例如，乘法和除法可以使用位运算符来实现
i*2 = i<<1
i*3 = (i<<1) + i;
i*10 = (i<<3) + (i<<1)
等等。
用say (i<<3)+(i<<1)乘10是否比直接用i*10更快?有没有什么输入是不能用这种方法乘或除的?
当前回答
I think in the one case that you want to multiply or divide by a power of two, you can't go wrong with using bitshift operators, even if the compiler converts them to a MUL/DIV, because some processors microcode (really, a macro) them anyway, so for those cases you will achieve an improvement, especially if the shift is more than 1. Or more explicitly, if the CPU has no bitshift operators, it will be a MUL/DIV anyway, but if the CPU has bitshift operators, you avoid a microcode branch and this is a few instructions less.
I am writing some code right now that requires a lot of doubling/halving operations because it is working on a dense binary tree, and there is one more operation that I suspect might be more optimal than an addition - a left (power of two multiply) shift with an addition. This can be replaced with a left shift and an xor if the shift is wider than the number of bits you want to add, easy example is (i<<1)^1, which adds one to a doubled value. This does not of course apply to a right shift (power of two divide) because only a left (little endian) shift fills the gap with zeros.
在我的代码中，这些乘/除2和2的幂运算被大量使用，因为公式已经很短了，每条可以消除的指令都可以获得很大的收益。如果处理器不支持这些位移操作符，就不会有增益，也不会有损失。
Also, in the algorithms I am writing, they visually represent the movements that occur so in that sense they are in fact more clear. The left hand side of a binary tree is bigger, and the right is smaller. As well as that, in my code, odd and even numbers have a special significance, and all left-hand children in the tree are odd and all right hand children, and the root, are even. In some cases, which I haven't encountered yet, but may, oh, actually, I didn't even think of this, x&1 may be a more optimal operation compared to x%2. x&1 on an even number will produce zero, but will produce 1 for an odd number.
再深入一点，如果x和3是0，我就知道4是这个数的因数，x%7是8，以此类推。我知道这些情况可能有有限的效用，但很高兴知道你可以避免模运算而使用按位逻辑运算，因为按位运算几乎总是最快的，而且对编译器来说不太可能是模糊的。
我在很大程度上发明了密集二叉树的领域，所以我预计人们可能不会理解这个评论的价值，因为很少有人想只对2的幂进行因数分解，或者只对2的幂进行乘/除。
其他回答
有些优化编译器无法做到，因为它们只适用于减少的输入集。
下面是c++示例代码，可以执行更快的除法，执行64位“乘倒数”。分子和分母都必须低于某个阈值。注意，它必须被编译为使用64位指令才能比普通除法更快。
#include <stdio.h>
#include <chrono>
static const unsigned s_bc = 32;
static const unsigned long long s_p = 1ULL << s_bc;
static const unsigned long long s_hp = s_p / 2;
static unsigned long long s_f;
static unsigned long long s_fr;
static void fastDivInitialize(const unsigned d)
{
s_f = s_p / d;
s_fr = s_f * (s_p - (s_f * d));
}
static unsigned fastDiv(const unsigned n)
{
return (s_f * n + ((s_fr * n + s_hp) >> s_bc)) >> s_bc;
}
static bool fastDivCheck(const unsigned n, const unsigned d)
{
// 32 to 64 cycles latency on modern cpus
const unsigned expected = n / d;
// At least 10 cycles latency on modern cpus
const unsigned result = fastDiv(n);
if (result != expected)
{
printf("Failed for: %u/%u != %u\n", n, d, expected);
return false;
}
return true;
}
int main()
{
unsigned result = 0;
// Make sure to verify it works for your expected set of inputs
const unsigned MAX_N = 65535;
const unsigned MAX_D = 40000;
const double ONE_SECOND_COUNT = 1000000000.0;
auto t0 = std::chrono::steady_clock::now();
unsigned count = 0;
printf("Verifying...\n");
for (unsigned d = 1; d <= MAX_D; ++d)
{
fastDivInitialize(d);
for (unsigned n = 0; n <= MAX_N; ++n)
{
count += !fastDivCheck(n, d);
}
}
auto t1 = std::chrono::steady_clock::now();
printf("Errors: %u / %u (%.4fs)\n", count, MAX_D * (MAX_N + 1), (t1 - t0).count() / ONE_SECOND_COUNT);
t0 = t1;
for (unsigned d = 1; d <= MAX_D; ++d)
{
fastDivInitialize(d);
for (unsigned n = 0; n <= MAX_N; ++n)
{
result += fastDiv(n);
}
}
t1 = std::chrono::steady_clock::now();
printf("Fast division time: %.4fs\n", (t1 - t0).count() / ONE_SECOND_COUNT);
t0 = t1;
count = 0;
for (unsigned d = 1; d <= MAX_D; ++d)
{
for (unsigned n = 0; n <= MAX_N; ++n)
{
result += n / d;
}
}
t1 = std::chrono::steady_clock::now();
printf("Normal division time: %.4fs\n", (t1 - t0).count() / ONE_SECOND_COUNT);
getchar();
return result;
}
用say (i<<3)+(i<<1)乘10是否比直接用i*10更快?
