V629. Bit shifting of the 32-bit value with a subsequent expansion to the 64-bit type. Consider inspecting the expression.
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;
Note that the V629 diagnostic doesn't refer to 64-bit errors. By 64-bit errors those cases are meant when the 32-bit version of a program works correctly, while the 64-bit version doesn't.
The case we consider here causes an error both in the 32-bit and 64-bit versions. That's why the V629 diagnostic refers to general analysis rules.
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:
You can look at examples of errors detected by the V629 diagnostic.