最小的数据类型 - 我可以定义一位变量吗？ [重复]

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Is it possible to create a data type of length one bit in C 8 answers

是否可以在C 8答案中创建长度为一位的数据类型

I need only one bit to represent my data - 1 or 0. What is the best way to do so in C? The "normal" data types are too large.

我只需要一位来表示我的数据 - 1或0.在C中这样做的最佳方法是什么? “正常”数据类型太大。

5 个解决方案

#1

You could create

你可以创造

typedef struct foo
{
    unsigned x:1;
} foo;

Where you have told the compiler that you'll only be using one bit of x.

你告诉编译器你只使用一位x的地方。

But due to structure packing arrangements (the C standard is intentionally flexible in order that compilers can optimise according to the machine architecture), it may well turn out that this still occupies as much space in memory as a regular unsigned and an array of foos doesn't have to be bitwise contiguous.

但是由于结构打包安排(C标准是有意灵活的,以便编译器可以根据机器架构进行优化),很可能会发现这仍然占用了内存中与常规无符号和一系列foos无关的空间。不必按位连续。

#2

If you really want, you can create a structure with a member variable , bit-fielded to 1 bit.

如果您真的想要,可以使用成员变量创建一个结构,位为1位。

Remember, the data type of the member variable needs to be unsigned, as you need to store 0 and 1.

请记住,成员变量的数据类型需要是无符号的,因为您需要存储0和1。

#3

If you don't need millions of these flags or have extremely limited memory constraints, the best way is definitively an int.

如果您不需要数百万个这样的标志或者内存限制非常有限,那么最好的方法肯定是int。

This is because an int normally corresponds to the natural word size of your platform and can, properly aligned, be accessed quickly. The machine reads a word at a time anyways and using the single bits requires masking and shifting, that costs time. On your typical PC with gigabytes of RAM, this would be just silly.

这是因为int通常对应于平台的自然字大小,并且可以快速访问正确对齐的内容。无论如何,机器一次读取一个字,并且使用单个位需要屏蔽和移位,这需要花费时间。在具有千兆字节RAM的典型PC上,这将是愚蠢的。

If memory consumption really is an issue, there are bitfield structures.

如果内存消耗确实存在问题,则存在位域结构。

#4

The portable way is the definition of a variable which individual bits are used as flags.

可移植方式是变量的定义,其中各个位用作标志。

    #define FLAG_FOO 0
    #define FLAG_BAR 1

    // in case platform does not support uint8_t
    typedef unsigned char uint8_t;

    uint8_t flags;

    void flag_foo_set()
    {
        flags |= (1 << FLAG_FOO);
    }

    void flag_foo_clr()
    {
        flags &= ~(1 << FLAG_FOO);
    }

    uint8_t flag_foo_get()
    {
        return flags & (1 << FLAG_FOO);
    }

While this can seem superfluos compared to C bit fields. It is portable to basically every ANSI C compiler.

虽然与C位字段相比,这似乎是超级流行的。它几乎可以移植到每个ANSI C编译器。

#5

Generally, the smallest addressable chunk of data in C is a byte. You can not have a pointer to a bit, so you can not declare a variable of 1 bit size. But as Sourav Ghosh already pointed out, you can declare bitfields, where one bit is accessed directly.

通常,C中最小的可寻址数据块是一个字节。您不能指向某个位,因此您无法声明1位大小的变量。但正如Sourav Ghosh已经指出的那样,你可以声明位域,直接访问一位。

#1