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For i 0 why is i i equal to 0

19 September 2026 · 9 min read

For i 0 why is i i equal to 0

Understanding the intricacies of programming languages often involves delving into seemingly simple operations that yield unexpected results. One such scenario arises when dealing with the statement i += i++ when i is initially set to 0. Many programmers, especially those new to languages like C++, Java, or JavaScript, are puzzled by why this operation results in i remaining 0, rather than incrementing to 1 or some other value. This behavior stems from the order of operations and how the post-increment operator interacts with the assignment operator. In this article, we’ll dissect this peculiar phenomenon, exploring the underlying mechanisms that govern this outcome, and provide a clear explanation of why for i = 0, why is (i += i++) equal to 0?

Dissecting the Post-Increment Operator

The post-increment operator (i++) is a fundamental concept in many programming languages. It’s crucial to understand that this operator performs two distinct actions: it returns the current value of the variable i, and then it increments the value of i by 1. The key is that the value returned is the original value before the increment. This behavior is often the source of confusion, particularly when combined with other operators like +=. To illustrate, consider this simple example:

If we have i = 5 and then execute j = i++, the variable j will be assigned the value 5 (the original value of i), and then i will be incremented to 6. This sequence of events is critical to understanding the behavior of i += i++ when i starts at 0. The post-increment operator’s effect is delayed until after its original value is used in the larger expression. According to research from the University of Cambridge, understanding operator precedence is key to avoiding common programming errors. Cambridge Computer Laboratory offers resources on this subject.

The post-increment operator is especially useful in loop constructs and other situations where you need to use the current value of a variable and then immediately increment it. However, its behavior can lead to unexpected results if not carefully considered in more complex expressions.

Analyzing i += i++ When i = 0

Let’s break down what happens when we execute i += i++ when i is initially 0. This statement is essentially shorthand for i = i + (i++). The expression i++ first evaluates to the current value of i, which is 0. Then, i is incremented to 1. However, the addition operation i + (i++) uses the original value of i++, which was 0. Therefore, the expression becomes i = i + 0. Since i was incremented to 1 after the i++ evaluation but before the final assignment, we are essentially reassigning it to 1 + 0, which equals 1. However, most compilers will optimize this operation. In many implementations, especially in older versions of C++ and Java, the assignment happens before the increment can take full effect in the overall expression due to the way temporary variables are handled.

The featured snippet optimized paragraph: The core reason i += i++ evaluates to 0 when i initially equals 0 is due to the combination of post-increment behavior and the assignment operation. The post-increment operator returns the original value (0) for the += operation, and then increments i. The += operation effectively adds 0 to the original value of i (which is still considered 0 at that point in the calculation), thus resulting in i being assigned the value 0. This is a direct consequence of the order in which the compiler evaluates the expression. This nuance is important to grasp for all developers.

Here’s a breakdown of the steps:

  1. i is initialized to 0.
  2. The expression i += i++ is encountered.
  3. i++ evaluates to 0 (the original value of i), and i is then incremented to 1.
  4. The expression becomes i = i + 0.
  5. i is assigned the value of i + 0, which, depending on the compiler’s optimization, might result in i being set back to 0, or remain as 1 for a very short time before being overwritten.

Compiler Optimizations and Undefined Behavior

It’s important to note that the behavior of expressions like i += i++ can be highly dependent on the specific compiler and the language standard being used. In some cases, the behavior might be considered “undefined,” meaning that the compiler is free to generate any code it sees fit, and the result can vary from one compilation to another. This is because modifying the same variable multiple times within a single expression without intervening sequence points can lead to ambiguity about the order in which the modifications are applied. The concept of “sequence points” defines the points in the code’s execution where all side effects of previous evaluations are guaranteed to be complete, and no side effects of subsequent evaluations have yet taken place. The absence of a clear sequence point between the post-increment and the addition introduces uncertainty.

Modern compilers often implement optimizations that can further complicate the analysis. For instance, a compiler might recognize that the incremented value of i is immediately overwritten by the assignment, and therefore optimize away the increment altogether. This is especially true if the compiler’s optimization level is set to prioritize speed over strict adherence to the source code’s apparent intent. Due to these complexities, it’s generally best practice to avoid writing code that relies on the specific behavior of such expressions. According to coding standards from organizations like MISRA, such constructs should be avoided to ensure code predictability and maintainability. MISRA Guidelines offer detailed advice on avoiding undefined behavior.

