分配对变量的引用 [英] Assigning a reference to a variable
问题描述
如果我有一个结构，例如：
struct Settings {
public string str;
public double doub1 ，doub2;
public int i1;
}
如何将i1的引用分配给变量？我可以通过将它作为ref int
传递给函数来轻松完成此操作，但有没有办法将它传递给函数？
11月26日下午4:44，Jon < .wrote：
如果我有结构，例如：
struct Settings {
public string str;
public double doub1，doub2;
public int i1;
}
如何将i1的引用分配给变量？我可以通过将它作为ref int
传递给函数来轻松完成此操作，但有没有办法在不将其传递给函数的情况下执行此操作？
编号如果你想让一个类型表现得像一个引用类型，你应该
使它成为一个引用类型。
Jon
Jon写道：
如果我有结构，例如：
struct Settings {
public string str;
public double doub1，doub2;
public int i1;
}
如何将i1的引用分配给变量？我可以通过将它作为ref int
传递给函数来轻松完成此操作，但有没有办法在不将其传递给函数的情况下执行此操作？
你可以使用不安全的代码来做，但我不建议这样做。如果您使用不安全的代码，则语法与C ++非常相似。
设置集=新设置（）;
sets.i1 = 5;
int * x =& sets.i1;
* x = 6;
控制台.WriteLine（sets.i1.ToString（））;
-
Tom Porterfield
谢谢你的支持回复Jon + Tom。
我来自C背景，但我真的不想使用不安全的代码。
如果你想让一个类型表现得像一个引用类型，你应该把它作为一个引用类型来启动
with。如何将整数作为引用类型？
Jon
" Jon" < .wrote in message news：Oq ************* @ TK2MSFTNGP06.phx.gbl ...
如果我有一个结构，例如：
struct Settings {
public string str;
public double doub1，doub2;
public int i1 ;
}
如何将i1的引用分配给变量？我可以通过将它作为ref int
传递给函数来轻松完成此操作，但有没有办法在不将其传递给函数的情况下执行此操作？
If I have a structure, for example:
struct Settings{
public string str;
public double doub1, doub2;
public int i1;
}
How do I assign the reference of i1 to a variable? I can easily do this by passing it as a ref int
to a function, but is there a way to do this without passing it to a function?
On Nov 26, 4:44 pm, "Jon" <.wrote:If I have a structure, for example:
struct Settings{
public string str;
public double doub1, doub2;
public int i1;
}
How do I assign the reference of i1 to a variable? I can easily do this by passing it as a ref int
to a function, but is there a way to do this without passing it to a function?No. If you want a type to behave like a reference type, you should
make it a reference type to start with.
Jon
Jon wrote:If I have a structure, for example:
struct Settings{
public string str;
public double doub1, doub2;
public int i1;
}
How do I assign the reference of i1 to a variable? I can easily do this by passing it as a ref int
to a function, but is there a way to do this without passing it to a function?You can do it using unsafe code, but I wouldn''t recommend that. If you
use unsafe code, the syntax is very similar to C++.
Settings sets = new Settings();
sets.i1 = 5;
int *x = &sets.i1;
*x = 6;
Console.WriteLine(sets.i1.ToString());
--
Tom Porterfield
Thanks for your replies Jon + Tom.
I''m from a C background, but I don''t really want to use unsafe code.
"If you want a type to behave like a reference type, you should make it a reference type to start
with." How do I make an integer a reference type?
Jon
"Jon" <.wrote in message news:Oq*************@TK2MSFTNGP06.phx.gbl...
If I have a structure, for example:
struct Settings{
public string str;
public double doub1, doub2;
public int i1;
}
How do I assign the reference of i1 to a variable? I can easily do this by passing it as a ref int
to a function, but is there a way to do this without passing it to a function?
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