I read the page Passing Arrays Using ref and out (C# Programming Guide) and was wondering why we would need to define an array parameter as a ref parameter when it is already a reference type. Won't changes in the callee function be reflected in the caller function?
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问题:
回答1:
Won't changes in the callee function be reflected in the caller function?
Changes to the contents of the array would be reflected in the caller method - but changes to the parameter itself wouldn't be. So for example:
public void Foo(int[] x)
{
// The effect of this line is visible to the caller
x[0] = 10;
// This line is pointless
x = new int[] { 20 };
}
...
int[] original = new int[10];
Foo(original);
Console.WriteLine(original[0]); // Prints 10
Now if we changed Foo
to have a signature of:
public void Foo(ref int[] x)
and changed the calling code to:
Foo(ref original);
then it would print 20.
It's very important to understand the difference between a variable and the object that its value refers to - and likewise between modifying an object and modifying a variable.
See my article on parameter passing in C# for more information.
回答2:
If you only plan to change the contents of the array, then you're correct. However, if you plan on changing the array itself, then you must pass by reference.
For example:
void foo(int[] array)
{
array[0] = 5;
}
void bar(int[] array)
{
array = new int[5];
array[0] = 6;
}
void barWithRef(ref int[] array)
{
array = new int[6];
array[0] = 6;
}
void Main()
{
int[] array = int[5];
array[0] = 1;
// First, lets call the foo() function.
// This does exactly as you would expect... it will
// change the first element to 5.
foo(array);
Console.WriteLine(array[0]); // yields 5
// Now lets call the bar() function.
// This will change the array in bar(), but not here.
bar(array);
Console.WriteLine(array[0]); // yields 1. The array we have here was never changed.
// Finally, lets use the ref keyword.
barWithRef(ref array);
Console.WriteLine(array[0]); // yields 5. And the array's length is now 6.
}