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Fast Algorithm to check the numbers 2^n

, 12 Dec 2002
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Optimised code for valid data length to FFT


Some applications of signal processing, like Fast Fourier Transform (FFT) need the length of sampled-data input equal to 2^n (where n : integer, n=1,2,3,...), and requires fast testing for this number. This test must be done in a very short time especially in real-time applications operated in DSP cart (as in TMS320C6xxx DSP). Here is an optimized code snippet for this purpose.

Source code

The following function return TRUE if the number x has the form 2^n

bool CheckInputToFFT(int x)
    return (!(x & (x-1)));

That's all, which means each of the numbers (2,4,8,16,...2^n) gives zero when ANDed with previous number!


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Comments and Discussions

GeneralC# Trick PinmemberKK Adams30-Jun-08 17:12 
GeneralRe: C# Trick PinmemberMember 150696612-Jun-09 10:36 
GeneralMod - Works for n = 0 Pinmemberst0le18-Feb-08 20:04 
GeneralFFT in Visual C++ PinsussAnonymous7-Jul-03 2:18 
GeneralQuick Mod for 2^0 and others PinmemberColin Davies27-Dec-02 15:16 
GeneralRe: Quick Mod for 2^0 and others Pinmembernutty21-Mar-06 7:23 
GeneralThanks... PinmemberRocky1010-Dec-02 4:36 
GeneralHacker's Delight by Warren PinmemberGeorge V. Reilly9-Dec-02 16:58 
GeneralMake it slightly faster... PinmemberKevinHall5-Dec-02 19:08 
GeneralHmmm.... PinmemberNitron5-Dec-02 18:38 

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