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V696. The 'continue' operator will terminate 'do { ... } while (FALSE)' loop because the condition is always false.
The analyzer has detected code that may be misleading The continue operator in the do { ... } while(0) loop terminates the loop instead of continuing it.
This is what the standard says about it:
§6.6.2 in the standard: "The continue statement (...) causes control to pass to the loop-continuation portion of the smallest enclosing iteration-statement, that is, to the end of the loop." (Not to the beginning.)
Thus, after calling the continue operator, the (0) condition is checked, and the loop terminates because the condition is false.
The example:
int i = 1;
do {
std::cout << i;
i++;
if(i < 3) continue;
std::cout << 'A';
} while(false);A developer would expect the program to print 12A, but it will actually print 1.
Even if the code was intentionally written this way, it is better to revise it. For example, you can use the break operator:
int i=1;
do {
std::cout << i;
i++;
if(i < 3) break;
std::cout << 'A';
} while(false);The code looks clearer now: the loop will terminate if the (i < 3) condition is true. The analyzer will not issue the warning for this code.
If the code is incorrect, it needs to be rewritten. There are no specific recommendations for this—everything depends on the code execution logic. For example, if you want to print 12A, it is better to write the following code:
for (i = 1; i < 3; ++i)
std::cout << i;
std::cout << 'A';This diagnostic is classified as:
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You can look at examples of errors detected by the V696 diagnostic. |
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