Arduino 中不需要的符号扩展 [英] Unwanted sign extension in Arduino
问题描述
我正在尝试在 Arduino 中实现逻辑右移(即避免符号扩展)，并在阅读 Arduino BitShift 指南后(https://www.arduino.cc/en/Reference/Bitshift)，它表明将无符号变量向右移动，不会导致符号扩展:
I am trying to achieve logical right shift in Arduino(i.e. avoid sign extension), and after reading the Arduino BitShift guide (https://www.arduino.cc/en/Reference/Bitshift), it suggests that shifting unsigned variables to the right, wont cause sign extension:
当你将 x 右移 y 位 (x >> y) 时，x 中的最高位是a 1，行为取决于 x 的确切数据类型.如果 x 是类型int，最高位为符号位，判断x是否为否定与否，正如我们上面讨论的那样.在这种情况下，标志由于深奥的历史原因，位被复制到低位:
When you shift x right by y bits (x >> y), and the highest bit in x is a 1, the behavior depends on the exact data type of x. If x is of type int, the highest bit is the sign bit, determining whether x is negative or not, as we have discussed above. In that case, the sign bit is copied into lower bits, for esoteric historical reasons:
int x = -16; // binary: 1111111111110000
int y = x >> 3; // binary: 1111111111111110 This behavior, called sign extension, is often not the behavior you want. Instead, you may
希望从左边移入零.事实证明，正确的无符号整数表达式的移位规则不同，因此您可以使用一种类型转换，用于抑制从左侧复制的内容.
wish zeros to be shifted in from the left. It turns out that the right shift rules are different for unsigned int expressions, so you can use a typecast to suppress ones being copied from the left.
在我的测试中，它不是这样工作的:
In my tests, it doesn't work like that:
Serial.print( ((uint32_t)(1<<15)) >> 15, BIN);
打印:
11111111111111111
11111111111111111
这意味着，正在进行符号扩展.我也尝试了那里的建议示例，结果相同.
Which means, there's sign extension going on. I've also tried the suggested example from there, with the same results.
我做错了吗?也可以做一个移位并强制操作是逻辑的而不是算术的吗?
Am I doing something wrong? Also Is it possible to do a shift and force the operation to be logical instead of arithmetic?
推荐答案
首先，我认为您遗漏了一些关键信息:看起来您必须使用带有 16 位 int 类型(例如，Arduino Uno).
First, I think you left out some key information: it looks like you must be using an Arduino board with a 16-bit int type (e.g., Arduino Uno).
这里的问题是整数提升在 C 和 C++ 中是如何工作的.当您将 16 位有符号整数文字 1<<15 转换为 32 位无符号整数文字时，将执行以下步骤:
The problem here is how integer promotion works in C and C++. When you cast a 16-bit signed integer literal, 1<<15, to a 32-bit unsigned integer literal, it will take the following steps:
- 您要从 16 位转换到 32 位，因此它会首先将您现有的文字扩展到 32 位.由于它是有符号文字，因此首先将其符号扩展为 32 位有符号值.
- 既然操作数具有与所需类型相同的位宽，编译器会将其强制转换为 32 位无符号整数类型.
我没有一台 16 位机器来测试这个，但我可以用这个测试程序在我的 64 位笔记本电脑上复制相同的行为:
I don't have a 16-bit machine sitting around to test this on, but I can duplicate the same behavior on my 64-bit laptop with this test program:
#include <stdio.h>
#include <inttypes.h>
int main(int argc, const char* argv[]) {
printf("%" PRIx64 "\n", ((uint64_t)(1 << 31)));
// prints ffffffff80000000
printf("%" PRIx64 "\n", ((uint64_t)(1 << 31)) >> 31);
// prints 1ffffffff
return 0;
}
所以你可以在这个程序中看到，不是 >>> 操作在做不需要的符号扩展，而是从 32 位有符号整数到 64 位无符号整数的转换整数，因为演员表实际上是这样的:
So you can see in this program, it's not the >> operation that's doing the unwanted sign extension, but rather the cast from a 32-bit signed integer to a 64-bit unsigned integer, because the cast actually acts like this:
int32_t → int64_t → uint64_t
如果你想避免额外的符号扩展，你应该从一个无符号文字开始(正如一些程序员在他的评论中建议的那样)，或者转换为相同宽度的无符号类型.其中任何一个都应该有效:
If you want to avoid the extra sign extensions, you should either start with an unsigned literal (as Some programmer dude suggested in his comment), or make a cast to the same-width unsigned type. Any of these should work:
Serial.print( ((uint32_t)(1u<<15)) >> 15, BIN);
Serial.print( ((uint16_t)(1<<15)) >> 15, BIN);
Serial.print( ((uint32_t)(uint16_t)(1<<15)) >> 15, BIN);
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