为什么将运算符％称为“模数"?而不是“剩余"运算符操作员? [英] Why is operator% referred to as the "modulus" operator instead of the "remainder" operator?
问题描述
今天在工作中，我与一位同事进行了有趣的讨论.当他发生以下事情时，他感到惊讶:
Today at work I had an interesting discussion with one of my coworkers. He was surprised when he had the following happen to him:
assert(-1 % 10 == -1) //Expecting 9
因此，当他来问我这个问题时，我告诉他很好，这是有道理的.将-1除以10时，您得到0，剩下的-1.但是他的论点是模运算符应该坚持永远是积极的"模型.我做了一点研究，发现他指的模数看起来像这样:
So when he came to ask me about it, I told him "well, that makes sense. When you divide -1 by 10, you get 0 with -1 remaining. His argument however was that the modulus operator is supposed to hold true to the "always positive" model. I did a little research and found that the modulus he was referring to looks like this:
让q为a和n的整数商.令r为余数.然后:
Let q be the integer quotient of a and n. Let r be the remainder. Then:
a = n * q + r
a = n * q + r
I 使用的定义似乎是模数的Knuth版本，即:
The definition I was using, however, appears to be the Knuth version of modulus, which is:
让q为n的底数.令r为余数.然后:
Let q be the floor of a divided by n. Let r be the remainder. Then:
r = a-n * q
r = a - n * q
所以，我的问题是，为什么它以FORTRAN标准(以及后来的C标准)结尾而使模数运算符截断为0?在我看来，称其为模数"而不是余数"是不恰当的(在数学上，答案实际上应该是9).这与硬件的划分方式有关吗?
So, my question is why it ended up in the FORTRAN standard (and subsequently the C-standard) to have the modulus operator truncate toward 0? It seems like a misnomer to me to call it "modulus" and not "remainder" (In math, the answer really should be 9). Is this related to how hardware is doing the division?
供参考:
- 有关Modulus的维基百科
- 模数"运算符上的MSDN条目
(是的，我意识到它是针对VS2003的...我目前仍坚持使用它.Sadface) - 模运算符更改
- 不要假设正余数...
- Wikipedia on Modulus
- MSDN entry on the "modulus" operator
(Yes, i realize its for VS2003...I'm stuck with it currently. Sadface) - Modulus operator changes
- Don't assume positive remainder...
TLDR；硬件是模数运算符向0截断的原因吗?
推荐答案
在我看来，称其为模数"而不是余数"是不恰当的(在数学上，答案实际上应该是9).
It seems like a misnomer to me to call it "modulus" and not "remainder" (In math, the answer really should be 9).
C将其称为％运算符，并将其结果称为余数. C ++从C复制它.两种语言都没有将其称为模运算符.这也解释了为什么余数为负的原因:因为/运算符会截断为0，并且(a / b) * b + (a % b)应该等于a.
C calls it the % operator, and calls its result the remainder. C++ copies this from C. Neither language calls it the modulus operator. This also explains why the remainder is negative: because the / operator truncates towards 0, and (a / b) * b + (a % b) should equal a.
编辑:DavidRodríguez正确地指出C ++ 确实定义了一个模板类std::modulus，该类称为operator%.我认为，该类的名字很差.稍微挖掘一下，它是从STL继承而来的，它现在已经被命名. STL的下载内容为:"STL是在SGI MIPSproTM C ++ 7.0、7.1、7.2和7.2.1上开发的."据我所知，实际上没有编译器和硬件，MIPSpro会将其划分给CPU和处理器. MIPS硬件将截断为0，这意味着std::modulus始终被错误命名.
Edit: David Rodríguez rightly points out that C++ does define a template class std::modulus, which calls operator%. In my opinion, that class is poorly named. Digging a little bit, it is inherited from STL where it was already named as it is now. The download for STL says "The STL was developed on SGI MIPSproTM C++ 7.0, 7.1, 7.2, and 7.2.1.", and as far as I can tell without actually having the compiler and hardware, MIPSpro passes the division to the CPU and MIPS hardware truncates to 0, which would mean std::modulus has always been misnamed.
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