C++ 中的 Uint32 和 unsigned int 有什么区别？
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What is the difference between an Uint32 and an unsigned int in C++?
提问by user1511417
Is it the same? If no: what is the difference? If yes then why do you need this type?
是一样的吗？如果不是：有什么区别？如果是，那么你为什么需要这种类型？
回答by Matteo Italia
uint32_t(or however pre-C++11 compilers call it) is guaranteed to be a 32-bit unsigned integer; unsigned intis whatever unsigned integer the compiler likes best to call unsigned int, as far as it meets the requirements of the standard (which demands for it a 0-65535 minimum range).
uint32_t（或者无论 C++11 之前的编译器如何称呼它）保证是 32 位无符号整数；unsigned int是编译器最喜欢调用的任何无符号整数unsigned int，只要它满足标准的要求（要求它的最小范围为 0-65535）。
Like int, unsigned inttypically is an integer that is fast to manipulate for the current architecture (normally it fits into a register), so it's to be used when a "normal", fast integer is required.
像int,unsigned int通常是一个整数，可以快速处理当前体系结构（通常它适合寄存器），因此在需要“正常”、快速整数时使用它。
uint32_t, instead, is used when you need an exact-width integer, e.g. to serialize to file, or when you require that exact range or you rely on unsigned overflow to happen exactly at 2^32-1.
uint32_t，相反，当你需要一个精确宽度的整数时使用，例如序列化到文件，或者当你需要那个精确范围或你依赖无符号溢出恰好发生在2^32-1。
For example, on a 16 bit-processor unsigned intwill typically be 16 bits wide, while uint32_twill haveto be 32 bits wide.
例如，在一个16位的处理器unsigned int将典型地是16个比特宽，而uint32_t将具有成为32个位宽。
Incidentally, as pointed out by @Marc Glisse, while unsigned intis always present, uint32_tis not mandatory - a particular compiler implementation may not provide it. This is mostly because not all platforms can easily provide such a type (typically DSPs with weird word sizes).
顺便说一句，正如@Marc Glisse所指出的，虽然unsigned int始终存在，但uint32_t不是强制性的 - 特定的编译器实现可能不提供它。这主要是因为并非所有平台都可以轻松提供这种类型（通常是具有奇怪字长的 DSP）。