Best Practices and Alternatives

Given the potential for confusion and undefined behavior, it’s generally recommended to avoid using expressions like i += i++. Instead, it’s much clearer and safer to separate the increment and assignment operations into distinct statements. This eliminates any ambiguity about the order in which the operations are performed and makes the code easier to understand and maintain.

Here are some reasons to avoid this construct:

  • It leads to code that is difficult to read and understand.
  • It can result in unexpected behavior due to compiler optimizations.
  • It violates coding standards that promote clarity and maintainability.

Instead of i += i++, consider using the following alternatives:

  • i = i + 1; // Increment i
  • i++; // Increment i (separate statement)

These alternatives are much more explicit and leave no room for interpretation. They also avoid the potential for undefined behavior that can arise from modifying the same variable multiple times within a single expression. For a deeper dive into coding best practices, consider resources from organizations like Codingame, which offer interactive coding challenges and tutorials.

FAQ

Why doesn't i += i++ simply increment i to 1?
Because the post-increment operator first returns the original value of i (which is 0) for the += operation, and then increments i. The addition uses the original 0 value.
Is the behavior of i += i++ the same across all programming languages?
No, the behavior can vary depending on the language standard and the compiler's implementation. Some languages might define the behavior more precisely than others.
What does "undefined behavior" mean in this context?
It means that the compiler is free to generate any code it sees fit, and the result can vary from one compilation to another. It's best to avoid code that leads to undefined behavior.
In summary, the expression for i = 0, why is (i += i++) equal to 0? illustrates a subtle but important aspect of programming languages: the interaction between operators and the order in which operations are performed. The post-increment operator, combined with the assignment operator, can lead to unexpected results if not carefully understood. The complexities of compiler optimizations and the potential for undefined behavior further emphasize the importance of writing clear and maintainable code. By avoiding ambiguous expressions and opting for more explicit alternatives, you can ensure that your code behaves as intended and is easy to understand by others. This is especially crucial in collaborative projects where clarity and predictability are paramount.

If you found this explanation helpful, consider exploring other articles on operator precedence, compiler optimizations, and coding best practices. Understanding these fundamental concepts can significantly improve your programming skills and help you avoid common pitfalls. Check out our other articles on advanced programming topics to deepen your knowledge and become a more proficient developer.

Question & Answer :
Take the following code (usable as a Console Application):

static void Main(string[] args) { int i = 0; i += i++; Console.WriteLine(i); Console.ReadLine(); }

The result of i is 0. I expected 2 (as some of my colleagues did). Probably the compiler creates some sort of structure that results in i being zero.

The reason I expected 2 is that, in my line of thought, the right hand statement would be evaluated first, incrementing i with 1. Than it is added to i. Since i is already 1, it is adding 1 to 1. So 1 + 1 = 2. Obviously this is not what’s happening.

Can you explain what the compiler does or what happens at runtime? Why is the result zero?

Some-sort-of-disclaimer: I’m absolutely aware you won’t (and probably shouldn’t) use this code. I know I never will. Nevertheless, I find it is interesting to know why it acts in such a way and what is happening exactly.

This:

int i = 0; i += i++

Can be seen as you doing (the following is a gross oversimplification):

int i = 0; i = i + i; // i=0 because the ++ is a postfix operator and hasn't been executed i + 1; // Note that you are discarding the calculation result

What actually happens is more involved than that - take a look at MSDN, 7.5.9 Postfix increment and decrement operators:

The run-time processing of a postfix increment or decrement operation of the form x++ or x– consists of the following steps:

  • If x is classified as a variable:

    • x is evaluated to produce the variable.
    • The value of x is saved.
    • The selected operator is invoked with the saved value of x as its argument.
    • The value returned by the operator is stored in the location given by the evaluation of x.
    • The saved value of x becomes the result of the operation.

Note that due to order of precedence, the postfix ++ occurs before +=, but the result ends up being unused (as the previous value of i is used).

A more thorough decomposition of i += i++ to the parts it is made of requires one to know that both += and ++ are not atomic (that is, neither one is a single operation), even if they look like they are. The way these are implemented involve temporary variables, copies of i before the operations take place - one for each operation. (I will use the names iAdd and iAssign for the temporary variables used for ++ and += respectively).

So, a closer approximation to what is happening would be:

int i = 0; int iAdd = i; // Copy of the current value of i, for ++ int iAssign = i; // Copy of the current value of i, for += i = i + 1; // i++ - Happens before += due to order of precedence i = iAdd + iAssign;