Question

I am writing a for loop with multiple if statements.

Is it possible that if the if statement (or one part of it) in the for statement evaluates to false, then the loop does not exit but the integer to iterates increments by one and continues through the loop (I need functionality like the continue; keyword).

Example:

for (int i = 0; i <= Collection.Count && Collection[i].Name != "Alan"; i++)
{
    // If name is not Alan, increment i and continue the loop.
} 

Is this possible?

Thanks

Answer 1

You need functionality like the continue keyword – have you considered using the continue keyword, then?

Update: Your example code is hard to decipher the intention of.

for (int i = 0; i <= Collection.Count && Collection[i].Name != "Alan"; i++)
{
    // If name is not Alan, increment i.
} 

The for loop has three parts to it, separated by two semi-colons. The first part initializes the loop variable(s). The second part is an expression that is evaluated each time an iteration is about to start; if it is false, the loop terminates. The third part executes after each iteration.

So your loop above will exit at the first “Alan” it encounters, and also it will increment i every time it finishes an iteration. Finally, if there are no Alans, it will execute the last time with i equal to Collection.Count, which is one larger than the maximum valid index into the collection. So it will throw an exception for sure, as you try to access Collection[i] when i is out of range.

Maybe you want this:

foreach (var item in Collection.Where(i => i.Name != "Alan"))
{
    // item is not an "Alan"
}

You can think of the Where extension method as a way of filtering a collection.

If this seems obscure, you can achieve the same thing with the continue keyword (as you guessed):

foreach (var item in Collection)
{
    if (item.Name == "Alan")
        continue;

    // item is not an "Alan"
}

Or you can just put the code in the if‘s block:

foreach (var item in Collection)
{
    if (item.Name != "Alan")
    {
        // item is not an "Alan"
    }
}