Implement in_array in c++

Question

It's easy to find an item in a vector and we know how. It's also possible to find an item in an array if we know the length of the array. But starting from c++11 with the announcement of std:: end(), it looks like we can get the size of the array by calling its pointer, right?

So this code works:

int arr[] = {1,2,3,4,5};
int needle = 3;

std::find(begin(arr), end(arr), needle);
int* result = find(begin(arr), end(arr), needle);
bool exists = result!=end(arr);

However, if this converted to a function:

bool in_array(int arr[], int needle){
    int* result = std::find(begin(arr), end(arr), needle);
    return result!=end(arr);
}

Will not compile:

error: no matching function for call to ‘begin(int*&)’ int* result = std::find( std::begin(arr), std::end(arr), needle);

The goal is to make a working function to send any type of array to it, just like in_array of Php:

template <typename T>
T in_array(T arr[],T needle){
    T* result = std::find(begin(arr), end(arr), needle);
    return result!=end(arr);
}

Why does this code not work? Does C++ compiler store some kind of a local variable to determine the end of arr?

Matthieu Brucher · Accepted Answer · 2019-02-26 13:30:40Z

6

This line:

int arr[] = {1,2,3,4,5};

Doesn't create an array of type int[] but int[5]. You can check this by asking the size of the array, it will return sizeof(int)*5.

But for the function, you don't pass the size (see the error message after array decay), just a pointer, so there is no possibility to get end() for it.

What you can do is pass the size as well (not tested, but should be good):

template <typename T, int N>
bool in_array(T arr[N],T needle){
    T* result = std::find(begin(arr), end(arr), needle);
    return result!=end(arr);
}

edited Feb 26, 2019 at 13:30

answered Feb 26, 2019 at 13:24

Matthieu Brucher

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8 Comments

lubgr Over a year ago

Maybe const T& needle? There shouldn't be any need to assume that copying an array element is inexpensive. Also, the return value should be bool.

Matthieu Brucher Over a year ago

@lubgr agreed for the return type (just copied OP's code), not for the const&.

lubgr Over a year ago

@JafarAkhondali No, its not, have a look at overload #2 here

lubgr Over a year ago

Ok I see, though I'd insist on const& as this is a template, but now it comes to nitpicking, you have my upvote anyway :)

Matthieu Brucher Over a year ago

I agreed, it's a matter of taste. I worked with enough people that don't know the difference that I cannot tell them in which situation to use one or the other and they don't mess up :( For my code, I may use const& as you suggested indeed.

|

bruno · Accepted Answer · 2019-02-26 13:30:08Z

1

error: no matching function for call to ‘begin(int*&)’ int* result = std::find( std::begin(arr), std::end(arr), needle);

having

 int arr[] = {1,2,3,4,5};

the size/number of elements if known, but in

 bool in_array(int arr[], int needle){

the number of element is unknown, array is just known as a pointer to int

edited Feb 26, 2019 at 13:30

answered Feb 26, 2019 at 13:24

bruno

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