當它經歷一個循環時，fprintf不打印

Question

我試圖在文檔中打印某些內容，但它只適用於我試過的第一個值！我試圖自己弄清楚，但我不到一個小時就把這段代碼給我的教授，所以我有點絕望。 ：/

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

double euler(double v0, double *t0, double h, double (*f)(double)); 
double dv(double v); 

int main(){ 
    double vlim, t = 0; 
    double *pt = &t; 
    double h0 = 1.e-7; 

    vlim = euler(0, pt, h0, dv); 
    printf("O tempo e a velocidade limite para um passo de %.0e s sao respectivamente:\n", h0); 
    printf("%.0e s e %lf mm/s.\n\n", *pt, vlim); 

    double h = h0, v = 0; 
    do { 
     *pt = 0; 
     v = euler(0, pt, h, dv); 
     h /= 5.; 
    } while (fabs(v-vlim)/v > 1e-5); 

    printf("O passo necessario para atingir a convergencia na velocidade limite e' %.1e s.\n",h); 
    return 0; 
} 

double euler(double y, double *x, double h, double (*f)(double)){ 
    FILE *arq = fopen("bolha.dat","w+"); 
    double y0, eps = 1e-4; 
    do{ 
     y0 = y; 
     y = y0 + h*f(y0); // Metodo de Euler 
     *x += h; 
     fprintf(arq, "%.1e\t%.7lf\n", *x, y); 
    } while (fabs((y-y0)/y) > eps); 

    fclose(arq); 
    return y; 
} 

dv是不相關的，所以我沒有發佈它。它看起來像循環中的東西真的很愚蠢，因爲如果我把h0 = 2.e-8（這是我試圖打印的值）它適用於該值。

Answer 1

我可以通過在euler函數中創建一個參數來修復它，然後在循環之後再次在另一個文件上調用它。這解決了問題，但我仍然不明白爲什麼它不工作。

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

double euler(double v0, double *t0, double h, double (*f)(double), FILE *arquivo); 
double dv(double v); 

int main(){ 
    FILE *saida = fopen("bolhateste.dat","w+"); 
    double vlim, t = 0; 
    double *pt = &t; 
    double h0 = 1.e-7; 

    vlim = euler(0, pt, h0, dv, saida); 
    printf("O tempo e a velocidade limite para um passo de %.0e s sao respectivamente:\n", h0); 
    printf("%.0e s e %lf mm/s.\n\n", *pt, vlim); 

    double h = h0, v = 0; 
    do { 
     *pt = 0; 
     v = euler(0, pt, h, dv, saida); 
     h /= 5.; 
    } while (fabs(v-vlim)/v > 1e-5); 

    saida = fopen("bolha.dat","w+"); 
    euler(0, pt, h, dv, saida); 

    printf("O passo necessario para atingir a convergencia na velocidade limite e' %.1e s.\n",h); 
    return 0; 
} 

double euler(double y, double *x, double h, double (*f)(double), FILE *arquivo){ 
    double y0, eps = 1e-4; 
    do{ 
     y0 = y; 
     y = y0 + h*f(y0); // Metodo de Euler 
     *x += h; 
     fprintf(arquivo, "%.1e\t%.7lf\n", *x, y); 
    } while (fabs((y-y0)/y) > eps); 

    return y; 
}