Difference between ! and ~

Question

I was implementing a simple state machine as follows:

typedef void V_V_StateFunc(void);
static V_V_StateFunc *const fpCurrentStateFunc[STATE_ELEM_SIZE]={

    &fn1,/*0*/

};

void Execute_StateMachine(const U8 bCurrent_StateIndex)
{
    if(bStateIndex == (~bInvert_StateIndex))
    {
        if(bCurrent_StateIndex <= STATE_ELEM_SIZE)
        {
            fpCurrentStateFunc[bCurrent_StateIndex]();
        }
        else
        {
            /*this should never enter*/
        }   
    }
    else
    {
            /*this should never enter*/
    }
}

void Update_State(U8 bNewIndex)
{
    bStateIndex=bNewIndex; //bStateIndex & bInvert_StateIndex are globals
    bInvert_StateIndex=(~bNewIndex);
}

In the function Execute_StateMachine(), if I check bStateIndex == (~bInvert_StateIndex), the value is always 0 and nevers enter the if statement, though bStateIndex=0 and ~bInvert_StateIndex=0 (i.e 0 == 0, which is true). Why is this happening?

If I change bStateIndex == (!bInvert_StateIndex),it enters the if statement.

What is the difference between ~ and ! here?

Answer 1

the ! operator is the logical not operator; for example the expression

if (!variable)

is equal to:

if (variable == 0)

The ~ operator is the bitwise NOT operator; thus changing the value of each bit.

 char variable = 1;           // value: 00000001
 char variable2 = ~variable;  // value: 11111110 = 254  (as noted by @Lundin; value = 11111111 11111111 11111111 11111110 and then truncated to 11111110 upon assignment to char)