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Code Analyzer for C#

There is a newer version: 10.2.0.105762

Why is this an issue?
When using the postfix
increment operator, it is important to know that the result of the expression x++ is the value before the operation
x.
This means that in some cases, the result might not be what you expect:

   When assigning x++ to x, it’s the same as assigning x to itself, since the value is assigned before the
  increment takes place 
   When returning x++, the returning value is x, not x+1 

The same applies to the postfix and prefix decrement
operators.
How to fix it
To solve the issue in assignments, eliminate the assignment, since x\++ mutates x anyways.
To solve the issue in return statements, consider using the prefix
increment operator, since it works in reverse: the result of the expression ++x is the value after the operation,
which is x+1, as one might expect.
The same applies to the postfix and prefix decrement
operators.
Code examples
Noncompliant code example
int PickNumber()
{
  int i = 0;
  int j = 0;

  i = i++;      // Noncompliant: i is still 0
  return j--;   // Noncompliant: returns 0
}

Compliant solution
int PickNumber()
{
  int i = 0;
  int j = 0;

  i++;          // Compliant: i is incremented to 1
  return --j;   // Compliant: returns -1
}

Resources
Documentation

   Microsoft Learn - Arithmetic
  operators (C# reference) 

Articles & blog posts

   StackOverflow - "What is the difference between i and i in C#?" - Eric Lippert’s answer