__builtin_ctz(0)或__builtin_clz(0)有多不确定? [英] How undefined are __builtin_ctz(0) or __builtin_clz(0)?
问题描述
很长一段时间以来， gcc一直在提供内置位纠缠功能，尤其是尾随和前导0位的数量(也适用于long unsigned和long long unsigned，其后缀为l和ll):
For a long time, gcc has been providing a number of builtin bit-twiddling functions, in particular the number of trailing and leading 0-bits (also for long unsigned and long long unsigned, which have suffixes l and ll):
—内置功能:
int __builtin_clz (unsigned int x)
返回
x中前导0位的数量，从最高有效位开始
位置.如果x为0，则结果不确定.
Returns the
number of leading 0-bits in x, starting at the most significant bit
position. If x is 0, the result is undefined.
—内置功能:int __builtin_ctz (unsigned int x)
返回
x中的尾随0位的数量，从最低有效位开始
位置.如果x为0，则结果不确定.
Returns the
number of trailing 0-bits in x, starting at the least significant bit
position. If x is 0, the result is undefined.
但是，在我测试的每个在线(免责声明:仅x64)编译器上，结果都是clz(0)和ctz(0)都返回基础内置类型的位数，例如
On every online (disclaimer: only x64) compiler I tested, however, the result has been that both clz(0) and ctz(0) return the number of bits of the underlying builtin type, e.g.
#include <iostream>
#include <limits>
int main()
{
// prints 32 32 32 on most systems
std::cout << std::numeric_limits<unsigned>::digits << " " << __builtin_ctz(0) << " " << __builtin_clz(0);
}
实时示例 .
在std=c++1y模式下的最新Clang SVN干线使所有这些函数都放松了C ++ 14 constexpr，这使它们可以在SFINAE表达式中用于3 ctz/<周围的包装器函数模板. c15> unsigned，unsigned long和unsigned long long
The latest Clang SVN trunk in std=c++1y mode has made all these functions relaxed C++14 constexpr, which makes them candidates to use in a SFINAE expression for a wrapper function template around the 3 ctz / clz builtins for unsigned, unsigned long, and unsigned long long
template<class T> // wrapper class specialized for u, ul, ull (not shown)
constexpr int ctznz(T x) { return wrapper_class_around_builtin_ctz<T>()(x); }
// overload for platforms where ctznz returns size of underlying type
template<class T>
constexpr auto ctz(T x)
-> typename std::enable_if<ctznz(0) == std::numeric_limits<T>::digits, int>::type
{ return ctznz(x); }
// overload for platforms where ctznz does something else
template<class T>
constexpr auto ctz(T x)
-> typename std::enable_if<ctznz(0) != std::numeric_limits<T>::digits, int>::type
{ return x ? ctznz(x) : std::numeric_limits<T>::digits; }
这种黑客攻击的好处在于，为ctz(0)提供所需结果的平台可以省略测试x==0的额外条件(这似乎是微优化的，但是当您已经达到内置位旋转功能，可以有很大的不同)
The gain from this hack is that platforms that give the required result for ctz(0) can omit an extra conditional to test for x==0 (which might seem a micro-optimization, but when you are already down to the level of builtin bit-twiddling functions, it can make a big difference)
内置函数clz(0)和ctz(0)的家族如何定义?
How undefined is the family of builtin functions clz(0) and ctz(0)?
- 他们可以抛出
std::invalid_argument异常吗? - 对于x64，当前的gcc发行版是否会返回底层字体的大小?
- ARM/x86平台有什么不同吗(我无权对其进行测试)?
- 上面的SFINAE技巧是否是一种定义明确的方法来分离此类平台?
- can they throw an
std::invalid_argumentexception? - for x64, will they for the current gcc distro return the size of the underyling type?
- are the ARM/x86 platforms any different (I have no access to that to test those)?
- is the above SFINAE trick a well-defined way to separate such platforms?
推荐答案
不幸的是，即使x86-64实现也可能有所不同-与Intel的
Unfortunately, even x86-64 implementations can differ - from Intel's instruction set reference,BSF and BSR, with a source operand value of (0), leaves the destination undefined, and sets the ZF (zero flag). So the behaviour may not be consistent between micro-architectures or, say, AMD and Intel. (I believe AMD leaves the destination unmodified.)
较新的LZCNT和TZCNT指令并非无处不在.两者都仅在Haswell架构(适用于Intel)上存在.
The newer LZCNT and TZCNT instructions are not ubiquitous. Both are present only as of the Haswell architecture (for Intel).
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