How do I fix this warning - "control reaches end of non-void function [-Wreturn-type]"-CodePudding

During compiling, it shows this warning - control reaches end of non-void function [-Wreturn-type]. I googled and found that this warning shows when you don't return anything in the function. But I couldn't figure out where's the error in my code.

Here's my code:

#include <iostream>
#include <algorithm>
using namespace std;

int findUnique(int *a, int n){
    sort(a, a n);
    int i=0;
    while(i<n){
        if(a[i]==a[i 1]){
            i  = 2;
        }
        else{
            return a[i];
        }
    }
}

int main(){

    int t;
    cin >> t;

    while (t--){

        int size;
        cin >> size;
        int *input = new int[size];

        for (int i = 0; i < size;   i)
        {
            cin >> input[i];
        }

        cout << findUnique(input, size) << endl;
    }

    return 0;
}

CodePudding user response：

You have to know why this warning is shown to understand what to do about it, this warning is shown when your function has a return type but you haven't returned value from one or more exit points of a function. Now see in your function, you return a[i] but consider a situation where your code doesn't go in the else block at all. So after coming out of the while block. There is no return statement therefore compiler is throwing control reaches the end of non-void function [-Wreturn-type].

CodePudding user response：

The function returns nothing in case when the array does not contain a unique number or when the parameter n is equal to 0.

So the compiler issues the warning message.

Moreover the while loop can invoke undefined behavior when i is equal to n-1 due to using a non-existent element with the index n in this if statement

    if(a[i]==a[i 1]){

Also there is a logical error. The if statement

    if(a[i]==a[i 1]){
        i  = 2;
    }
    else{
        return a[i];
    }

does not guarantee that indeed a unique number will be returned.

Using your approach when it is allowed to change the original array by calling the algorithm std::sort the function can be defined for example the following way

size_t findUnique( int *a, size_t n )
{
    std::sort( a, a   n );

    size_t i = 0;

    bool unique = false;

    while ( !unique && i != n )
    {
        size_t j = i  ;

        while ( i != n && a[i] == a[j] ) i  ;

        unique = i - j == 1;
    }

    return unique ? i - 1 : i;
}

And in main the function can be called like

size_t pos = findUnique(input, size);

if ( pos != size )
{ 
   cout << input[pos] << endl;
}
else
{
    // output a message that there is no unique number
}

Pay attention to that your program produces multiple memory leaks. You need to free the allocated memory in each iteration of the while loop.

CodePudding user response：

The problem is in the function findUnique This function is supposed to return int no matter what although in your code you are returning an integer only under certain conditions

Here is a possible fix:

// return true if unique number found
// return false otherwise
bool findUnique(int *a, int n, int *unique){
    sort(a, a n);
    int i=0;
    while(i<n){
        if(a[i]==a[i 1]){
            i  = 2;
        }
        else{
            *unique = a[i];
            return true;
        }
    }

    return false;
}

Then in the main something like that:

int unique;
bool uniqueFound = findUnique(input, size &unique);
if (uniqueFound == true)
    cout << unique << endl;
else
    cout << "No unique number found" << endl;