I just got a seg fault in overloading the assignment operator for a class FeatureRandomCounts, which has _rects as its pointer member pointing to an array of FeatureCount and size rhs._dim, and whose other date members are non-pointers: FeatureRandomCounts & FeatureRandomCounts::operator=(const FeatureRandomCounts &rhs) { if (_rects) delete [] _rects; *this = rhs; // segment
我只是在为类FeatureRandomCounts重载赋值运算符时出现了seg错误，该类有一个_rects作为其指针成员，指向FeatureCount的数组并且大小为rhs._dim，其他日期成员是非指针： FeatureRandomCounts & FeatureRandomCounts::operator=(const FeatureRandomCounts &rhs) { if (_rects) delete [] _rects; *this = rhs; // segment fault _rects = new FeatureCount [rhs._dim]; for (int i = 0; i < rhs
The following have been proposed for an upcoming C++ project. C++ Coding Standards, by Sutter and Alexandrescu JSF Air Vehicle C++ coding standards The Elements of C++ Style Effective C++ 3rd Edition, by Scott Meyers Are there other choices? Or is the list above what be should used on a C++ project? Some related links Do you think a software company should impose developers a codin
以下是针对即将推出的C ++项目提出的建议。 C ++ Coding Standards，由Sutter和Alexandrescu撰写 JSF飞行器C ++编码标准 C ++风格的元素 Effective C ++ 3rd Edition，作者：Scott Meyers 还有其他选择吗？ 或者是上面的列表应该在C ++项目上使用什么？ 一些相关的链接 你认为一家软件公司应该向开发者强加一种编码风格吗？ https://stackoverflow.com/questions/66268/what-is-the-best-cc-coding-style-close
I want to see the results of a GET request. By my understanding, this code should do it. What am I doing wrong? void getDoc::on_pushButton_2_clicked() { manager = new QNetworkAccessManager(this); connect(manager, SIGNAL(finished(QNetworkReply*)), this, SLOT(replyFinished(QNetworkReply*))); manager->get(QNetworkRequest(QUrl("http://www.google.com"))); } void getDoc::replyFinish
我想查看GET请求的结果。 根据我的理解，这段代码应该这样做。 我究竟做错了什么？ void getDoc::on_pushButton_2_clicked() { manager = new QNetworkAccessManager(this); connect(manager, SIGNAL(finished(QNetworkReply*)), this, SLOT(replyFinished(QNetworkReply*))); manager->get(QNetworkRequest(QUrl("http://www.google.com"))); } void getDoc::replyFinished(QNetworkReply *reply) { qDe
I'm just trying to check the optimum approach to optimizing some basic routines. In this case I tried very simply example of multiplying 2 float vectors together: void Mul(float *src1, float *src2, float *dst) { for (int i=0; i<cnt; i++) dst[i] = src1[i] * src2[i]; }; Plain C implementation is very slow. I did some external ASM using AVX and also tried using intrinsics. These are
我只是试图检查优化一些基本例程的最佳方法。 在这种情况下，我尝试了将2个浮点数向量相乘的例子： void Mul(float *src1, float *src2, float *dst) { for (int i=0; i<cnt; i++) dst[i] = src1[i] * src2[i]; }; 平原C的实施非常缓慢。 我使用AVX做了一些外部ASM，并尝试使用内部函数。 这些是测试结果（时间越小越好）： ASM: 0.110 IPP: 0.125 Intrinsics: 0.18 Plain C++: 4.0 （使用MSVC 2013，SSE2编译，尝
I want to multiply the data stored in one xmm register with a single float value and save the result in a xmm register. I made a little graphic to explain it a bit better. As you see I got a xmm0 register with my data in it. For example it contains: xmm0 = |4.0|2.5|3.5|2.0| Each floating point is stored in 4 bytes. My xmm0 register is 128 bits, 16 bytes long. That works pretty good. N
我想将存储在一个xmm寄存器中的数据乘以一个浮点值并将结果保存在一个xmm寄存器中。 我做了一个小图片来解释它好一点。 正如你看到的，我有一个xmm0寄存器，里面有我的数据。 例如它包含： xmm0 = | 4.0 | 2.5 | 3.5 | 2.0 | 每个浮点都以4个字节存储。 我的xmm0寄存器是128位，长16字节。 这工作很好。 现在，我想将0.5存储在另一个xmm寄存器中，例如xmm1，并将此寄存器与xmm0寄存器相乘，以便将存储在xmm0中的每个
