表达式必须有指向类类型的指针 expression must have pointer-to-class type 表达式必须有指向类类型的指针
意思是 表达必须有由点到类的类型 请采纳 谢谢
待解决 悬赏分:1 - 离问题结束还有 3 IntelliSense: expression must have pointer-to-class type问题补充:匿名 2013-05-23 12:21:38 3智能感知:表达式必须具有指针到类类型 匿名 2013-05-23 12:23:18 3 IntelliSense :表示必须有尖对班的类型 匿名 2013-05-23 12:24:58 3 IntelliSense : 表...
USER\usartcmd.c(121): error: #109: expression must have (pointer-to-) function type USER\us...
C++中 Expression Must Have Class Type 错误的原因 当通常用于访问对象成员的点 (.) 运算符用于指向对象的指针时,会发生表达式必须具有类类型错误。 这样想吧。 您可以在指向对象的指针上使用点 (.) 运算符。 现在,像平常一样工作,点 (.) 运算符将尝试查找指针成员(字段或方法)。
aIf I just believe lt,there's nothing to it 如果我相信lt,没什么到它[translate] aCurrent of the thermal datasheet for SP1 is not in NDA website yet. 热量datasheet的潮流为SP1不在NDA网站。[translate] aExpression must have pointer type 表示必须有尖类型[translate]...
pd is a pointer to the start of the array. pd[i] is an element of the array. In this case, you would use the "." operator instead of the "->" operator. Why do you think there are 8 elements in the array? pd[i] has type pointer to PictureBoxArray. Why do you think it ...
This produces the error message "error: expression must have class type" with the finger firmly pointed at *MBox_[i] Later assignments to the dereferenced pointer produce the same message. I'm fairly sure that the dereferencing is correct. The editor kindly suggest...
Use class c2 { public: int i; }; class c1 { public: c2 func() { return c2(); } }; void main(void) { std::shared_ptr<c1> ptr = std::make_shared<c1>(); ptr->func()->i; } Bug: "expression must have a pointer type" is shown on the "ptr", whi...
Thus, can we say that the pointer is trying to find something that does not even exist? Absolutely yes, and that is why we get the error. Fix for the Expression Must Have Class Type Error in C++ We can fix theexpression must have class type errorin C++ in two ways. Let us discuss...