what does ~(uint32_t) mean in C?

Question

I'm reading a bit of C code in an OS kernel that says

x & ~(uint32_t)CST_IEc;

What does the ~() mean?

Answer 1

~() is actually two things:

1. (uint32_t) is a cast.
2. ~ is a bitwise complement operator.

Answer 2

You are interpreting the operator precedence incorrectly. The cast (uint32_t)CST_IEc is done first and ~ happens after that. Take a look at an operator precedence chart for help.

Answer 3

• The (uint32_t) bit is a cast to a type of unsigned int (32 bits),
• the ~ means bitwise not (or complement), so it reverses the bits in CST_IEc after it has been cast to uint32_t.

Answer 4

(uint32_t)CST_IEc; //casting CST_IEc to uint32_t

~( ) //taking one's complement

Answer 5

Isn't (uint32_t) a type cast?

~ is bitwise NOT

Answer 6

A few more parantheses to clearify evaluation order:

(x & (~((uint32_t)CST_IEc)))

First CST_IEc is casted into a uint32_t then it is bitwise negated with ~ before being bitwise anded with x through &.

Answer 7

You need to read the expression slightly differently:

(uint32_t)CST_IEc

This converts the value CST_IEc into a 32-bit unsigned integer.

~(uint32_t)CST_IEc;

The ~ then does a bit-wise inversion of the value; each one bit becomes a zero and each zero bit becomes a one.

The whole expression then does:

x & ~(uint32_t)CST_IEc;

This means that the result contains the bits in x except for the bits implied by the value of CST_IEc; those are zeroed.

So, if CST_IEc was, for sake of example, 0x0F00, and the input value of x was 0x12345678, the result would be 0x12345078.