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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 mislead the programmer. Not every programmer is aware that the continue operator in the "do { ... } while(0)" loop will terminate the loop instead of continuing it.
This is what the standard has to say 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 will be checked and the loop will terminate because the condition is false.
For example:
int i = 1;
do {
std::cout << i;
i++;
if(i < 3) continue;
std::cout << 'A';
} while(false);The programmer would expect the program to print "12A", but it will actually print "1".
Even if the code was written that way consciously, you'd better change it. For example, you may 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. You can see right away that the loop will terminate if the (i < 3) condition is true. Besides, the analyzer won't generate the warning on this code.
If the code is incorrect, it needs to be rewritten. I cannot give any precise recommendations about that; it all depends on the code execution logic. For instance, if you want to get "12A" printed, you'd better 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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