I have recently stumbled upon some strange behaviour in my equation solver, which made me ask myself if I really understood how move semantics and RVO work together. There are plenty of related questions on this forum, and I've also read many general explanations on this. But my problem seems to be quite specific so I hope someone will help me out. The involved struct is a bit complex al
我最近在我的方程求解器中偶然发现了一些奇怪的行为，这让我问自己是否真的理解了移动语义和RVO如何协同工作。 在这个论坛上有很多相关的问题，我也读过很多关于这个问题的一般性解释。 但我的问题似乎很具体，所以我希望有人能帮助我。 涉及的结构完全有点复杂，但至少可以打破这一点： struct Foo { Bar* Elements; Foo(void) : Elements(nullptr) { cout << "Default-constructing Foo objec
有neither A nor B语法？ While there isn't a built-in syntax to do this, I'd suggest you take a look at the list of supported logical operators and then carefully study De Morgan's laws. Sufficient knowledge in these two fields will allow you to write any logical statement in if–else if syntax. EDIT: To completely answer your question (although this has been done already in other an
有neither A nor B语法？ 虽然没有内置的语法来做到这一点，但我建议你看一下支持的逻辑运算符列表，然后仔细研究De Morgan的定律。 在这两个字段中的足够的知识将允许您在if-else语法中编写任何逻辑语句。 编辑：要完全回答你的问题（虽然这已经在其他答案中已经完成），你可以写一个既不 - 也不是这样的陈述： if (!A && !B) { DoStuff(); } 哦...你想要“不是”关键字？ VB添加不是关键字 （Newswire 8-19-200
Having read this interesting article on the results of intrinsic-guided optimization of SSE code in different C++ compilers I decided to do a test of my own, especially since the post is a few years old. I used MSVC which did so very poorly in the tests performed by the author of the post (although in the VS 2010 version) and decided to stick to a very basic scenario: packing some values into a
在阅读了关于不同C ++编译器中SSE代码内在引导优化结果的有趣文章之后，我决定对自己做一个测试，特别是自从这篇文章已经过去几年之后。 我使用MSVC，这在文章作者的测试中做得非常差（虽然在VS 2010版本中），并决定坚持一个非常基本的场景：将一些值打包到XMM寄存器中，并进行简单的操作，如添加。 在文章中，_mm_set_ps被翻译成一个奇怪的标量移动和解压缩指令序列，所以让我们看看： int _tmain(int argc, _TCHAR* argv[])
Many years ago, C compilers were not particularly smart. As a workaround K&R invented the register keyword, to hint to the compiler, that maybe it would be a good idea to keep this variable in an internal register. They also made the tertiary operator to help generate better code. As time passed, the compilers matured. They became very smart in that their flow analysis allowing them to m
很多年前，C编译器并不特别聪明。 作为一种解决方法，K＆R发明了register关键字，向编译器暗示，将这个变量保存在内部寄存器中可能是一个好主意。 他们还让高等运营商帮助生成更好的代码。 随着时间的推移，编译器逐渐成熟。 他们变得非常聪明，因为他们的流程分析使他们能够更好地决定登记在寄存器中的值，而不是你可能做的。 register关键字变得不重要。 由于别名问题，FORTRAN对于某些操作可能比C更快。 理论上仔细
I tried to compare the performance of inline assembly language and C++ code, so I wrote a function that add two arrays of size 2000 for 100000 times. Here's the code: #define TIMES 100000 void calcuC(int *x,int *y,int length) { for(int i = 0; i < TIMES; i++) { for(int j = 0; j < length; j++) x[j] += y[j]; } } void calcuAsm(int *x,int *y,int lengthOfAr
我试图比较内联汇编语言和C ++代码的性能，所以我编写了一个函数，它可以将两个2000大小的数组添加到100000次。 代码如下： #define TIMES 100000 void calcuC(int *x,int *y,int length) { for(int i = 0; i < TIMES; i++) { for(int j = 0; j < length; j++) x[j] += y[j]; } } void calcuAsm(int *x,int *y,int lengthOfArray) { __asm { mov edi,TIMES st
