I'm using gcov for the first time and I'm having a problem which is similar to the one reported in this thread. However, I wasn't able to solve my problem following the comments in that thread. I'm trying to measure the coverage of a concrete test case in KMyMoney-4.6.4, namely "testConstructor" in the test file "mymoneyaccounttest.cpp". This test case is u
我第一次使用gcov，并且遇到类似于此线程中报告的问题。 但是，在该主题中的评论之后，我无法解决我的问题。 我正在测试KMyMoney-4.6.4中具体测试用例的覆盖率，即测试文件“mymoneyaccounttest.cpp”中的“testConstructor”。 该测试用例使用以下方法，该方法位于头文件“mymoneyaccount.h”中，并具有可执行代码： const QString& institutionId(void) const { return m_institution; } 因此，我构建了程序和测试
On Mac OS, if I run clang --version , I get: Apple LLVM version 6.0 (clang-600.0.34.4) (based on LLVM 3.5svn) Target: x86_64-apple-darwin14.0.0 Thread model: posix Since LLVM official page suggests that the current LLVM version is 3.5, it seems that Apple clang versioning is distinct from the open-source LLVM. The output (based on LLVM 3.5svn) suggests that probably it is 3.5 on my machine as
在Mac OS上，如果我运行clang --version ，我会得到： Apple LLVM version 6.0 (clang-600.0.34.4) (based on LLVM 3.5svn) Target: x86_64-apple-darwin14.0.0 Thread model: posix 由于LLVM的官方页面显示当前的LLVM版本为3.5，因此Apple的铿锵版本似乎与开源LLVM不同。 输出(based on LLVM 3.5svn)表明，我的机器上可能也是3.5。 现在我到了这里，因为我想编译一个C ++ 14文件，使用g++ -std=c++14 main.cc 这个页面建
I'm running gcov over some C code with a switch statement. I've written test cases to cover every possible path through that switch statement, but it still reports a branch in the switch statement as not taken and less than 100% on the "Taken at least once" stat. Here's some sample code to demonstrate: #include "stdio.h" void foo(int i) { switch(i) { cas
我用switch语句在一些C代码上运行gcov。 我已经编写了测试用例来覆盖通过switch语句的每条可能路径，但它仍然在switch语句中报告没有采用的分支，并且在“采取至少一次”状态下报告的分支不到100％。 这里有一些示例代码来演示： #include "stdio.h" void foo(int i) { switch(i) { case 1:printf("an");break; case 2:printf("bn");break; case 3:printf("cn");break; default: prin
I am trying to use boost::interprocess to allocate a very simple data structure in shared memory but I cannot quite figure out how to use the boost interprocess allocators to perform the memory allocations/deallocations within the shared memory segment which I allocate as follows using namespace boost::interprocess; shared_memory_object::remove("MySharedMem"); mSharedMemory = std::make_unique<
我试图使用boost::interprocess在共享内存中分配一个非常简单的数据结构，但我不能完全弄清楚如何使用boost进程间分配器在我分配的共享内存段中执行内存分配/释放，如下所示 using namespace boost::interprocess; shared_memory_object::remove("MySharedMem"); mSharedMemory = std::make_unique<managed_shared_memory>( open_or_create, "MySharedMem", 65536); 我以前问过类似的问题，但不幸的是我从来没有得到
I'm trying to use Boost::Interprocess for shared memory. However, my program also requires the use of LLVM, and LLVM requires fno-rtti, while Boost::Interprocess requires RTTI. Specifically, I tried the first example here, and I got compilation errors: /usr/include/boost/interprocess/detail/in_place_interface.hpp:50:71: error: cannot use typeid with -fno-rtti : in_place_interface(::boost:
我正在尝试使用Boost :: Interprocess来共享内存。 但是，我的程序也需要使用LLVM，并且LLVM需要fno-rtti，而Boost :: Interprocess需要RTTI。 具体来说，我在这里尝试了第一个例子，并且遇到了编译错误： /usr/include/boost/interprocess/detail/in_place_interface.hpp:50:71：error：can not use typeid with -fno-rtti：in_place_interface（:: boost :: alignment_of :: value，sizeof（T），typeid（T ）。名称（））
