宿題、ポインタと関数呼び出しで立ち往生

Question

現在、ビルドと実行が行われます。しかし、私の問題は、ポインタ関数を呼び出そうとしているとき、私のmain関数で最初のprint文を出力しないことです。私の最終目標は、1から36までのf（）の値の表のように印刷することです。これらのコードの行に&と*を使用することは、宿題の一部です。

#include <stdio.h> 
#include <math.h> 


double pointer (double *bi); 
double f(double x); 
double trapz(int n, double a, double b); 
double simpsons(int m, double c, double d); 

int main() 
{ 
    int g, h, xii=0; 

    double e[35]; 
    for (xii=1; xii>36; xii++){ 
    printf("%d,%f", xii, pointer(&e)); 
    } 


    double x0 = 1, xn = 36, x1 = 1, xm =36; 
    int n1 = 35, n2 = 100, n3 = 10000, n4 = 35, n5 = 1000, n6 = 10000; 




    printf("For n = %d, the integral is %f\n", n1, trapz(n1, x0, xn)); 
    printf("For n = %d, the integral is %f\n", n2, trapz(n2, x0, xn)); 
    printf("For n = %d, the integral is %f\n", n3, trapz(n3, x0, xn)); 


    printf("for n = %d, the integral is %f\n", n4, simpsons(n4, x1, xm)); 
    printf("for n = %d, the integral is %f\n", n5, simpsons(n5, x1, xm)); 
    printf("for n = %d, the integral is %f\n", n6, simpsons(n6, x1, xm)); 

    printf("Enter the amount of intervals 'n' for the Trapezoidal rule and I'll give you an answer!\n"); 
    scanf("%d", &g); 
    printf("\nYour answer is: %f\n", trapz(g, x0, xn)); 

    printf("Enter the amount of intervals 'n' for the Simpson's rule and I'll give you an answer!\n"); 
    scanf("%d", &h); 
    printf("\nYour answer is: %f\n", simpsons(h, x1, xm)); 

    return 0; 
} 

double pointer (double *bi){ 

    int xi=0; 
    for (xi=1; xi>36; xi++){ 
     *bi=f(xi); 
    } 
return 0; 
} 

double f(double x)/*function that calculates f(x)*/ 
{ 

    double y; 

    y = 1000 * sin(5*x)/(x*x+7.68); 

    return y; 
} 

double trapz(int n, double a, double b) 
{ 
    int i; 
    double dx, x, sum; 
    dx = (b-a)/n; 
    sum=f(a)+f(b); 
    for (i=1; i<n; i++){ 
     x=a+dx*i; 
     sum+=2*f(x); 
    } 
    sum*=dx/2; 
    return sum; 
} 
double simpsons(int m, double c, double d) 
{ 
    int j; 
    double dz, z, sum2; 
    dz = (d-c)/m; 
    sum2 = f(c)+f(d); 
    for (j=1; j<m; j++){ 
     z=c+dz*j; 
     sum2 += 2*(j+j%2)*f(z); 
    } 
    sum2 *= dz/3; 
    return sum2; 
} 

アップデート：私はそれが今、印刷されたポインタ機能を、編集ではなく、右の数字

#include <stdio.h> 
#include <math.h> 


double pointer(double *bi, int ci); 
double f(double x); 
double trapz(int n, double a, double b); 
double simpsons(int m, double c, double d); 

int main() 
{ 
    int g, h, xii=0; 

    double e[35]; 
    for (xii=1; xii<36; xii++){ 
    printf("\t%d \t\t %f\n", xii, pointer(&e, xii));/*calls function and prints xii,f(xi). 1-36 for xi, and 1-36 for f(xi)*/ 
    } 


    double x0 = 1, xn = 36, x1 = 1, xm =36; 
    int n1 = 35, n2 = 100, n3 = 10000, n4 = 35, n5 = 1000, n6 = 10000; 




    printf("For n = %d, the integral is %f\n", n1, trapz(n1, x0, xn)); 
    printf("For n = %d, the integral is %f\n", n2, trapz(n2, x0, xn)); 
    printf("For n = %d, the integral is %f\n", n3, trapz(n3, x0, xn)); 


    printf("for n = %d, the integral is %f\n", n4, simpsons(n4, x1, xm)); 
    printf("for n = %d, the integral is %f\n", n5, simpsons(n5, x1, xm)); 
    printf("for n = %d, the integral is %f\n", n6, simpsons(n6, x1, xm)); 

    printf("Enter the amount of intervals 'n' for the Trapezoidal rule and I'll give you an answer!\n"); 
    scanf("%d", &g); 
    printf("\nYour answer is: %f\n", trapz(g, x0, xn)); 

    printf("Enter the amount of intervals 'n' for the Simpson's rule and I'll give you an answer!\n"); 
    scanf("%d", &h); 
    printf("\nYour answer is: %f\n", simpsons(h, x1, xm)); 

    return 0; 
} 

double pointer(double *bi, int ci){ 
    double ans; 
     *bi = (1000 * sin(5*ci))/(ci*ci+7.68); 
     ans = *bi; 
     return ans; 

/*function is supposed to return f(xi)*/ 
} 

double f(double x)/*function that calculates f(x)*/ 
{ 

    double y; 

    y = 1000 * sin(5*x)/(x*x+7.68); 

    return y; 
} 

double trapz(int n, double a, double b) 
{ 
    int i; 
    double dx, x, sum; 
    dx = (b-a)/n; 
    sum=f(a)+f(b); 
    for (i=1; i<n; i++){ 
     x=a+dx*i; 
     sum+=2*f(x); 
    } 
    sum*=dx/2; 
    return sum; 
} 
double simpsons(int m, double c, double d) 
{ 
    int j; 
    double dz, z, sum2; 
    dz = (d-c)/m; 
    sum2 = f(c)+f(d); 
    for (j=1; j<m; j++){ 
     z=c+dz*j; 
     sum2 += 2*(j+j%2)*f(z); 
    } 
    sum2 *= dz/3; 
    return sum2; 
} 

Answer 1

コンパイラの警告に注意して：

16:34: warning: passing argument 1 of 'pointer' from incompatible pointer type [-Wincompatible-pointer-types] 
    printf("%d,%f", xii, pointer(&e)); 
           ^
5:8: note: expected 'double *' but argument is of type 'double (*)[35]' 
double pointer (double *bi); 

eがへのポインタでありますポインタ関数内の期待されるdouble配列関数：

double e[35]; 

printf("%d,%f", xii, pointer(e)); 

このような計算の後

for (xii=0; xii<35; xii++){ 
printf("%d,%f", xii, pointer(e)); 
} 

：

はまた、あなたの意図は、電子のすべての要素を反復処理することが私のあなたの `main`機能で

double pointer (double *bi){ 

    int xi=0; 
    for (xi=0; xi<35; xi++){ 
     bi[xi] = f(xi); 
    } 
    return 0; 
}