This question already has an answer here: With arrays, why is it the case that a[5] == 5[a]? 17 answers The reason this is the case is because arr[n] == *(arr + n) . However, because addition is commutative, *(arr + n) == *(n + arr) . Thus, *(n + arr) == n[arr] == *(arr + n) == arr[n] . It might be worth mentioning that *(arr + n) is still a little misleading. In assembly it actually m
这个问题在这里已经有了答案： 对于数组，为什么会出现[5] == 5 [a]？ 17个答案 这是因为arr[n] == *(arr + n) 。 但是，因为加法是可交换的，所以*(arr + n) == *(n + arr) 。 因此， *(n + arr) == n[arr] == *(arr + n) == arr[n] 。 可能值得一提的是*(arr + n)仍然有点误导。 在汇编中，它实际上意味着*(arr + (n * s)) ，其中s是sizeof arr[0] ，但这是在封面之下，所以您不必担心它。 我不认为这个特殊的[ab
Possible Duplicate: In C arrays why is this true? a[5] == 5[a] Someone told me this... I didn't believe them at first but it does work. If x and y do not change throughout the code, why does this work: int x [5] = { 0,1,2,3,4}; int y = 3; if(x[y] == y[x]){ cout << "Why..." << endl; } How does x array's value in index y is = the x index's value's in array y
可能重复： 在C数组中，为什么这是真的？ a [5] == 5 [a] 有人告诉我这一点......我一开始并不相信它，但它确实有效。 如果x和y在整个代码中都没有改变，为什么这会起作用： int x [5] = { 0,1,2,3,4}; int y = 3; if(x[y] == y[x]){ cout << "Why..." << endl; } x数组在索引y中的值如何=数组y中的x索引值？ 但是没有y阵列。 它总是如此（对于普通的operator ==） a[i] --> *(a+i) --> *(i+
Possible Duplicate: In C arrays why is this true? a[5] == 5[a] How is it possible that this is valid C++? void main() { int x = 1["WTF?"]; } On VC++10 this compiles and in debug mode the value of x is 84 after the statement. What's going on? Array subscript operator is commutative. It's equivalent to int x = "WTF?"[1]; Here, "WTF?" is an array of 5 char
可能重复： 在C数组中，为什么这是真的？ a [5] == 5 [a] 这怎么可能是有效的C ++？ void main() { int x = 1["WTF?"]; } 在VC ++ 10上编译并且在调试模式下，语句后x的值是84。 这是怎么回事？ 数组下标运算符是可交换的。 它相当于int x = "WTF?"[1]; 在这里， "WTF?" 是一个包含5个char的数组（它包含空终止符）， [1]给我们第二个字符，即'T' - 隐式转换为int它给出值84。 Of
This question already has an answer here: With arrays, why is it the case that a[5] == 5[a]? 17 answers This doesn't look like a function. What is this? 5 answers It's perfectly legal. With pointer arithmetic, a[42] is equivalent to *(a + 42) , which (addition being commutative) is equivalent to *(42+ a) , which (by definition of [] ) is equivalent to 42[a] . So it's obscu
这个问题在这里已经有了答案： 对于数组，为什么会出现[5] == 5 [a]？ 17个答案 这看起来不像一个功能。 这是什么？ 5个答案 这是完全合法的。 使用指针运算， a[42]等价于*(a + 42) ，其中（加法是可交换的）相当于*(42+ a) ，其（根据[]的定义等价于42[a] 。 所以这是不明确的，但明确的。 数组运算符是可交换的。 a[n] == *(a + n) == *(n + a) == n[a] 这完全合法。 a[i]被定义为*(a+i) 。 所以42[a]
Possible Duplicate: In C arrays why is this true? a[5] == 5[a] 3["zdvnngfgnfg"]; It's equivalent to "zdvnngfgnfg"[3]; which is legal and means "take the address of that literal and add 3*sizeof(char) to it". Will have no effect anyway. Also see this very similar question. arr[i] is parsed as *(arr+i) which can be written as *(i+arr) and hence i[arr] Now &qu
可能重复： 在C数组中，为什么这是真的？ a [5] == 5 [a] 3 [ “zdvnngfgnfg”]; 这相当于 "zdvnngfgnfg"[3]; 这是合法的，并且意味着“采用该文字的地址并向其添加3*sizeof(char) ”。 反正无效。 也看到这个非常类似的问题。 arr [i]被解析为*（arr + i），它可以被写为*（i + arr），因此i [arr] 现在“strngjwdgd”是一个指向存储在只读位置的常量字符数组的指针。 所以它的作品！ 字符串文字（ array ）衰变