What does int i = *(int*)&x; mean?

Question

C++ / C#

float interesting_function(float x){
    float xhalf = 0.5f*x;
    int i = *(int*)&x;
    i = 0x5f3759df - (i>>1);
    x = *(float*)&i;
    x = x*(1.5f - xhalf*x*x);
    return x;
}

Can you help me figure out what the designation of variables in this format means?

Answer 1

&x- the address (pointer) is taken, at which the value of the variable x is located
(int*)- the designation that with the value of the address that was received earlier, you need to work as a inttype (although it was earlier float)
* - we get the value at this address, but it will no longer be float, a int. Which results in a icompletely different number. Only in the binary representation nothing has changed

Answer 2

*(int*)&x - take the address of the x variable and dereference it as the address of an int variable.