Passing an array to function without changing its value?

Question

I want to do something with array A in function B. In function B, some value of A is changed. I know that because A is a pointer so its values will be changed after B is executed. So that I have to use C which is a copy of A to make sure that A aren't changed after B is executed.

I am wonder that is there any other way that I don't have to use the array C?

Answer 1

如果不想更改其值，可以将其设置为const ：

void func( const int * intArray, size_t size) {...}

Answer 2

There are many ways to do it:

1. If you are using C++11, use std::array .
2. Use std::vector<int> to pass array. You can modify local copy and argument array won't change.
3. If you want not to modify value, pass it as const * or const reference to array. For ex const int *arr or const int (&arr)[10]

Answer 3

or you can write something like this:

void func (const int array[10])
{
   // you can work with you data here withot changing an array
}

Answer 4

If you want to pass a pointer to an array and you want to make sure that the array content is not modified inside the function, you can use const :

void DoSomething(const int * data, size_t size)

In C++, you may want to consider a container class like std::vector , and you can pass it using a reference to const (to avoid useless and potentially expensive deep-copies of vector):

void DoSomethingCppStyle(const std::vector<int>& data)

Note that in this case there is no need to pass a size parameter separately, since std::vector knows its own size (returned by std::vector::size() ).

Answer 5

It's not possible to pass an array by value in C++.

You either have to manually make a copy of it and "pass" that, or replace the array with a container that wraps an array such as std::array or std::vector .