C++ int8_t 与 char ; 哪个是最好的？

int8_t vs char ; Which is the best one?

提问by sokid

My question may be confuse you, I know both are different types (signed charand char), however my company coding guidelines specifies to use int8_tinstead of char.

我的问题可能会迷惑你，我知道，这两个是不同类型的（signed char和char），但我公司编码规范指定使用int8_t代替char。

So, I want to know, why I have to use int8_tinstead of chartype. Is there any best practices to use int8_t?

所以，我想知道，为什么我必须使用int8_t而不是char类型。是否有任何最佳做法可供使用int8_t？

回答by Mats Petersson

The use of int8_tis perfectly good for some circumstances - specifically when the type is used for calculations where a signed 8-bit value is required. Calculations involving strictly sized data [e.g. defined by external requirements to be exactly 8 bit in the result] (I used pixel colour levels in a comment above, but that really would be uint8_t, as negative pixel colours usually don't exist - except perhaps in YUV type colourspace).

在int8_t某些情况下，使用是非常好的 - 特别是当该类型用于需要有符号 8 位值的计算时。涉及严格大小数据的计算[例如由外部要求定义为结果中的 8 位]（我在上面的评论中使用了像素颜色级别，但确实如此uint8_t，因为通常不存在负像素颜色 - 除了可能在YUV 类型色彩空间）。

The type int8_tshould NOT be used as a replacement of charin for strings. This can lead to compiler errors (or warnings, but we don't really want to have to deal with warnings from the compiler either). For example:

该类型int8_t不应用作char字符串的in替换。这可能会导致编译器错误（或警告，但我们也不想处理来自编译器的警告）。例如：

int8_t *x = "Hello, World!\n";

printf(x);

may well compile fine on compiler A, but give errors or warnings for mixing signed and unsigned char values on compiler B. Or if int8_tisn't even using a chartype. That's just like expecting

可能在编译器 A 上编译得很好，但在编译器 B 上混合有符号和无符号字符值时会给出错误或警告。或者int8_t甚至不使用char类型。这就像期待

int *ptr = "Foo";

to compile in a modern compiler...

在现代编译器中编译...

In other words, int8_tSHOULD be used instead of charif you are using 8-bit data for caclulation. It is incorrect to wholesale replace all charwith int8_t, as they are far from guaranteed to be the same.

换句话说，如果您使用 8 位数据进行计算，int8_t则应使用 SHOULD 而不是char。这是不正确的批发更换所有char有int8_t，因为他们远离保证是相同的。

If there is a need to use charfor string/text/etc, and for some reason charis too vague (it can be signed or unsigned, etc), then usign typedef char mychar;or something like that should be used. (It's probably possible to find a better name than mychar!)

如果需要使用char字符串/文本/等，并且由于某种原因char太模糊（它可以是有符号或无符号等），那么typedef char mychar;应该使用usign或类似的东西。（可能找到比 更好的名字mychar！）

Edit: I should point out that whether you agree with this or not, I think it would be rather foolish to simply walk up to whoever is in charge of this "principle" at the company, point at a post on SO and say "I think you're wrong". Try to understand what the motivation is. There may be more to it than meets the eye.

编辑：我应该指出，无论您是否同意这一点，我认为简单地走到公司中负责这项“原则”的人面前，指着 SO 上的帖子说“我认为你错了”。尝试了解动机是什么。它可能比我们看到的更多。

回答by Stephen Canon

They simply make different guarantees:

他们只是做出不同的保证：

charis guaranteed to exist, to be at least 8 bits wide, and to be able to represent either all integers between -127 and 127 inclusive (if signed) or between 0 and 255 (if unsigned).

char保证存在，至少为 8 位宽，并且能够表示 -127 到 127（如果有符号）或 0 到 255（如果无符号）之间的所有整数。

int8_tis not guaranteed to exist (and yes, there are platforms on which it doesn't), but if it exists it is guaranteed to an 8-bit twos-complement signed integer type with no padding bits; thus it is capable of representing all integers between -128 and 127, and nothing else.

int8_t不保证存在（是的，有些平台不存在），但如果存在，则保证为 8 位二进制补码有符号整数类型，没有填充位；因此它能够表示 -128 到 127 之间的所有整数，仅此而已。

When should you use which? When the guarantees made by the type line up with your requirements. It is worth noting, however, that large portions of the standard library require char *arguments, so avoiding charentirely seems short-sighted unless there's a deliberate decision being made to avoid usage of those library functions.

什么时候应该使用哪个？当类型所做的保证符合您的要求时。然而，值得注意的是，标准库的大部分内容都需要char *参数，因此char完全避免似乎是短视的，除非有意做出避免使用这些库函数的决定。

回答by Pete Becker

int8_tis onlyappropriate for code that requires a signed integer type that is exactly8 bits wide and should not compile if there is no such type. Such requirements are far more rare than the number of questions about int8_tand it's brethren indicates. Most requirements for sizes are that the type have at leasta particular number of bits. signed charworks just fine if you need at least 8 bits; int_least8_talso works.

int8_t是仅适合于该要求有符号整数类型，它是代码恰好宽8位，并且如果不存在这样的类型不应编译。这样的要求远比关于int8_t它的兄弟们所指出的问题数量少得多。对大小的大多数要求是该类型至少具有特定数量的位。signed char如果您需要至少 8 位，则工作正常；int_least8_t也有效。

回答by Sneftel

int8_tis specified by the C99 standard to be exactly eight bits wide, and fits in with the other C99 guaranteed-width types. You should use it in new code where you want an exactly 8-bit signed integer. (Take a look at int_least8_tand int_fast8_ttoo, though.)

int8_t由 C99 标准指定为正好 8 位宽，并且适合其他 C99 保证宽度类型。您应该在需要精确 8 位有符号整数的新代码中使用它。（不过int_least8_t，int_fast8_t也看看and 。）

charis still preferred as the element type for single-byte character strings, just as wchar_tshould be preferred as the element type for wide character strings.

char仍然首选作为单字节字符串的元素类型，就像wchar_t应该首选作为宽字符串的元素类型一样。