所以我的代码有效，只是出于某种原因，我的代码总是运行两个 if 语句（两个 y 方程，无论我为第一个 fprintf 问题输入哪个数字）。此外，t,y 列总是比 t,y2 列长得多（编辑，即如果我输入 funt=1,t0=0,t1=10,inity=1,dt=0.1，我的 t,y 列有 100行和我的 t,yexact 列有 10 行）。此外，t,y 部分的 t 列对于第一个 if 语句（y 方程）全为零，对于第二个 if 语句（y 方程）全为一个数字。任何帮助将不胜感激。
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
double rk4(double(*f)(double,double),double,double,double);
double user_fun_1(double,double);
double user_fun_2(double,double);
int main (void) {
int funct;
double *y,t,y2;
double t0,t1,dt,inity,C;
int i,n;
fprintf(stdout, "Which function do you wish to integrate?\n(1)f(t,y)=t\n(2)f(t,y)=-y\n");
fscanf(stdin,"%d",&funct);
fprintf(stdout,"What is the initial time?\n");
fscanf(stdin,"%lf",&t0);
fprintf(stdout,"What is the final time?\n");
fscanf(stdin,"%lf",&t1);
fprintf(stdout,"What is the initial value of y?\n");
fscanf(stdin,"%lf",&inity);
fprintf(stdout,"what is the step size?\n");
fscanf(stdin,"%lf",&dt);
if (dt>(t1-t0)){
printf("The stepsize has to be less than or equal to the total time\n");
return EXIT_FAILURE;}
n = 1 + (t1-t0)/dt;
y = (double *)malloc(n*sizeof(double));
if (funct=1){
C=inity-(t0*t0*0.5);
printf("%s%25s\n","t","y");
for (y[0]=inity,i=1;i<n;i++){
y[i]=rk4(user_fun_1,dt,t0 + dt * (i-1),y[i-1]);
printf("%lf%25lf\n",t,y[i]);}
printf("%s%25s\n","t","y exact");
for(i=0;i<n;i+=t1){
t=t0+dt*i;
y2=(t*t*0.5) + C;
printf("%lf%25lf\n",t,y2);}}
if (funct=2){
C=inity/(exp(-t0));
printf("%s%25s\n","t","y");
for(y[0]=inity,i=1;i<n;i++){
y[i]=rk4(user_fun_2,dt,t0+dt*(i-1),y[i-1]);
printf("%lf%25lf\n",t,y[i]);}
printf("%s%25s\n","t","y exact");
for(i=t0;i<n;i+=t1){
t=t0+dt*i;
y2=C*exp(-t);
printf("%lf%25lf\n",t,y2);}}
free(y);
return EXIT_SUCCESS;}
double user_fun_1(double t,double y){
return t;}
double user_fun_2(double t, double y){
return -y;}
double rk4(double(*f)(double, double), double dt,double t,double y){
double k1,k2,k3,k4;
k1=dt*f(t,y);
k2=dt*f(t + dt/2,y + k1/2);
k3=dt*f(t + dt/2,y + k2/2);
k4=dt*f(t + dt,y + k3);
return y + (k1 + 2*k2 + 2*k3 +k4)/6;}