目前它将构建并运行。但我的问题是,当我试图调用指针函数时,它不会打印出我的main函数中的第一个print语句。我的最终目标是让它像1-36的f()值一样打印。使用&和*在这几行代码中是家庭作业的一部分。
#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;
}
更新:我编辑了现在正在打印的指针功能但不是正确的数字
#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;
}
答案 0 :(得分:0)
注意编译器警告:
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
是指向指针函数内预期双数组的指针:
double e[35];
printf("%d,%f", xii, pointer(e));
另外,你的意图是迭代e:
的所有元素for ( xii=0; xii<35; xii++){
printf("%d,%f", xii, pointer(e));
}
经过这样的计算:
double pointer (double *bi){
int xi=0;
for ( xi=0; xi<35; xi++){
bi[xi] = f(xi);
}
return 0;
}