clear and test bit in single statement in C

I would like to clear, and test a particular bit in single statement, for instance to clear bit 27 and test the same bit, and wanted to return the status for same:

{
    status = number & BIT(27);

    if (status)
        number & ~BIT(27);

    return number & BIT(27);
}

Is there way to do it using Ternary operator ?

The code fragment does not modify number so it will return the bit value but not clear it. You probably meant this instead:

#define BIT(n)  (1U << (n))  // assuming this definition for BIT or similar

{
    status = number & BIT(27);

    if (status)
        number &= ~BIT(27);

    return status;
}

You can indeed achieve this in a single expression using the ternary operator and a comma expression:

    return (number & BIT(27)) ? (number &= ~BIT(27), BIT(27)) : 0;

or this alternative:

    return (number & BIT(27)) ? (number ^= BIT(27), BIT(27)) : 0;

There is a sequence point between the evaluation of the test expression and the evaluation of the second or third expression, so it is OK to modify number this way as a side effect.

This approach may allow you to encapsulate the test and clear trick inside an expression macro, but be aware that this macro would evaluate number twice. You might consider an inline function taking the address of number:

static inline int test_and_clear27(int *number) {
    if (*number & BIT(27)) {
        *number &= ~BIT(27);
        return BIT(27);
    } else {
        return 0;
    }
}

Or without branches:

static inline int test_and_clear27(int *number) {
    int status = *number & BIT(27);
    *number &= ~BIT(27);
    return status;
}

These methods are not thread safe: it cannot be used to implement semaphores as the test and clear operations are separate and do not happen atomically.

Your current code does not do what your description claims. Instead, it is equivalent to

return number & BIT(27);

— A nice, concise, single statement.

However, assuming that you actually do wanted to clear bit 27, the following code does that (the if in your code is unnecessary):

status = number & BIT(27);
number &= ~ BIT(27);
return status;

There’s really no good reason to rewrite this as a single statement (except by encapsulating it inside a function¹). But, just for argument’s sake, it is possible. chqrlie has already posted an answer, but if you only use status as a boolean flag you can shorten the code and remove the conditional operator:

return (number & BIT(27)) && (number &= ~ BIT(27), 1);

Is this better than the three-statement solution? Absolutely not! On the contrary: it introduces a branch into the logic which, if anything, makes this code slower.

¹ And as per Ian Abbott’s comment the Linux kernel provides just such a function: test_and_clear_bit(), but be careful since it requires to have a variable of unsigned long type and any casting to it is a very bad idea.