Delay outputting 0's

Can someone please tell me what is wrong here. delay has a buffer (del) with a pointer *p, but is outputting zero (0) every sample.

(Fixed Delay with 3 secs)

float delay(float *sig, float dtime, float *del, int *p, int vecsize, float sr) {

    int dt;
    float out;
    dt = (int) (dtime*sr);// dtime = 3开发者_如何学编程secs, sr=44100.0

    for(int i=0; i<vecsize; i++){
        out = del[*p]; 
        del[*p] = sig[i];
        sig[i] = out;
        *p = (*p != dt-1 ? *p+1 : 0);
    }

    return *sig;
}

Your delay() function appears to work fine, so the error must be in how you called it. See the following example program:

#include <stdio.h>

float delay(float *sig, float dtime, float *del, int *p, int vecsize, float sr)
{

    int dt;
    float out;
    dt = (int) (dtime*sr);// dtime = 3secs, sr=44100.0

    for(int i=0; i<vecsize; i++){
        out = del[*p];
        del[*p] = sig[i];
        sig[i] = out;
        *p = (*p != dt-1 ? *p+1 : 0);
    }

    return *sig;
}

#define DTIME 3
#define SAMPLERATE 44100
#define VECSIZE (10 * SAMPLERATE)

int main()
{
    float del[DTIME * SAMPLERATE] = { 0.0 };
    int del_p = 0;
    float sig[VECSIZE] = { -5.0, -4.0, -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, 4.0, 5.0 };
    int i, j;

    delay(sig, DTIME, del, &del_p, VECSIZE, SAMPLERATE);

    for (i = 0; i < VECSIZE && sig[i] == 0.0; i++)
        ;

    for (j = i; j < VECSIZE && j < i + 10; j++)
        printf("sig[%d] = %f\n", j, sig[j]);

    return 0;
}

The output produced is as expected:

sig[132300] = -5.000000
sig[132301] = -4.000000
sig[132302] = -3.000000
sig[132303] = -2.000000
sig[132304] = -1.000000
sig[132305] = 0.000000
sig[132306] = 1.000000
sig[132307] = 2.000000
sig[132308] = 3.000000
sig[132309] = 4.000000

Note that you should really be passing dt directly to the delay() function, rather than calculating it from sr and dtime - as far as the delay() function is concerned, it's the delay in number of samples (and hence the assumed size of the del[] delay line array) that matters. You'd have to have calculated that value in the caller anyway, to appropriately size the del[] array.

You're returning *sig always. That's always going to be the first element of the sig array. sig[0] will be equal to del[*p] because you're the first time through the loop (when i=0) you're doing:

out = del[*p];
del[*p] = sig[0]; // doesn't affect sig
sig[0] = out;

That's the only time sig[0] is affected. And then when you return *sig you're returning sig[0].

So I would expect this function would always return whatever del[*p] is.