在C ++ / CLI中通过引用传递参数,因此重新分配会影响调用者 [英] Pass an argument by reference in C++/CLI so re-assignment affects the caller
问题描述
这可能不是一个困难的问题,但我总是有点困惑,如何处理String类型作为Visual C ++中的参数。我有以下函数:
Probably this is not a difficult question, but I am always a little bit confused on how to treat String type as an argument in Visual C++. I have the following to functions:
void function_1(String ^str_1)
{
str_1 = gcnew String("Test");
}
void function_2()
{
String ^str_2 = nullptr;
function_1(str_2);
}
调用 function_1
, str_2
仍然等于 null
,但我想实现的是 str_2
等于 Test
。那么,如何实现 str_1
的内容传递给 function_2
?
After calling function_1
, str_2
is still equal to null
, but what I want to achieve is that str_2
is equal to Test
. So, how can I achieve that the content of str_1
is passed to function_2
?
感谢您的任何建议。
推荐答案
使用跟踪参考:
void function_1(String ^%str_1)
{
str_1 = gcnew String("Test");
}
说明:传递 String ^
就像传递一个指针一样。仅对引用的本地副本进行更改。 String ^%
就像传递一个引用一样,就像你在调用一个应该改变原始指针的函数时传递一个指针一样。
Explanation: Passing String ^
is like passing a pointer. Changes are only made to the local copy of the reference. String ^%
is like passing a reference to a reference... just as you would pass a pointer to a pointer when calling a function that should change the original pointer.
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