它可能在您的机器上，也可能不在您的机器上——如果您关心的话，请在您的实际使用情况中进行测量。
一个案例研究——从486到core i7
Benchmarking is very difficult to do meaningfully, but we can look at a few facts. From http://www.penguin.cz/~literakl/intel/s.html#SAL and http://www.penguin.cz/~literakl/intel/i.html#IMUL we get an idea of x86 clock cycles needed for arithmetic shift and multiplication. Say we stick to "486" (the newest one listed), 32 bit registers and immediates, IMUL takes 13-42 cycles and IDIV 44. Each SAL takes 2, and adding 1, so even with a few of those together shifting superficially looks like a winner.
如今，随着酷睿i7的出现:
(来自http://software.intel.com/en-us/forums/showthread.php?t=61481)
整数加法的延迟为1个周期，整数乘法的延迟为3个周期。您可以在“Intel®64 and IA-32架构优化参考手册”的附录C中找到延迟和吞吐量，该手册位于http://www.intel.com/products/processor/manuals/。
(来自英特尔的宣传)
使用SSE，酷睿i7可以同时发出加法和乘法指令，导致每个时钟周期有8个浮点运算(FLOP)的峰值速率
That gives you an idea of how far things have come. The optimisation trivia - like bit shifting versus * - that was been taken seriously even into the 90s is just obsolete now. Bit-shifting is still faster, but for non-power-of-two mul/div by the time you do all your shifts and add the results it's slower again. Then, more instructions means more cache faults, more potential issues in pipelining, more use of temporary registers may mean more saving and restoring of register content from the stack... it quickly gets too complicated to quantify all the impacts definitively but they're predominantly negative.
源代码中的功能vs实现
More generally, your question is tagged C and C++. As 3rd generation languages, they're specifically designed to hide the details of the underlying CPU instruction set. To satisfy their language Standards, they must support multiplication and shifting operations (and many others) even if the underlying hardware doesn't. In such cases, they must synthesize the required result using many other instructions. Similarly, they must provide software support for floating point operations if the CPU lacks it and there's no FPU. Modern CPUs all support * and <<, so this might seem absurdly theoretical and historical, but the significance thing is that the freedom to choose implementation goes both ways: even if the CPU has an instruction that implements the operation requested in the source code in the general case, the compiler's free to choose something else that it prefers because it's better for the specific case the compiler's faced with.
示例(使用假设的汇编语言)
source literal approach optimised approach
#define N 0
int x; .word x xor registerA, registerA
x *= N; move x -> registerA
move x -> registerB
A = B * immediate(0)
store registerA -> x
...............do something more with x...............
像exclusive or (xor)这样的指令与源代码没有关系，但是用自身进行xor-ing会清除所有的位，所以它可以用来将一些东西设置为0。暗示内存地址的源代码可能不需要使用任何内存地址。
These kind of hacks have been used for as long as computers have been around. In the early days of 3GLs, to secure developer uptake the compiler output had to satisfy the existing hardcore hand-optimising assembly-language dev. community that the produced code wasn't slower, more verbose or otherwise worse. Compilers quickly adopted lots of great optimisations - they became a better centralised store of it than any individual assembly language programmer could possibly be, though there's always the chance that they miss a specific optimisation that happens to be crucial in a specific case - humans can sometimes nut it out and grope for something better while compilers just do as they've been told until someone feeds that experience back into them.
因此，即使移动和添加在某些特定的硬件上仍然更快，那么编译器编写者可能已经准确地计算出什么时候它既安全又有益。
可维护性
如果你的硬件改变了，你可以重新编译，它会查看目标CPU并做出另一个最佳选择，而你不太可能想要重新审视你的“优化”或列出哪些编译环境应该使用乘法，哪些编译环境应该移位。想想10多年前编写的所有非2位移位的“优化”，现在它们在现代处理器上运行时减慢了它们所使用的代码……!
