V564. The '&' or '|' operator is applied to bool type value. You've probably forgotten to include parentheses or intended to use the '&&' or '||' operator.

The analyzer detected a potential error: operators '&' and '|' handle bool-type values. Such expressions are not necessarily errors but they usually signal misprints or condition errors.

Consider this sample:

int a, b;
#define FLAG 0x40
if (a & FLAG == b)

This example is a classic one. A programmer may be easily mistaken in operations' priorities. It seems that computing runs in this sequence: "(a & FLAG) == b". But actually it is "a & (FLAG == b)". Most likely, it is an error.

The analyzer will generate a warning here because it is odd to use the '&' operator for variables of int and bool types.

If it turns out that the code does contain an error, you may fix it the following way:

if ((a & FLAG) == b)

Of course, the code might appear correct and work as it was intended. But still you'd better rewrite it to make it clearer. Use the && operator or additional brackets:

if (a && FLAG == b)
if (a & (FLAG == b))

The V564 warning will not be generated after these corrections are done while the code will get easier to read.

Consider another sample:

#define SVF_CASTAI 0x00000010
if ( !ent->r.svFlags & SVF_CASTAI ) {

Here we have an obvious error. It is the "!ent->r.svFlags" subexpression that will be calculated at first and we will get either true of false. But it does not matter: whether we execute "true & 0x00000010" operation or "false & 0x00000010" operation, the result will be the same. The condition in this sample is always false.

This is the correct code:

if ( ! (ent->r.svFlags & SVF_CASTAI) )

Note. The analyzer will not generate the warning if there are bool-type values to the left and to the right of the '&' or '|' operator. Although such code does not look too smart, still it is correct. Here is a code sample the analyzer considers safe:

bool X, Y;
if (X | Y)
{ ... }

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