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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T07:58:53+00:00 2026-05-16T07:58:53+00:00

Possible Duplicate: John Carmack’s Unusual Fast Inverse Square Root (Quake III) I came across

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Possible Duplicate:
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 different result.

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  1. Editorial Team
    Editorial Team
    2026-05-16T07:58:54+00:00Added an answer on May 16, 2026 at 7:58 am

    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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