值得庆幸的是，像GCC这样的优秀编译器通常可以在启用任何优化时用直接乘法替换一系列位移位和算术(即. ...main(…){return (argc << 4) + (argc << 2) + argc;} -> imull $ 21,8 (%ebp)， %eax)所以重新编译可能有帮助，即使不修复代码，但这是不保证的。
实现乘法或除法的奇怪位移代码远不能表达您在概念上试图实现的目标，因此其他开发人员会对此感到困惑，而困惑的程序员更有可能引入错误或删除一些必要的东西，以努力恢复表面上的理智。如果你只做那些不明显的事情，但它们确实是有实际好处的，然后好好记录它们(但不要记录其他直观的东西)，每个人都会更快乐。
通解和部分解
如果你有一些额外的知识，比如你的int将只存储值x, y和z，那么你可能能够制定出一些指令，适用于这些值，并更快地得到你的结果，而不是编译器没有洞察，需要一个实现，适用于所有int值。例如，考虑你的问题:
乘法和除法可以使用位运算符实现…
你演示了乘法，那除法呢?
int x;
x >> 1; // divide by 2?
根据c++标准5.8:
-3—E1 >> E2为E1位右移E2位位置。如果E1为无符号类型，或者E1为有符号类型且值为非负值，则结果值为E1的商除以2的E2次方的积分部分。如果E1具有符号类型和负值，则结果值是由实现定义的。
因此，当x为负时，位移位有一个实现定义的结果:在不同的机器上可能不会以相同的方式工作。但是，/工作起来更容易预测。(它也可能不是完全一致的，因为不同的机器可能有不同的负数表示，因此即使构成表示的位数相同，范围也不同。)
You may say "I don't care... that int is storing the age of the employee, it can never be negative". If you have that kind of special insight, then yes - your >> safe optimisation might be passed over by the compiler unless you explicitly do it in your code. But, it's risky and rarely useful as much of the time you won't have this kind of insight, and other programmers working on the same code won't know that you've bet the house on some unusual expectations of the data you'll be handling... what seems a totally safe change to them might backfire because of your "optimisation".
有没有什么输入是不能用这种方法乘或除的?
是的……如上所述，负数在被位移“分割”时具有实现定义的行为。
如果在gcc编译器上比较x+x, x*2和x<<1语法的输出，那么在x86汇编中会得到相同的结果:https://godbolt.org/z/JLpp0j
push rbp
mov rbp, rsp
mov DWORD PTR [rbp-4], edi
mov eax, DWORD PTR [rbp-4]
add eax, eax
pop rbp
ret
因此，您可以认为gcc足够聪明，可以独立于您输入的内容确定自己的最佳解决方案。
I think in the one case that you want to multiply or divide by a power of two, you can't go wrong with using bitshift operators, even if the compiler converts them to a MUL/DIV, because some processors microcode (really, a macro) them anyway, so for those cases you will achieve an improvement, especially if the shift is more than 1. Or more explicitly, if the CPU has no bitshift operators, it will be a MUL/DIV anyway, but if the CPU has bitshift operators, you avoid a microcode branch and this is a few instructions less.
I am writing some code right now that requires a lot of doubling/halving operations because it is working on a dense binary tree, and there is one more operation that I suspect might be more optimal than an addition - a left (power of two multiply) shift with an addition. This can be replaced with a left shift and an xor if the shift is wider than the number of bits you want to add, easy example is (i<<1)^1, which adds one to a doubled value. This does not of course apply to a right shift (power of two divide) because only a left (little endian) shift fills the gap with zeros.
在我的代码中，这些乘/除2和2的幂运算被大量使用，因为公式已经很短了，每条可以消除的指令都可以获得很大的收益。如果处理器不支持这些位移操作符，就不会有增益，也不会有损失。
Also, in the algorithms I am writing, they visually represent the movements that occur so in that sense they are in fact more clear. The left hand side of a binary tree is bigger, and the right is smaller. As well as that, in my code, odd and even numbers have a special significance, and all left-hand children in the tree are odd and all right hand children, and the root, are even. In some cases, which I haven't encountered yet, but may, oh, actually, I didn't even think of this, x&1 may be a more optimal operation compared to x%2. x&1 on an even number will produce zero, but will produce 1 for an odd number.
再深入一点，如果x和3是0，我就知道4是这个数的因数，x%7是8，以此类推。我知道这些情况可能有有限的效用，但很高兴知道你可以避免模运算而使用按位逻辑运算，因为按位运算几乎总是最快的，而且对编译器来说不太可能是模糊的。
我在很大程度上发明了密集二叉树的领域，所以我预计人们可能不会理解这个评论的价值，因为很少有人想只对2的幂进行因数分解，或者只对2的幂进行乘/除。
据我所知，在一些机器上，乘法运算可能需要16到32个机器周期。是的，根据机器类型，位移运算符比乘除运算符快。
然而，某些机器确实有它们的数学处理器，其中包含乘法/除法的特殊指令。
