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V1019. Compound assignment expression i…
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V1019. Compound assignment expression is used inside condition.

08 Jui 2018

The analyzer has detected a compound assignment operator used inside an expression of type 'bool'.

Consider the following example:

if (adj_number == (size_t)(roving->adj_count - 1) &&
    (total_quantity += quantity_delta) < 0) 
{
    /* ... */
}

In this case, the 'total_quantity' variable is changed inside the conditional expression of the 'if' statement. What makes this code even more suspicious is the fact that 'total_quantity' will change only if the condition to the left of the '&&' operator is true. This is probably a typo, and the programmer must have intended to use the addition operator '+' rather than the compound assignment operator '+=':

if (adj_number == (size_t)(roving->adj_count - 1) &&
  (total_quantity + quantity_delta) < 0) 
{
    /* ... */
};

Even if the initial snippet is correct, it is still strongly recommended that you avoid writing complex expressions like that. Their logic is quite complicated, and programmers could easily make a mistake when modifying such code.

The analyzer is not always able to tell if the code detected by this diagnostic is really faulty or if the programmer was simply trying to make it shorter. So, we reviewed a lot of open-source projects and singled out a few programming patterns where such constructs are harmless. To reduce the number of false positives, we set the diagnostic to keep silent in the following cases:

  • The left operand of the compound assignment operator is a pointer;
  • The compound assignment operator is part of a macro;
  • The compound assignment operator is inside a loop body.

If you get too many false positives on your project, you can disable this diagnostic or use the false-positive suppression means.

This diagnostic is classified as:

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