为什么我们在 C 中指定 size_t 数组大小而不是仅使用整数？这样做的好处是什么？
Why do we specify size_t array size in C rather than just using integers? And what are the advantages to do so?
我开始学习 C，我的一个家庭作业要求我创建一个数组
size_t const N_MAX = 10; double v1[N_MAX];
我知道 size_t 代表 v1 数组长度，但我仍然不明白 size_t 类型在这里有什么贡献，而不是使用 int N_MAX?
size_t can store the maximum size of a theoretically possible object
of any type (including array).
size_t is commonly used for array indexing and loop counting. Programs
that use other types, such as unsigned int, for array indexing may
fail on, e.g. 64-bit systems when the index exceeds UINT_MAX or if it
relies on 32-bit modular arithmetic.
(source)
这意味着 size_t 保证能够在您编译的任何平台上保持任何 size/index 计数。
它通常是根据您编译的目标平台来定义的。例如Windows SDK中包含的intsafe.h header定义如下：
#ifdef _WIN64
typedef __int64 ptrdiff_t;
typedef unsigned __int64 size_t;
#else
typedef int ptrdiff_t;
typedef unsigned int size_t;
#endif
直接使用 size_t 意味着您不必担心在为不同的体系结构（例如 x86）重新编译时更改用于索引或保持 object 大小的数据类型与 x86-64 相比）。
编辑：正如@Eric Postpischil 在评论中提到的那样，C 标准的实际措辞与我从 cppreference 链接的解释不同。
查看 the standard 我们可以看到关于 size_t 的一些提及：
6.5.3.4 The sizeof and alignof operators
[...]
5 The value of the result of both operators is implementation-defined,
and its type (an unsigned integer type) is size_t, defined in
<stddef.h> (and other headers)
[...]
7.19 Common definitions <stddef.h>
[...]
size_t
which is the unsigned integer type of the result of the sizeof operator;
size_t 是 sizeof 运算符返回值的类型这一事实是我们可以假设“size_t 可以存储最大大小的原因理论上可能 object 任何类型”，但实际上在标准中没有任何地方强制要求。该标准还说明了以下关于 sizeof 的内容：
The sizeof operator yields the size (in bytes) of its operand, which
may be an expression or the parenthesized name of a type. The size is
determined from the type of the operand. The result is an integer. If
the type of the operand is a variable length array type, the operand
is evaluated; otherwise, the operand is not evaluated and the result
is an integer constant
我的解释（这可能是错误的）是你不能有一个 object 尺寸不适合 size_t因为在那种情况下你不能有一个有效的 sizeof 操作员。
编辑 2：
我将在这里引用 Eric Postpischil：
There are simply limits as to what we can do with computers, and
sometimes calculations go out of bounds. It is generally easy on
modern hardware to define size_t so that sizeof always works, and you
obviously cannot malloc a larger object, but you might construct one
with a static array, in some esoteric C implementation with 16-bit
size_t but 22-bit addresses.
我开始学习 C，我的一个家庭作业要求我创建一个数组
size_t const N_MAX = 10; double v1[N_MAX];
我知道 size_t 代表 v1 数组长度，但我仍然不明白 size_t 类型在这里有什么贡献，而不是使用 int N_MAX?
size_t can store the maximum size of a theoretically possible object of any type (including array).
size_t is commonly used for array indexing and loop counting. Programs that use other types, such as unsigned int, for array indexing may fail on, e.g. 64-bit systems when the index exceeds UINT_MAX or if it relies on 32-bit modular arithmetic.
(source)
这意味着 size_t 保证能够在您编译的任何平台上保持任何 size/index 计数。
它通常是根据您编译的目标平台来定义的。例如Windows SDK中包含的intsafe.h header定义如下：
#ifdef _WIN64
typedef __int64 ptrdiff_t;
typedef unsigned __int64 size_t;
#else
typedef int ptrdiff_t;
typedef unsigned int size_t;
#endif
直接使用 size_t 意味着您不必担心在为不同的体系结构（例如 x86）重新编译时更改用于索引或保持 object 大小的数据类型与 x86-64 相比）。
编辑：正如@Eric Postpischil 在评论中提到的那样，C 标准的实际措辞与我从 cppreference 链接的解释不同。
查看 the standard 我们可以看到关于 size_t 的一些提及：
6.5.3.4 The sizeof and alignof operators
[...]
5 The value of the result of both operators is implementation-defined, and its type (an unsigned integer type) is size_t, defined in <stddef.h> (and other headers)
[...]
7.19 Common definitions <stddef.h>
[...]
size_t which is the unsigned integer type of the result of the sizeof operator;
size_t 是 sizeof 运算符返回值的类型这一事实是我们可以假设“size_t 可以存储最大大小的原因理论上可能 object 任何类型”，但实际上在标准中没有任何地方强制要求。该标准还说明了以下关于 sizeof 的内容：
The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant
我的解释（这可能是错误的）是你不能有一个 object 尺寸不适合 size_t因为在那种情况下你不能有一个有效的 sizeof 操作员。
编辑 2：
我将在这里引用 Eric Postpischil：
There are simply limits as to what we can do with computers, and sometimes calculations go out of bounds. It is generally easy on modern hardware to define size_t so that sizeof always works, and you obviously cannot malloc a larger object, but you might construct one with a static array, in some esoteric C implementation with 16-bit size_t but 22-bit addresses.