C语音程序无法运行？

Question

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
 
/* run this program using the console pauser or add your own getch, system("pause") or input loop /
 
float f(float x)
{
       return  pow(x,4)+2pow(x,3)+pow(x,2)-5;
}
 
 
int main(int argc, char *argv[])
{
       double a,b;
       double m,n,i,sum,root;
       double esp=1e-8;
       printf("Please enter the root interval[a,b]:");
       scanf("%lf %lf",&a,&b);
       i=0;
       if(f(a)*f(b)<0)
       {
              while(b-a>esp)
              {
                     m=(a+b)/2;
                     sum=f(m);
                     if (sum==0)
                     {
                            i++;
                break;
            }
                     else if (f(a)*f(m)<0)
                     {
                         m=b;
                         i++;
                     }
                     else
                     {
                         m=a;
                         i++;
                     }
              }
       }
       else
       {
          printf("There are no roots in this interval.\n");
       }
         
       root=(a+b)/2;
       printf("Number of iterations is %d\n",i);
       printf("The approximate root of the equation is %lf\n",root);
      
      
      
       system("PAUSE");
       return 0;
}

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
float f(float x)
{
       return  pow(x,4)+2*pow(x,3)+pow(x,2)-5;
}
 
int main(int argc, char *argv[])
{
       double a,b;
       double m,n,i,sum,root;
       double esp=1e-8;
       printf("Please enter the root interval[a,b]:");
       scanf("%lf %lf",&a,&b);
       i=0;
       if(f(a)*f(b)<0)
       {
              while(b-a>esp)
              {
                     m=(a+b)/2;
                     sum=f(m);
                     if (sum==0)
                     {
                            i++;
                break;
            }
                     else if (f(a)*f(m)<0)
                     {
                         m=b;
                         i++;
                     }
                     else
                     {
                         m=a;
                         i++;
                     }
              }
       }
       else
       {
          printf("There are no roots in this interval.\n");
       }
         
       root=(a+b)/2;
       printf("Number of iterations is %d\n",i);
       printf("The approximate root of the equation is %lf\n",root);
       system("PAUSE");
       return 0;
}