C error: "Called error is not a function or function pointer"

Question

For my homework in programming, I have to write a program that calculates the integral of a function

main.c file:

    #include <stdio.h>
    #include <math.h>
    #include "integrand.h"
    #define M 64
    #define N 64
    #define dif 0.01
    #define max 8 

    float (*pf)(float,float);

    void integrate(float pf(float x,float y)){
    int i,j,k=0,l=1; 
    float a,b,c,d,dx,dy,dA,x[i],y[j],r=0,r_old;

    printf("Enter the integration bounds a,b,c,d separated by commas: \n\n");scanf("%f,%f,%f,%f",&a,&b,&c,&d); 

    do{
    r_old = r;

    dx = (b - a)/(l*M);
    dy = (d - c)/(l*N);
    dA = dx * dy;

    for(i=0;i<(l*M)-1;i++){
        x[i] = a + (i + 0.5)*dx;
    }

    for(j=0;j<(l*N)-1;j++){
        y[j] = c + (j + 0.5)*dy;
    }

    for(i=0;i<(l*M)-1;i++){
        for(j=0;j<(l*N)-1;j++){
            r+=(((*pf)(x[i],y[j]))*dA);
            }}

    k++;
    l*=2;
    }while(k<max || abs(r-r_old)<=dif*abs(r));

    printf("Integral = %f",r);

    return;
    }

    int main(void)
    {
    float x,y;
    int q;

    printf("Choose an integrand:\n\n (1) f_1(x,y) = x^3 + 3y^2*x + 7x^2*y + 10 \n\n (2) f_2(x,y) = cos(0.1*x^5 + y^2) - sin(y*x)^5 + log(|x-y|+1) + 2 \n\n (3) f_3(x,y) = sin(x^2+y^2) + exp(-50*(x^2+y^2))\n\n (4) f_4(x,y) = f_3(f_1(x,y),f_2(x,y))\n\n (5) QUIT \n\n");

    do{
    scanf("%d",&q);
    switch(q){
        case 1:
            pf = f_1;
            integrate((&pf)(x,y));
            break;
                case 2:
            pf = f_2;
            integrate((&pf)(x,y));
            break;
        case 3:
            pf = f_3;
            integrate((&pf)(x,y));
            break;
        case 4:
            pf = &f_4;
            integrate((&pf)(x,y));
            break;

    }}while(q!=5);

    return 0;
    }

integrand.c file:

    #include <math.h>

    float f_1(float x, float y){return (pow(x,3) + 3*pow(y,2)*x + 7*pow(x,2)*y + 10);}

    float f_2(float x, float y){return (cos(0.1*pow(x,5)+pow(y,2)) - pow(sin(y*x),5) + log(fabs(x-y)+1) + 2);}

    float f_3(float x, float y){return (sin(pow(x,2)+pow(y,2)) + pow(M_E,-50*(pow(x,2)+pow(y,2))));}

    float f_4(float x, float y){return (f_3(f_1(x,y),f_2(x,y)));}

integrand.h file:

    #ifndef INTEGRAND_H
    #define INTEGRAND_H

    extern float f_1(float x, float y);
    extern float f_2(float x, float y);
    extern float f_3(float x, float y);
    extern float f_4(float x, float y);

    #endif // INTEGRAND_H

The error in the title shows up in this part in all four cases:

    integrate((&pf)(x,y));

I can't figure out where exactly the issue lies, I've tried swapping the & for * or deleting it entirely, but that only leads to an error saying that the argument type is incompatible for argument 1 of 'integrate'

The task says I have to do this with a function pointer, so that won't work directly. integrate(pf); helped fix the issue though, thanks! But my new problem is that the program crashes... — pavus
– pavus, Commented Jan 4, 2018 at 11:22
@MichaelWalz what I'm saying is that I can't use integrate(f_1); because of my task saying that I have to do it with a function pointer — pavus
– pavus, Commented Jan 4, 2018 at 11:29

Sergey Kalinichenko · Accepted Answer · 2018-01-04 11:20:12Z

1

pf is a function pointer, you need to pass it to integrate without any additional parameters. Since you call the function the same way in all four branches, move the call to integrate outside switch:

do{
    scanf("%d", &q);
    switch(q){
        case 1: pf = f_1; break;
        case 2: pf = f_2; break;
        case 3: pf = f_3; break;
        case 4: pf = f_4; break;
        default: pf = NULL; break;
    }
    if (pf) {
        integrate(pf);
    }
} while (q != 5);

This pattern of switch use could also be replaced with an array.

answered Jan 4, 2018 at 11:20

Sergey Kalinichenko

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