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V1104. Priority of the 'M' operator is higher than that of the 'N' operator. Possible missing parentheses.
The analyzer has detected a potential error: the priority of bitwise shift operations is higher than the priority of '&', '|', and '^' bitwise operations. As a result, the expression may yield a completely different result than a programmer expected.
Let's take a look at an erroneous code example:
unsigned char foo(unsigned char byte2, unsigned char disp)
{
disp |= byte2 & 0b10000000 >> 6;
return disp;
}According to the operator precedence rules in C and C++, the expression is evaluated as follows:
( disp |= ( byte2 & ( 0b10000000 >> 6 ) ) )The bitwise shift of the '0b10000000' mask to the right looks suspicious in this case. Most likely, the programmer expected the result of the bitwise AND to shift by 6.
Here's the correct code:
unsigned char f(unsigned char byte2, unsigned char disp)
{
disp |= (byte2 & 0b10000000) >> 6;
return disp;
}The general recommendation: If the operator precedence in a complex expression is not clear, it is better to wrap part of the expression in parentheses (CERT EXP00-C, ES.41 CppCoreGuidelines). Even if the parentheses turn out to be redundant, it's okay. The code will be easier to understand and less prone to errors.
If you think the analyzer issued a false positive, you can either suppress it with the '//-V1104' comment or wrap the expression in parentheses:
// first option
disp |= byte2 & 0b10000000 >> 6; //-V1104
// second option
disp |= byte2 & (0b10000000 >> 6);Was this page helpful?
