Some textbooks describe pre-increment/decrement and post-increment/decrement as follows:
Note: Within an expression, the increment/decrement operators may behave differently as part of the operation. If the increment/decrement operator is placed before the variable, the increment or decrement operation is completed before the variable participates in the expression; if the operator is placed after the variable, the increment or decrement operation is completed after the variable participates in the expression.
Is this description correct?
The answer is: no! However, this may be the understanding of most people. Below is a detailed analysis of why this description is incorrect.
In C language, the behavior of the pre-increment (++) and pre-decrement (–) operators is as follows: the value of the sub-expression ++x is equal to x+1, and the side effect is that the value of x is updated to x+1; the value of the sub-expression –x is equal to x-1, and the side effect is that the value of x is updated to x-1. The post-increment (++) and post-decrement (–) first use the current value of x as the value of the sub-expression, and then the side effect updates x to x+1 or x-1.
All side effects (the update of x and the completion of assignment) must be completed before the next sequence point (such as the end of a statement (;) or in a comma expression (,), logical operators && or ||, etc.), but the C standard does not specify the exact timing of side effects.
For example, for
int x = 1; y = ++x + 1;The value of the sub-expression ++x is 2 (i.e., x+1).
The value of the expression ++x + 1 is 2 + 1 = 3.
The value of the assignment expression y = ++x + 1 is 3, assigning 3 to y (side effect).
The side effects (x updated to 2, y assigned to 3) are completed before the sequence point at the end of the statement (;).
After the statement, x = 2, y = 3.
Note: In the same expression, if a variable is modified multiple times (such as through ++x, x++ or assignment), and there are no sequence points separating these operations, it will lead toundefined behavior. For example,
int x = 1; y = ++x + x;The C standard only guarantees that the value of ++x is x + 1. The side effect (x update) is completed before the next sequence point. The value of ++x is 2, but the second access to x may occur before or after the side effect of ++x is completed, leading to y potentially being 3 (2 + 1) or 4 (2 + 2), which isundefined behavior.
Comparison with post-increment.
int x = 1; int y = x++ + 1;The value of the sub-expression x++ is the current value of x (1).
Side effect: x is updated to 2, completed before the sequence point (;).
The expression x++ + 1 evaluates to 1 + 1 = 2.
Assignment y = 2 (side effect).
After the statement, x = 2, y = 2.
Conclusion
Clearly distinguish between the value of the sub-expression ++x (x + 1) and the side effect (x update).
Emphasize that side effects are completed before the sequence point, but the exact timing is undefined.
The above examples clearly illustrate the handling of evaluation and side effects.
Be reminded to avoid undefined behavior (such as modifying a variable multiple times in an expression without sequence points separating them).
Based on the above discussion, the description in textbooks should be modified as follows: (The description is a bit verbose but comprehensive)
In C language expressions, the behavior of increment (++) and decrement (–) operators depends on their pre or post position.
Pre-increment/decrement (such as ++x or –x): The value of the sub-expression ++x is equal to x+1, and the value of –x is equal to x-1, with the side effect updating x to x+1 or x-1.
Post-increment/decrement (such as x++ or x–): The value of the sub-expression x++ or x– is equal to the current value of x, with the side effect updating x to x+1 or x-1.
All side effects (such as the update of x) must be completed before the next sequence point (such as the end of a statement (;), in a comma expression (,), logical operators && or ||, etc.), but the C standard does not specify the exact timing of side effects.
For example, for int x = 1; int y = ++x + 1;
The value of the sub-expression ++x is 2 (i.e., x+1), the expression ++x + 1 evaluates to 3, and is assigned to y.
The side effects (x updated to 2, y assigned to 3) are completed before the sequence point at the end of the statement (;).
After the statement, x = 2, y = 3.
Note: In the same expression, if a variable is modified multiple times (such as through ++x, x++ or assignment), and there are no sequence points separating these operations, it will lead toundefined behavior. For example, the result of the statement y = ++x + x; is unpredictable. Such writing should be avoided to ensure clear code behavior.