WebOct 30, 2015 · 解决问题的最终办法，就是显式地告诉编译器，T::bar是一个类型名。 这就必须用typename关键字 1 template 2 void foo ( const T& t) 3 { 4 // 声明一个指向某个类型为T::bar的对象的指针 5 typename T::bar * p; 6 } 这样，编译器就确定了T::bar是一个类型名，p也就自然地被解释为指向T::bar类型的对象的指针了 （以上来源网 …Web"qualified name is not allowed" . 我只导入了 boost 库。 我已经忙了几个小时这些错误，如果有人能告诉我这个错误的可能原因是什么，那将是一个很大的帮助。 最佳答案 确保您使用 C++17 进行编译，因为您的 header 使用嵌套的命名空间说明符 (例如 namespace Utils::iterators { ... } )。 这可以通过 -std=c++17 来完成GCC/clang 的标志，或 …
type name not allowed - C++ Forum - cplusplus.com
WebJul 22, 2013 · Member access operators only allow data members, member functions, and enumerators on the right. Type names are not allowed. (the standardese for this is "If E2 is a nested type, the expression E1.E2 is ill-formed" in § 5.2.5 [expr.ref]/4) Now if you replace that with Graph::edge, which names a type, it will compile. Remove typedef from typedef struct Address * next. Also, you can simply declare a pointer *Start like so; Adressbook *Start = NULL; You also need to allocate it memory. Start->next = NULL; will give a SEGV fault. Here; Start = (Adressbook *) malloc (sizeof (Adressbook)); Then you can make assignments to its members. Share Improve this answer Follow green flash sanibel
C++ Error: Type Name is Not Allowed - Stack Overflow
WebFeb 25, 2014 · That is true, but not what I want for the job that I'm doing. My reasons are not yours, so of course y ou might not see the point, but that's not the point. The point is that the IDE is responding inappropriately. Thanks for your interest. WebFeb 21, 2013 · You can't include the typename when calling a function. Change to: if (age>59) senior (pAge); else everyoneElse (pAge); Share Improve this answer Follow … WebC++ 错误 : Type Name is Not Allowed 标签 c++ 我正在尝试学习指针参数中的新类(class)，我想让函数 senior 和 everyoneElse 接受指针 x，但是当我尝试使用指针 pAge …green flash sanibel island