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我必须编写一个程序,使用辛普森方法和梯形方法计算函数的积分。它在我使用 MinGW 的计算机上运行良好,但是当我尝试在我的 uni 计算机上编译它时,我得到:

It = 0 //should be 0.954499
*pn = 0 //should be 18
Is = 0 //should be 0.954500
*pn = 0 //should be 6

这是我想出的(对不起,变量和注释是葡萄牙语,我回家后会修复它):

积分.h:

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

#define eps 1.e-6
#define kmax 20

double trapezio(double (*f)(double x), double a, double b, int *n);
double simpson(double (*f)(double x), double a, double b, int *n);

积分.c:

#include "integral.h"
#define xmin -2
#define xmax 2

double f(double x);

int main(){
    double It,Is;
    int n = 0;
    int *pn = NULL;

    pn = &n;
    It = trapezio(f,xmin,xmax,pn)/sqrt(2*M_PI);
    printf("Pelo metodo dos trapezios a integral vale aproximadamente %lf\n", It);
    printf("O numero de iteracoes usadas foi %d\n\n",*pn);

    *pn = 0;
    Is = simpson(f,xmin,xmax,pn)/sqrt(2*M_PI);
    printf("Pelo metodo de simpson a integral vale aproximadamente %lf\n", Is);
    printf("O numero de iteracoes usadas foi %d\n",*pn);

    return 0;
}

double f(double x){
    return exp(-0.5*x*x); // Funcao que sera integrada
}

梯形图.c:

#include "integral.h"

double trapezio(double (*f)(double x), double a, double b, int *n){
    double To, Tk;
    double soma;
    int i, k = 1;
    Tk = 0.5*(f(a) - f(b))*(b - a);

    while (fabs((Tk-To)/To) > eps && k < kmax){
        soma = 0; // Resetando variavel soma
        To = Tk; // To e' T(k - 1), caso o loop se repita o ultimo Tk vira To
        for (i = 1 ; i <= (pow(2,k)-1) ; i += 2) soma += f(a + i*(b - a)/pow(2.,k));
        Tk = 0.5*To + soma*(b - a)/pow(2.,k);
        k++;
        *n += 1;
    }

    return Tk;
}

辛普森.c:

#include "integral.h"

double simpson(double (*f)(double x), double a, double b, int *n){
    double So, Sk = 0;
    double somaimp, somapar;
    int i, k = 1;

    while (fabs((Sk-So)/So) > eps && k < kmax){
        somaimp = 0;
        somapar = 0;
        So = Sk; // So e' S(k - 1)
        for (i = 1; i <= (pow(2,k)-1); i += 2) somaimp += f(a + i*(b - a)/pow(2.,k));
        for (i = 2; i <= (pow(2,k)-2); i += 2) somapar += f(a + i*(b - a)/pow(2.,k));
        Sk = (b - a)*(f(a) + 4*somaimp + 2*somapar + f(b))/(3*pow(2.,k));
        k++;
        *n += 1;
    }

    return Sk;
}

编辑:我忘了提到如果我把指针拿出来,梯形可以工作,但辛普森仍然返回 0。

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1 回答 1

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在中,您在循环中使用它之前trapezio()永远不会初始化:Towhile

double To, Tk;
/* ... no assignment to To ... */
while (fabs((Tk-To)/To) > eps && k < kmax){

这意味着它将以未定义的方式运行,并且可能根本不会进入循环。

于 2013-06-19T12:27:10.927 回答