In C-style increment semantics, the expression ++i is functionally equivalent to which assignment update?

Introduction / Context:
Operators ++ and -- are syntactic sugar in C, C++, Java, and many other languages for common increment/decrement operations. Recognizing their equivalence to explicit assignment helps avoid off-by-one errors and clarifies pre- vs post-increment behavior in expressions.

Given Data / Assumptions:

• Variable i is numeric (typically integer).
• We focus on the effect on i itself, not on expression value in larger contexts.
• Pre-increment (++i) updates i before using it in any surrounding expression.

Concept / Approach:

The core semantics of ++i are to increase the stored value of i by one unit. As a stand-alone statement (++i;) this is equivalent to i = i + 1;. In compound expressions, ++i yields the updated value, while i++ yields the previous value.

Step-by-Step Solution:

Verification / Alternative check:

Consider i = 5; ++i makes i = 6. Using i = i + 1 also makes i = 6. In a context like j = ++i;, j becomes 6; with j = i++; j becomes 5 though i becomes 6 afterward. The update magnitude remains +1.

Why Other Options Are Wrong:

• i = i + 2: Adds two, not one.
• i = i − 1: Decrement, opposite effect.
• i = i + i + i: Triples i, not an increment.
• i = 1 + 1/i: Unrelated arithmetic.

Common Pitfalls:

Confusing pre-increment with post-increment in expressions; assuming ++i always equals i++ in value position—only the side effect on i matches (+1).

Final Answer:

i = i + 1