实际上没有这样的事情。通常，std::size_t表示当前体系结构上的无符号本机整数大小。即16位、32位或64位，但并不总是如此，就像这个答案的评论中指出的那样。

至于所有其他内置类型，它实际上取决于编译器。以下是摘自最新c++标准的当前工作草案的两段摘录:

There are five standard signed integer types : signed char, short int, int, long int, and long long int. In this list, each type provides at least as much storage as those preceding it in the list. For each of the standard signed integer types, there exists a corresponding (but different) standard unsigned integer type: unsigned char, unsigned short int, unsigned int, unsigned long int, and unsigned long long int, each of which occupies the same amount of storage and has the same alignment requirements.

如果您愿意，您可以静态(编译时)断言这些基本类型的sizeof。如果假设的大小发生变化，它会提醒人们考虑移植您的代码。

1)文章“64位程序开发中被遗忘的问题”中的表N1

2)“数据模型”

如前所述，大小应该反映当前的体系结构。如果你想知道当前编译器是如何处理的，你可以在limits.h中取一个峰值。

有标准。

C90标准要求

sizeof(short) <= sizeof(int) <= sizeof(long)

C99标准要求

sizeof(short) <= sizeof(int) <= sizeof(long) <= sizeof(long long)

这是C99的规格。第22页详细介绍了不同整型的大小。

下面是Windows平台的int类型大小(位):

Type           C99 Minimum     Windows 32bit
char           8               8
short          16              16
int            16              32
long           32              32
long long      64              64

如果你关心可移植性，或者你想要类型的名称反映大小，你可以查看头文件<inttypes.h>，其中有以下宏:

int8_t
int16_t
int32_t
int64_t

Int8_t保证为8位，int16_t保证为16位，以此类推。

如果你需要固定大小的类型，使用类型像uint32_t(32位无符号整数)定义在stdint.h。它们在C99中指定。

c++标准是这样说的:

3.9.1,§2:

There are five signed integer types : "signed char", "short int", "int", "long int", and "long long int". In this list, each type provides at least as much storage as those preceding it in the list. Plain ints have the natural size suggested by the architecture of the execution environment (44); the other signed integer types are provided to meet special needs. (44) that is, large enough to contain any value in the range of INT_MIN and INT_MAX, as defined in the header <climits>.

结论:这取决于您使用的是哪种体系结构。其他任何假设都是错误的。