How to check if all values in array are the same?

user_10613167 0
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Lets say I have an array short int check[10] = {1,1,1,1,1,1,1,1,1};.
I want to check if all the element are the same.
I can't find the answer in stackoverflow nor google, but I've come across this code in C++.

bool aresame(int a[], int n)
{
    int i;
    unordered_map<int, int> m;

    for (i = 0; i < n; i++)
    {
        m[a[i]]++;
    }
    if (m.size() == 1)
    {
        return true;
    }
    else
    {
        return false;
    }
}

Adjust a bit, and the result is huge error.
My attempt is usingif's but that's very un-professional.
Might as well know, is there are any other ways to do it?

if-statement c arrays
5 Answers

As Gerhardh pointed out in the comments, there is nothing unprofessional about using if. This code should work:

#include <stdbool.h>

bool are_same(int *arr, unsigned int len)
{
    for (int i = 1; i < len; ++i)
        if (arr[0] != arr[i])
            return false;
    return true;
}

You can call the function are_same like this:

int arr[] = {1, 1, 1, 1, 1};
unsigned int len = sizeof(arr) / sizeof(int);
printf("The elements in the array are %s.\n",
       are_same(arr, len) ? "all the same" : "not all the same");
user_149470440

if is perfectly fine, there is nothing unprofessional about it.

I should note that in short int check[10] = {1,1,1,1,1,1,1,1,1}; only 9 elements are 1, the last element will be initialized to 0, so this check will always be false, if you omit the size i.e. check[] = {1,1,1... you won't have this problem because the size of the array will be deduced by the number of elements in the initializer.

#include <stdio.h>
#include <stdbool.h>

bool aresame(short int a[], size_t n) // added value to check
{
    for (size_t i = 1; i < n; i++)
    {
        if(a[i] != a[0])
            return false; // if a different value is found return false
    }
    return true; // if it reaches this line, all the values are the same
}

int main()
{
    short int check[]={1,1,1,1,1,1,1,1,1};
    printf("%s", aresame(check, sizeof check / sizeof *check) ? "true" : "false");
}

Live demo

user_68659320

If you do not like if statements then try this:

bool aresame(int a[], int n) {
    int i = 0;
    while(i<n && a[i]==a[0]) 
        i++;
    return i == n;
}

No need to use extra local storage, just loop until you see an element that is not the same. If you reach the end, everything is fine. Otherwise not.

See here: https://godbolt.org/z/8r6YK6W34

user_130764480

Just for completeness here's a recursive version (no explicit ifs) :

bool aresame(int a[],int n){
    return (n <= 1) || (a[0] == a[n-1] && aresame(a, n-1));
}
user_4489320

Here is a quick and dirty if-less implementation assuming two's complement without padding bits:

#include <stdbool.h>
#include <string.h>

bool are_same(const int *arr, size_t n) {
    return n == 0 || !memcmp(arr, arr + 1, (n - 1) * sizeof(*arr));
}

You can generalize this method to check if the array contains a repeating sequence of length r:

#include <stdbool.h>
#include <string.h>

bool is_repeating(const int *arr, size_t n, size_t r) {
    return n <= r || !memcmp(arr, arr + r, (n - r) * sizeof(*arr));
}
user_45932670
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