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V1093. The result of the right shift op…
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V1093. The result of the right shift operation will always be 0. The right operand is greater than or equal to the number of bits in the left operand.

Jan 11 2023

The analyzer detected a pointless action: the left operand is shifted bitwise to the right. The shift is done by so many bits that the result is always zero.

Look at the example:

void Metazone_Get_Flag(unsigned short* pFlag, int index)
{
  unsigned char* temp = 0;
  unsigned char flag = 0;

  if (index >= 8 && index < 32)
  {
    temp = (u8*)pFlag;
    flag = (*temp >> index) & 0x01;  // <=
  }
  // ....
}

If you look closely at the 'if' statement's condition, you can notice that the value of the 'index' variable always stays within the range [8 .. 31]. The 'temp' pointer points to an object of the 'unsigned char' type. During the shift operation, the left operand of the 'unsigned char' type is converted to 'int' due to integral promotion. The higher bits are filled with zero values. Therefore, if you shift to the to the right by more bits than there were before the conversion, the result of the operation will be 0.

The code above is meaningless and most likely contains a logic error or a typo.

Note

This warning may be issued for macros that are expanded for corner cases. In other words, such macros don't have an error, and 0 will be the expected result of an expression. If you write such code and don't want the analyzer to issue warnings on it, you can suppress them with a special comment. The comment should contain the name of your macro and the number of the diagnostic rule:

//-V:YOUR_MACRO_NAME:1093

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