The order in which operands in an expression are evaluated is undefined in C except at the sequence points.
Evaluation of an expression may produce side effects. At specific points in the execution sequence called sequence points, all side effects of previous evaluations have completed, and no side effects of subsequent evaluations have yet taken place.
The following are the sequence points defined by C99:
Between the previous and next sequence point an object can only have its stored value modified once by the evaluation of an expression. Additionally, the prior value can be read only to determine the value to be stored.
This rule means that statements such as
i = i + 1; |
are allowed, while statements like
i = i++; |
are not allowed because they modify the same value twice.
In this example, the order of evaluation of the operands to + is undefined.
a = i + b[++i]; |
If i was equal to 0 before the statement, this statement may result in the following outcome:
a = 0 + b[1]; |
Or it may legally result in the following outcome:
a = 1 + b[1]; |
As a result, programs cannot safely rely on the order of evaluatoin of operands between sequence pionts.
These examples are independend on the order of evaluation of the operands and can only be interpreted in one way.
++i; a = i + b[i]; |
Or alternatively:
a = i + b[i+1]; ++i; |
There is no ordering of subexpressions implied by the assignment operator, so the behavior of these statements is undefined.
i = ++i + 1; a[i++] = i; |
These statements are allowed by the standard.
i = i + 1; a[i] = i; |
The order of evaluation of arguments to a function is undefined.
func(i++, i++); |
This solution is appropriate when the programmer intends for both arguments to func() to be equivalent.
i++; func(i, i); |
This solution is appropriate when the programmer intends for the second argument to be one greater than the first.
j = i; j++; func(i, j); |
Component |
Value |
|---|---|
Severity |
2 (medium) |
Likelihood |
2 (probable) |
Remediation cost |
2 (medium) |