The analyzer has detected a potential error in an expression containing a shift operation: a 32-bit value is shifted in the program. The resulting 32-bit value is then explicitly or implicitly cast to a 64-bit type.
Consider an example of incorrect code:
unsigned __int64 X; X = 1u << N;
This code causes undefined behavior if the N value is higher than 32. In practice, it means that you cannot use this code to write a value higher than 0x80000000 into the 'X' variable.
You can fix the code by making the type of the left argument 64-bit.
This is the correct code:
unsigned __int64 X; X = 1ui64 << N;
The analyzer will not generate the warning if the result of an expression with the shift operation fits into a 32-bit type. It means that significant bits don't get lost and the code is correct.
This is an example of safe code:
char W = 7; long long Q = W << 10;
The code works in the following way. At first, the 'W' variable is extended to the 32-bit 'int' type. Then a shift operation is performed and we get the value 0x00001C00. This number fits into a 32-bit type, which means that no error occurs. At the last step this value is extended to the 64-bit 'long long' type and written into the 'Q' variable.
This diagnostic is classified as: