Error: control may reach end of non-void function

Question

I cannot figure out why this error is happening: error: control may reach end of non-void function

Here is the code:

bool search(int value, int values[], int n) {

    if (n < 1) {
        return false;
    }   

    for (int i = 0; i < n; i++) {
        if (values[i] == value) {
            return true;
            break;
        }
        else { 
            return false;
        }
    }    
}

I understand that the error means that the function may reach the end without returning anything, but I cannot figure out how that might happen.

Answer 1

Your code is equivalent to

return (n > 0 && values [0] == value);

Either you are in the habit of writing very simple things in an excessively complicated way, or that code doesn't do what you want it to do.

Answer 2

You are getting this error because if your for loop breaks due to breaking condition i < n; then it don't find any return statement after for loop (see the below, I mentioned in code as comment).

for (int i = 0; i < n; i++){
    if (values[i] == value){
        return true;
        break;
    }
    else{ 
        return false;
    }
}
  // here you should add either return true or false     
}

If for loop break due to i >= n then control comes to the position where I commented and there is no return statement present. Hence you are getting an error "reach end of non-void function in C".

Additionally, remove break after return statement. if return executes then break never get chance to execute and break loop.

   return true;  -- it returns from here. 
    break;  -- " remove it it can't executes after return "

Check your compiler should give you a warning - 'unreachable code'.

Answer 3

That compiler warning is not correct. Anyway, there is a bigger problem with your code:

bool search(int value, int values[], int n) {

    if (n < 1) {
        return false;
    }   

    for (int i = 0; i < n; i++) {
        if (values[i] == value) {
            return true;
            break;
        }
        else {            // !
            return false; // ! <-- Here is the mistake.
        }                 // !
    }    
}

This code only checks values[0] == value and then always returns. It's happening because of that else {return false;}.

You should write it this way:

bool search(int value, int values[], int n) {

    if (n < 1) {
        return false;
    }   

    for (int i = 0; i < n; i++) {
        if (values[i] == value) {
            return true;
            // break;  <- BTW, it's redundant.
        }
    }    
    return false;
}

Now, function checks entire values array and then returns false if there was no matches. But if it found a match, it will instantly return true witout checking other elements.

Also, compiler will not emit a warning for this code.

Answer 4

Some folks will probably hate this, but....

bool search(int value, int values[], int n) {

   if (n < 1) {
      return false;
   }   

   bool ret = false;
   for (int i = 0; i < n; i++) {
      if (values[i] == value) {
         ret = true;
         break;
      }
   }    
   return ret;
}