What is the difference between '* (<type> *) & x' and 'x'?

Question

What's the difference between

int i = 123; int k; k = *(int *) &i; cout << k << endl; //Output: 123 

and

 int i = 123; int k; k = i; cout << k << endl; //Output: 123 

Both of them give the same conclusion, but is there any difference?

(I found the first Quake3 code snippet of the fast reverse square root)

Answer 1

In Q3:

 float Q_rsqrt( float number ) { long i; float x2, y; const float threehalfs = 1.5F; x2 = number * 0.5F; y = number; i = * ( long * ) &y; // evil floating point bit level hacking i = 0x5f3759df - ( i >> 1 ); // what the fuck? y = * ( float * ) &i; y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration // y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed return y; } 

As I understand it, you are interested in the following line:

  i = * ( long * ) &y; 

y is a float , and i is a long . So this is a reinterpretation for a floating point bitmap as an integer bitmap.

Answer 2

It all depends on whether i object or a primitive type. If it is an object, operator* can be overloaded, which gives a different general meaning.

Answer 3

No, there is no difference, both assignments essentially copy the int bit from the memory storing i into memory k .

Sometimes such tricks are used when the types of the source and destination variables are different, but it's just int to int .

I think that a smart enough compiler should generate the same code for both versions.