Execution order of conditions in C# If statement

The && and || operators short-circuit. That is:

1) If && evaluates its first operand as false, it does not evaluate its second operand.

2) If || evaluates its first operand as true, it does not evaluate its second operand.

This lets you do null check && do something with object, as if it is not null the second operand is not evaluated.

You should use:

  if (employees != null && employees.Count > 0)
  {
        string theEmployee = employees[0];
  }

&& will shortcircuit and employees.Count will not be execucted if employees is null.

In your second example, the application will throw an exception if employees is null when you attempt to Count the elements in the collection.

http://msdn.microsoft.com/en-us/library/2a723cdk(v=vs.71).aspx

The conditions are checked left to right. The && operator will only evaluate the right condition if the left condition is true.

Section 5.3.3.24 of the C# Language Specification states:

5.3.3.24 && expressions

For an expression expr of the form expr-first && expr-second:

· The definite assignment state of v before expr-first is the same as the definite assignment state of v before expr.

· The definite assignment state of v before expr-second is definitely assigned if the state of v after expr-first is either definitely assigned or “definitely assigned after true expression”. Otherwise, it is not definitely assigned.

· The definite assignment state of v after expr is determined by:

o If the state of v after expr-first is definitely assigned, then the state of v after expr is definitely assigned.

o Otherwise, if the state of v after expr-second is definitely assigned, and the state of v after expr-first is “definitely assigned after false expression”, then the state of v after expr is definitely assigned.

o Otherwise, if the state of v after expr-second is definitely assigned or “definitely assigned after true expression”, then the state of v after expr is “definitely assigned after true expression”.

o Otherwise, if the state of v after expr-first is “definitely assigned after false expression”, and the state of v after expr-second is “definitely assigned after false expression”, then the state of v after expr is “definitely assigned after false expression”.

o Otherwise, the state of v after expr is not definitely assigned.

So this makes it clear that expr-first is always evaluated and if true then, and only then, expr-second is also evaluated.