Why is the Output of Multiple Post Increments in a C Expression Unpredictable?

Behavior of Post Increment in C

In C , post increment (e.g., i ) increments a variable but returns the original value. Understanding its behavior is crucial in complex expressions.

Consider the following code:

<code class="cpp">int i = 5;

cout << i++ << i-- << ++i << --i << i << endl;</code>

This statement evaluates the expression i i-- i --i i before outputting the result. However, the order of evaluation is undefined, leading to unpredictable output (e.g., "45555").

Let's break down the sequence point rule:

• Sequence Points: In C , certain points in the code force the evaluation of pending operations. These points include semicolons (;), commas (,), and the end of the statement.
• Undefined Behavior: In the above code, there are unsequenced side effects on the scalar variable i due to post increment operations. This results in undefined behavior.

Example:

<code class="cpp">int x = 20, y = 35;

x = y++ + y + x++ + y++;</code>

This expression evaluates in the following order:

1. y : Increments y to 36 and returns 35 (the original value of y).
2. x : Increments x to 21 and returns 20 (the original value of x).
3. y : Increments y to 37 and returns 36 (the original value of y).
4. x : Increments x to 22 and returns 21 (the original value of x).

Therefore, the final value of x is 126 (35 36 20 21), while y is 37.

Conclusion:

Post increment in C can lead to undefined behavior when used in unsequenced expressions. It's essential to understand sequence points and avoid side effects on the same variable within an unsequenced context.

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