointers have 16 bits, the other 32 bits. intptr_t and uintptr_t were only defined much later in C99.

Answer 2

That, at least potentially, depends on which compiler you are using.

As a data point from the 16-bit era, the Borland Turbo C++ 3.0 manual has this to say:

The integer type of the result of sizeof is size_t , defined as unsigned int in stddef.h.

(emphasis in original.)

The size of unsigned int was 16 bits, independent of the memory model.

(None of the other types you asked about existed yet, or, if they did, were not supported by this particular compiler.)