为什么这种情况总是如此？ （pktNum！= invPktNum）

Question

When I compile the following code, my compiler complains that the following line is always true. I think I may have a faulty understanding of the != operator...

if (pktNum != ~invPktNum) {
    return 1;
}

I'm trying to verify that invPktNum is indeed the inverse of pktNum. If not, exit immediately, otherwise proceed as normal.

I've checked and pktNum an unsigned char that is 0x01 and invPktNum is an unsigned char that is 0xFE at the time of comparison.

Can anybody enlighten me? Thanks in advance!

Answer 1

In C, values in most expressions that are of a type narrower than int are promoted to a wider type before the calculation takes place. If int is wide enough to hold all the values of the narrower type, then it is promoted to int ; otherwise it is promoted to unsigned int .

In this case, int is wide enough to hold all the values of your unsigned char , so your values are promoted to int .

pktNum (and thus, the promoted pktNum ) can have a value between 0 and 255 inclusive. This is the value that will be used on the left hand side of the != operator.

invPktNum can likewise have a value between 0 and 255 inclusive. This value will be promoted to int , and then bitwise-negated. The result of this bitwise negation will always be a negative number, since the sign bit will be negated. This is the value that will be used on the right hand side of the != operator.

No negative number can ever be equal to the promoted pktNum , so the condition is always true.

To perform the calculation that you actually want to, you need to mask out the lower eight bits after the negation:

if (pktNum != (~invPktNum & 0xff)) {
    return 1;
}

Or alternatively, you can just negate the bits you're interested in:

if (pktNum != (invPktNum ^ 0xff)) {
    return 1;
}

Answer 2

You're looking at 8-bit values. I bet that pktNum and invPktNum are 32-bit values, so you're comparing 0x000000fe to 0xfffffffe.

Answer 3

Perhaps the simple type independent test is:

if (0 == (pktNum & pkNum)) {

Which is equal to:

if (0 == (0xfe & 0x01)) {

Quickly tested with http://codepad.org/nziOGYJG seems to work with multiple signed and unsigned long / int types.

Out of interest anybody have any comments?

Answer 4

Assuming the following:

invPktNum = ~pktNum

then your comparison is equivalent to:

if (pktNum != ~(~pktNum)) {

or

if (pktNum != pktNum) {

which is always false. Unless invPktNum has a different definition in your code which is not shown.