If condition understanding problem in c with syntax [closed]

Question

if (!(c & ~0xFF))
    c |= 0x0700;

Here what not than do what. Not able to understand this statement

[Sorry but my mind was just gone blur after writing too much code that i was not able to understand so a simple thing sorry for wasting your all times] Sorry once again

Answer 1

The behavior of the code is:

• If c, widened to at least an int, has no bits above the first eight set, turn attempt to turn on the three bits that are on in 0x0700 (bits 8, 9, and 10).

Equivalently:

• If 0 <= c && c < 256, turn attempt to on the three bits that are on in 0x0700.

For simplicity, we assume an ordinary C implementation using two’s complement and a char narrower than an int. Then, in detail:

• 0xFF has value 255. In its binary representation, the low eight bits are on, and all higher bits are off. The type of this constant is int.
• ~0xFF inverts each bit. The low eight bits are off, and all higher bits are on.
• c & ~0xFF performs a bitwise AND, but first, if c is narrower than an int, it converts c to an int. (Technically this depends on the ranks of the types, but “narrower” suffices in ordinary C implementations.) This conversion is important because it may act to extend the sign bit of c, discussed below. If c is an unsigned int or is wider than an int, then ~0xFF is converted to a wider type. Because ~0xFF is a negative value and two’s complement is used, the result of this conversion still has all of its high bits set.
• In the result of the bitwise AND, the low eight bits are off, and the only high bits set are those that were set in c (after conversion, if performed).
• !(c & ~0xFF) tests this result. It is true if and only its operand is zero, meaning no bits are set in c & ~0xFF.

If c is negative, then, after conversion to an int if necessary, it will have its sign bit set, so there will be a bit above the first eight set, so the test will be false. Otherwise, any conversion to an int will not turn on any bits, so the test will be true if and only if c has no bits above the first eight set. Therefore, the test is equivalent to “c is greater than or equal to 0 and is less than 256.”

c |= 0x0700 performs a bitwise OR of c with 0x0700 and stores the result in c. If c is wide enough, the bits that are on in 0x0700 will be turned on in c. If c is not wide enough but is unsigned, the bits that do not fit will be ignored. If c is not wide enough but is signed, an implementation-defined conversion will be performed (or a signal will be raised), and the result could be anything.

So, what this code does is, if 0 <= c && c < 256, turn on the bits that are on in 0x0700, subject to the exceptional conditions described above.

Answer 2

   if (!(c & ~0xFF)) c |= 0x0700;

~0xFF is the same as 0xF...FFFFF00 or all bits 1 except the 8 least-significant ones.

(c & 0xF...FFFFF00) is the most significant bits of c followed by 8 0 bits

!(c & 0xFFFFFF00) is either 0 (false) or 1 (true). It's false if the thing inside the parenthesis is not zero, true otherwise.

So, the full expression sets bits 8, 9, and 10 of c if the first most-significant bits of the initial c were all 0 (same as if ((c / 256) == 0) or if ((c >> 8) == 0)).