Bitwise operators ( ~ , & , | and ^ ) operate on the bitwise representation of their promoted operands. Can such operations cause undefined behavior? For example, the ~ operator is defined this way in the C Standard: 6.5.3.3 Unary arithmetic operators The result of the ~ operator is the bitwise complement of its (promoted) operand (that is, each bit in the result is set if and only if
按位运算符（ ~ ， & ， |和^ ）对其升级的操作数的按位表示进行操作。 这样的操作会导致未定义的行为吗？ 例如， ~运算符在C标准中以这种方式定义： 6.5.3.3一元算术运算符 ~运算符的结果是其（提升的）操作数的按位补数（即，当且仅当转换的操作数中的相应位未设置时，结果中的每个位才被设置）。 整数提升在操作数上执行，并且结果具有提升类型。 如果升级类型是无符号类型，则表达式~E等于该类型中表示的最大值
I am reading about sets representing as bits at following location http://www.brpreiss.com/books/opus4/html/page395.html class SetAsBitVector : public Set { typedef unsigned int Word; enum { wordBits = bitsizeof (Word) }; Array<Word> vector; public: SetAsBitVector (unsigned int); // ... }; SetAsBitVector::SetAsBitVector (unsigned int n) : Set (n),
我正在阅读关于在以下位置表示为位的集合 http://www.brpreiss.com/books/opus4/html/page395.html class SetAsBitVector : public Set { typedef unsigned int Word; enum { wordBits = bitsizeof (Word) }; Array<Word> vector; public: SetAsBitVector (unsigned int); // ... }; SetAsBitVector::SetAsBitVector (unsigned int n) : Set (n),
I have a part of code that contains the following functions: void Keyboard(int key) { switch (key) { case GLFW_KEY_A: m_controlState |= TDC_LEFT; break; case GLFW_KEY_D: m_controlState |= TDC_RIGHT; break; case GLFW_KEY_W: m_controlState |= TDC_UP; break; case GLFW_KEY_S: m_controlState |= TDC_DOWN; break; default: Test::Keyboard(key); } } void KeyboardUp( int key) {
我有一部分代码包含以下功能： void Keyboard(int key) { switch (key) { case GLFW_KEY_A: m_controlState |= TDC_LEFT; break; case GLFW_KEY_D: m_controlState |= TDC_RIGHT; break; case GLFW_KEY_W: m_controlState |= TDC_UP; break; case GLFW_KEY_S: m_controlState |= TDC_DOWN; break; default: Test::Keyboard(key); } } void KeyboardUp( int key) { switch (key) { case GLF
This question already has an answer here: Understanding the bitwise AND Operator 4 answers The bitwise AND operator is a single ampersand: &. A handy mnemonic is that the small version of the boolean AND, &&, works on smaller pieces (bits instead of bytes, chars, integers, etc). In essence, a binary AND simply takes the logical AND of the bits in each position of a number in bin
这个问题在这里已经有了答案： 了解按位与运算符4的答案 按位AND运算符是一个单符号：＆。 一个方便的助记符是，布尔AND &&的小版本在小块（比特而不是字节，字符，整数等）上工作。 本质上，二进制AND只是以二进制形式将数字的每个位置中的位进行逻辑AND。 例如，使用一个字节（char类型）： EX。 01001000 & 10111000 = -------- 00001000 第一个数字的最高有效位是0，所以我们知道结果的最高有效位
This question already has an answer here: Understanding the bitwise AND Operator 4 answers It will result in an entire integer. When you use the bitwise and, each bit of the two values are and'ed together, and each bit in the result is set accordingly. The result will be the same number of bits as the values being and'ed together. This is assuming you're using two integer vari
这个问题在这里已经有了答案： 了解按位与运算符4的答案 这将导致整个整数。 当你使用按位，并且两个值的每一位都被连在一起，并且结果中的每一位都被相应地设置。 结果将与数值在一起的位数相同。 这假设你正在使用两个整数变量。 假设mask是一个指向整数类型的指针，编译器将执行以下操作： 以整数形式访问mask “数组”的第i个元素 确保两个操作数的大小相同（这可能涉及位扩展或截断） 做一点点AND 如果前一