C++ 变量类型后的“&”含义
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"&" meaning after variable type
提问by vico
Possible Duplicate:
What are the differences between pointer variable and reference variable in C++?
What's the meaning of * and & when applied to variable names?
Trying to understand meaning of "&" in this situation
&在这种情况下试图理解“ ”的含义
void af(int& g)
{
g++;
cout<<g;
}
If you call this function and pass variable name - it will act the same like normal void(int g). I know, when you write >hat means you are passing address of variable g. But what does it means in this sample?
如果你调用这个函数并传递变量名 - 它会像正常一样工作void(int g)。我知道，当你写&g这意味着你正在传递变量的地址g。但它在这个样本中意味着什么？
回答by Luchian Grigore
It means you're passing the variable by reference.
这意味着您正在通过引用传递变量。
In fact, in a declaration of a type, it means reference, just like:
事实上，在一个类型的声明中，它的意思是引用，就像：
int x = 42;
int& y = x;
declares a reference to x, called y.
声明对 的引用x，称为y。
回答by cegfault
The &means that the function accepts the address(or reference) to a variable, instead of the valueof the variable.
这&意味着函数接受变量的地址（或引用），而不是变量的值。
For example, note the difference between this:
例如，注意这两者之间的区别：
void af(int& g)
{
g++;
cout<<g;
}
int main()
{
int g = 123;
cout << g;
af(g);
cout << g;
return 0;
}
And this (without the &):
而这个（没有&）：
void af(int g)
{
g++;
cout<<g;
}
int main()
{
int g = 123;
cout << g;
af(g);
cout << g;
return 0;
}
