xcode 没有匹配的调用函数（期望引用指针而不是指针）
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No matching function for call (expects reference to pointer instead of pointer)
提问by Michael Chinen
I get the error from xcode (3.2.4)/gcc(4.0):
我从 xcode (3.2.4)/gcc(4.0) 得到错误：
/Users/admin/scm/audacity/mac/../src/toolbars/DeviceToolBar.cpp: In member function 'void DeviceToolBar::ShowInputDialog()':
/Users/admin/scm/audacity/mac/../src/toolbars/DeviceToolBar.cpp:817: error: no matching function for call to 'DeviceToolBar::ShowComboDialog(wxChoice*&, wxString)'
/Users/admin/scm/audacity/mac/../src/toolbars/DeviceToolBar.h:74: note: candidates are: void DeviceToolBar::ShowComboDialog(wxChoice*, wxString&)
So it looks like it expects a reference to a pointer in ShowComboDialog, but I don't know why as the signatures are clearly normal pointers. Furthermore if it was expecting a reference to a pointer the way I am calling it should work. This is the first error, and there are no special warnings before it.
所以看起来它希望引用 ShowComboDialog 中的一个指针，但我不知道为什么，因为签名显然是正常的指针。此外，如果它期望对指针的引用，我调用它的方式应该可以工作。这是第一个错误，在此之前没有特殊警告。
Also, this compiles in MSVC 2008 express. Please give me a clue.
此外，这在 MSVC 2008 express 中编译。请给我一个线索。
//in the class def
//(only relevant portions included
class DeviceToolBar:public ToolBar {
public:
DeviceToolBar();
virtual ~DeviceToolBar();
void ShowInputDialog();
private:
void ShowComboDialog(wxChoice *combo, wxString &title);
wxChoice *mInput;
};
//in the cpp file
void DeviceToolBar::ShowInputDialog()
{
ShowComboDialog(mInput, wxString(_("Select Input Device")));
}
void DeviceToolBar::ShowComboDialog(wxChoice *combo, wxString &title)
{
//...
}
回答by Daniel Gallagher
The problem is not the first parameter; its the second. You're passing in a temporary wxString, but the function is expecting a reference. C++ will automatically convert a temporary to a constreference, but it cannot convert it to a reference. You need to make ShowComboDialogtake a const reference as its second parameter.
问题不在于第一个参数；它的第二个。您正在传入一个临时wxString，但该函数需要一个引用。C++ 会自动将临时文件转换为const引用，但不能将其转换为引用。您需要将ShowComboDialogconst 引用作为其第二个参数。
回答by James McNellis
Your ShowComboDialogtakes a wxStringby non-const reference and you are trying to pass a temporary object as an argument to this parameter. You can only bind const references to temporary objects.
您ShowComboDialog采用wxString非常量引用，并且您正尝试将临时对象作为参数传递给此参数。您只能将 const 引用绑定到临时对象。
You either need to change ShowComboDialogto take its second argument either by value (wxString) or by const reference (const wxString&) or you need to create a variable for the wxStringthat you create when you call the function and then pass (a reference to) that variable instead.
您要么需要更改ShowComboDialog为通过值 ( wxString) 或通过常量引用 ( const wxString&)获取其第二个参数，要么需要为wxString调用函数时创建的变量创建一个变量，然后改为传递（引用）该变量。
