C++ “错误：请求'a'中的成员'size'，它是指针类型”但我不认为它是一个指针
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"error: request for member ‘size’ in ‘a’, which is of pointer type" but I didn't think it was a pointer
提问by Jamie Twells
So, I thought I was trying to do something simple, but apparently not...
所以，我以为我正在尝试做一些简单的事情，但显然不是......
I wrote this function so I could extend it later and have a quick way to give the user a menu when required by going menu(mystrings):
我写了这个函数，所以我可以稍后扩展它，并有一个快速的方法在需要时为用户提供一个菜单menu(mystrings)：
int menu(string a[]) {
int choice(0);
cout << "Make a selection" << endl;
for(int i=0; i<a.size(); i++) {
cout << i << ") " << a[i] << endl;
}
cin >> choice;
cout << endl;
return choice;
}
But for some reason I get:
但出于某种原因，我得到：
main.cpp: In function ‘int menu(std::string*)':
main.cpp:38:12: error: request for member ‘size' in ‘a', which is of pointer type ‘std::string* {aka std::basic_string<char>*}' (maybe you meant to use ‘->' ?)
int n = a.size();
when I try compiling. Could anyone translate that error for me and explain what ->is, thank you.
当我尝试编译时。任何人都可以为我翻译该错误并解释是什么->，谢谢。
回答by yizzlez
You are passing an array of stringsand trying to call size()on the array. Arrays degenerate to pointers when passed to a function, which explains your error.
您正在传递一个数组strings并尝试调用size()该数组。数组在传递给函数时退化为指针，这解释了您的错误。
The ->operator, or "arrow operator" (name I use), is just shorthand for (*obj).func(). This is useful if you have a pointer to a class object. Example:
该->经营者，或“箭经营者”（名字我使用），只是简写(*obj).func()。如果您有一个指向类对象的指针，这很有用。例子：
string *s = &someotherstring;
s->size(); //instead of (*s).size(), saves keystrokes
