C++ 成员引用基类型“int”不是结构或联合
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C++ Member Reference base type 'int' is not a structure or union
提问by Alex
I'm running into a Problem in my C++ Code.
我在我的 C++ 代码中遇到了一个问题。
I have a union StateValue:
我有一个工会StateValue：
union StateValue
{
int intValue;
std::string value;
};
and a struct StateItem
和一个结构 StateItem
struct StateItem
{
LampState state;
StateValue value;
};
I have a method which goes through a vector of type StateItem
我有一个方法，它通过一个类型的向量 StateItem
for(int i = 0; i < stateItems.size(); i++)
{
StateItem &st = stateItems[i];
switch (st.state)
{
case Effect:
result += std::string(", \"effect\": ") + st.value.value;
break;
case Hue:
result += std::string(", \"hue\": ") + st.value.intValue.str();
break;
case On:
result += std::string(", \"on\": ") + std::string(st.value.value);
break;
default:
break;
}
}
In the case HueI get the following Compiler error:
如果Hue我收到以下编译器错误：
Member reference base type 'int' is not a structure or union
I can′t understand the problem here. Can anyone of you please help me?
我无法理解这里的问题。你们中的任何人都可以帮助我吗？
采纳答案by Mike Seymour
You're trying to call a member function on intValue, which has type int. intisn't a class type, so has no member functions.
你试图调用一个成员函数intValue，它有型int。int不是类类型，因此没有成员函数。
In C++11 or later, there's a handy std::to_stringfunction to convert intand other built-in types to std::string:
在 C++11 或更高版本中，有一个方便的std::to_string函数可以将int其他内置类型转换为std::string：
result += ", \"hue\": " + std::to_string(st.value.intValue);
Historically, you'd have to mess around with string streams:
从历史上看，您必须处理字符串流：
{
std::stringstream ss;
ss << st.value.intValue;
result += ", \"hue\": " + ss.str();
}
回答by Juan Ramirez
Member reference base type 'int' is not a structure or union
Member reference base type 'int' is not a structure or union
intis a primitive type, it has no methods nor properties.
int是原始类型，它没有方法或属性。
You are invoking str()on a member variable of type intand that's what the compiler is complaining about.
您正在调用str()类型的成员变量，int这就是编译器所抱怨的。
Integers cannot be implicitly converted to string, but you can used std::to_string()in C++11, lexical_castfrom boost, or the old-slow approach of the stringstream.
整数不能隐式转换为字符串，但您可以std::to_string()在 C++11、lexical_castfromboost或stringstream.
std::string to_string(int i) {
std::stringstream ss;
ss << i;
return ss.str();
}
or
或者
template <
typename T
> std::string to_string_T(T val, const char *fmt ) {
char buff[20]; // enough for int and int64
int len = snprintf(buff, sizeof(buff), fmt, val);
return std::string(buff, len);
}
static inline std::string to_string(int val) {
return to_string_T(val, "%d");
}
And change the line to:
并将该行更改为：
result += std::string(", \"hue\": ") + to_string(st.value.intValue);
回答by scraatz
Your intvalue is no object. It has no member functions. You could use sprintf() or itoa() to convert it to a string.
您的 intvalue 不是对象。它没有成员函数。您可以使用 sprintf() 或 itoa() 将其转换为字符串。
回答by Yves Daoust
intValueis an int, it has no methods.
intValue是一个int，它没有方法。
