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

Question

为了我的编程作业，我必须编写一个计算函数积分的程序

main.c 文件:

    #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;
    }

积分函数.c文件:

    #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)));}

积分函数.h文件:

    #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

标题中的错误在所有四种情况下都出现在这部分:

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

我无法弄清楚问题到底出在哪里，我试过将 & 换成 * 或完全删除它，但这只会导致错误，指出参数类型与“集成”的参数 1 不兼容

Answer 1

pf 是一个函数指针，你需要将它传递给 integrate 而不需要任何额外的参数。由于您在所有四个分支中以相同的方式调用函数，因此将调用移至 switch 之外的 integrate:

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);

这种 switch 的使用模式也可以用数组代替。