为什么不是C具有二进制文字？ [英] Why doesn't C have binary literals?

问题描述

我经常希望我可以做在C是这样的：

I am frequently wishing I could do something like this in c:

val1 |= 0b00001111; //clear high nibble
val2 &= 0b01000000; //set bit 7
val3 |= ~0b00010000; //clear bit 5

有了这个语法似乎是一个非常有用的除了到C没有缺点，我能想到的，它似乎像在那里位变换是相当普遍的一种低层次的语言很自然的事。

Having this syntax seems like an incredibly useful addition to C with no downsides that I can think of, and it seems like a natural thing for a low level language where bit-twiddling is fairly common.

编辑：我看到其他一些伟大的选择，但他们都土崩瓦解时，有一个更复杂的面具。例如，如果章是一个微控制器控制I / O引脚的寄存器，我想设置引脚2，3，7高，同时我可以写 =章的0x46; ，但我不得不花费10秒想着它（我很可能不得不每次我读到后，这些code时候再花费10秒一个不看它一两天），或者我可以写 =章（1＆LT;＆LT; 1）| （1 <<; 2）| （1 LT; 6;）; 但我个人认为这是比只是写不太清楚`章= 0b01000110;'我可以同意，它不超过8位很好地扩展或但也许16位架构。这并不是说我曾经需要做一个32位掩码。

I'm seeing some other great alternatives but they all fall apart when there is a more complex mask. For example, if reg is a register that controls I/O pins on a microcontroller, and I want to set pins 2, 3, and 7 high at the same time I could write reg = 0x46; but I had to spend 10 seconds thinking about it (and I'll likely have to spend 10 seconds again every time I read those code after a not looking at it for a day or two) or I could write reg = (1 << 1) | (1 << 2) | (1 << 6); but personally I think that is way less clear than just writing `reg = 0b01000110;' I can agree that it doesn't scale well beyond 8 bit or maybe 16 bit architectures though. Not that I've ever needed to make a 32 bit mask.

推荐答案

据的理由国际标准 - 编程语言ç§6.4.4.1的整型常量的

According to Rationale for International Standard - Programming Languages C §6.4.4.1 Integer constants

添加二进制常量的建议被拒绝，由于缺乏precedent和效用不足。

A proposal to add binary constants was rejected due to lack of precedent and insufficient utility.

这不是标准C，但海湾合作委员会支持它由 0B 或 0B ：

It's not in standard C, but gcc supports it as an extension, prefixed by 0b or 0B:

 i = 0b101010;

请参阅这里细节。

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