What does this C idiom mean? [duplicate]

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John Carmack’s Unusual Fast Inverse Square Root (Quake III)

I came across this piece of code a blog recently - it is from the Quake3 Engine. It is meant to calculate the inverse square root fast using the Newton-Rhapson method.

float InvSqrt (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;
}

What is the reason for doing int i = *(int*)&x;? Doing int i = (int) x; instead gives a completely diffe开发者_开发问答rent result.

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int i = *(int*)&x; doesn't convert x to an int -- what it does is get the actual bits of the float x, which is usually represented as a whole other 4-byte value than you'd expect.

For reference, doing this is a really bad idea unless you know exactly how float values are represented in memory.

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int i = *(int*)&x; says "take the four bytes which make up the float value x, and treat them as if they were an int." float values and int value are stored using completely different methods (e.g. int 4 and float 4.0 have completely different bit patterns)

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The number that ends up in i is the binary value of the IEEE floating point representation of the number in x. The link explains what that looks like. This is not a common C idiom, it's a clever trick from before the SSE instructions got added to commercially available x86 processors.