Question

I’ve been looking around on the other threads somehow related to this, but somehow I just don’t get it…

I want to do some FFT on a set of values I have evaluated and wrote this program to first read the values and save them to an array of size n.

int main () {
    // some variables and also a bit of code to read the 'messwerte.txt'

printf("Geben sie an wieviele Messwerte ausgelesen werden sollen: ");
scanf("%d", &n);
double werte[n]; //Array der "fertigen" Messwerte
in = fopen ("messwerte.txt","r");
double nul[n]; //Array von nullen

int logN = 14;
l=FFT(logN,&werte,&nul);
}

In the same file I also do the FFT with the help of this program:

double FFT (int logN, double *real, double *im) //logN is base 2 log(N) {
// blabla FFT calculation
}

However, when I compile I always get this error:

gcc FFT.c -lm
FFT.c: In function ‘main’:
FFT.c:94:2: warning: passing argument 2 of ‘FFT’ from incompatible pointer type [enabled by default]
FFT.c:4:8: note: expected ‘double *’ but argument is of type ‘double (*)[(unsigned int)(n)]’
FFT.c:94:2: warning: passing argument 3 of ‘FFT’ from incompatible pointer type [enabled by default]
FFT.c:4:8: note: expected ‘double *’ but argument is of type ‘double (*)[(unsigned int)(n)]’

Since this is my first time programming, I really don’t know what is wrong with my code. Will I have to set more flags for the compiler or stuff like that (because I had to do this -lm stuff or it wouldn’t compile and said something like pow not found or so)?

Also I was made aware that there might be a difference when writing on a Windows or a Linux machine, and I am using Linux, Lubuntu 12.10 32-bit if it’s a problem of the OS.

Answer 1

l=FFT(logN,&werte,&nul);
           ^      ^

Drop ampersands from that line.

---

The problem is that the & operator in this context produces an expression with a different type than what FFT expects. FFT expects a pointer to a double and &werte produces a pointer to an array of N elements. So, in order to make FFT happy, just pass werte which will quietly decay to a pointer to the first element.

For more information on pointers to arrays, there’s a C FAQ.