I am having a boost interprocess vector which contains string (boost::interprocess::basic_string) as values kept in shared memory, I am getting this error in long run include/boost/interprocess/mem_algo/rbtree_best_fit.hpp:1346: void boost::interprocess::rbtree_best_fit<MutexFamily, VoidMutex, MemAlignment>::priv_deallocate(void*) [with MutexFamily = boost::interprocess::mutex_family; Voi
我有一个包含字符串（boost :: interprocess :: basic_string）的boost进程间向量作为保存在共享内存中的值，我长期得到这个错误 include/boost/interprocess/mem_algo/rbtree_best_fit.hpp:1346: void boost::interprocess::rbtree_best_fit<MutexFamily, VoidMutex, MemAlignment>::priv_deallocate(void*) [with MutexFamily = boost::interprocess::mutex_family; VoidPointer = boost::interprocess::offset_ptr<
Please consider following struct: struct ThingThatWillGoInSharedMemory { boost::optional<int> opt_value; }; I'm using boost::interprocess to create the shared memory area. My understanding of boost::optional was that it was a discriminated union rather than a nullable pointer. As a counter-example things like std::map and std::vector which use the heap need an explicit allocator
请考虑以下结构： struct ThingThatWillGoInSharedMemory { boost::optional<int> opt_value; }; 我使用boost :: interprocess来创建共享内存区域。 我对boost :: optional的理解是它是一个有区别的联合而不是可空指针。 作为一个例子，像std :: map和std :: vector这些使用堆的东西需要一个显式的分配器来在进程间内存中使用它们，但是boost :: optional，我相当肯定不会使用堆并且相当于写作： struct ThingThat
I have a thread in process 1 create a boost::interprocess::managed_shared_memory segment. In this segment I allocate a boost::interprocess::deque using a custom allocator and I create a boost::interprocess::interprocess_mutex and 2 boost::interprocess::interprocess_condition variables using the default allocator. I use the find_or_construct method to create these. I have another process (proc
我有一个线程正在创建一个boost :: interprocess :: managed_shared_memory段。 在这一部分，我使用自定义分配器分配boost :: interprocess :: deque，并使用默认分配器创建boost :: interprocess :: interprocess_mutex和2个boost :: interprocess :: interprocess_condition变量。 我使用find_or_construct方法来创建这些。 我有另一个进程（进程2），它使用我在进程2中打开的boost :: interprocess :: managed_shared_mem
I am trying to create a unordered_map in shared memory using Boost Interprocess library. Here, is the code, which I'm trying to use (taking examples from Boost Interprocess documentation): #include <boost/interprocess/managed_shared_memory.hpp> #include <boost/interprocess/allocators/allocator.hpp> #include <functional> #include <boost/functional/hash.hpp> #include &
我正在尝试使用Boost Interprocess库在共享内存中创建一个unordered_map。 这里是我正在尝试使用的代码（从Boost Interprocess文档中获取示例）： #include <boost/interprocess/managed_shared_memory.hpp> #include <boost/interprocess/allocators/allocator.hpp> #include <functional> #include <boost/functional/hash.hpp> #include <boost/unordered_map.hpp> #include <iostream>
I have a set of templates/functions that allow me to print a tuple/pair assuming that each type in the tuple/pair has operator<< defined for it. Unfortunately, due to 17.4.3.1, it is illegal to add my operator<< overloads to std . Is there another way to get ADL to find my operator<< ? If not, is there any actual harm in wrapping my overload in namespace std{} ? The code fo
我有一套模板/函数，允许我打印一个元组/对，假定元组/对中的每个类型都有operator<<为其定义。 不幸的是，由于17.4.3.1，将我的operator<< overloads添加到std是非法的。 是否有另一种方法让ADL找到我的operator<< ？ 如果没有，在namespace std{}中包装我的重载是否有任何实际的危害？ 任何感兴趣的代码:(我使用gcc-4.5） namespace tuples { using ::std::tuple; using ::std::make_tuple; usin