问题描述
我有这样的地图：
std :: map< string，CObject> ObjectList;
我有这样的函数：
CObject * NewObject（char * Name，CArg * Arg）
{
std :: string key = Name;
ObjectList [key] = CObject（Name，Arg）;
return& ; ObjectList [Name];
}
我试图编译它抱怨没有默认构造函数
for CObject
所以我将构造函数声明更改为：
CObject :: CObject（char * Name =" Default"，CArg * Arg = NULL）;
只是因为我没有有效的CArg指针就无法创建它。
当我调用NewObject时它会编译崩溃，我看到它的调试器
似乎首先用正确的参数调用构造函数，然后
再次调用它而没有参数（在这种情况下使我的程序崩溃） ，
但我不明白为什么。
如果有人能解释我不知道的是什么站在这里....
另一个快速的问题，如果名称确实& ObjectList [Name]返回NULL
与地图中的键不匹配？
感谢advanec。
Well I have a map like this :
std::map <string, CObject> ObjectList;
I have a function like this :
CObject* NewObject( char* Name, CArg* Arg)
{
std::string key = Name;
ObjectList[ key] = CObject( Name, Arg);
return &ObjectList[Name];
}
I tried to compile and it complained that there was no default constructor
for CObject
So I changed my constructor declaration to :
CObject::CObject( char* Name = "Default", CArg* Arg = NULL);
Just to sse because I can''t create it without a valid CArg pointer.
it compiles when I call NewObject it crashes, with the debugger I saw it
seemed to call the constructor with the correct arguments first, and then
call it again with no arguments (which in this case makes my program crash),
but I don''t understant why.
If someon ecould explain what I don''t understand here....
An other quick question, does &ObjectList[ Name] return NULL if Name does
not match a key in the map ?
Thanks in advanec.
推荐答案
否。它插入一个新对象并返回对它的引用。 RTFM。
Victor
No. It inserts a new object and returns a reference to it. RTFM.
Victor
不，在这种情况下，operator []会在地图中插入一个新的键值对，并且
你获取指向默认构造的CObject实例的指针。
-
David Hilsee
No, in that case, operator[] inserts a new key-value pair into the map and
you get a pointer to the default-constructed CObject instance.
--
David Hilsee
再次，这应该读作复制品。
-
David Hilsee
Again, that should read "to a copy of the".
--
David Hilsee
这篇关于std :: map问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！