#include<stdio.h>
#include "stdlib.h"
#include "math.h"
int n;
int agaus(a,b,n)
double a[],b[];    
  {
      int *js,l,k,i,j,is,p,q;
    double d,t;
    js=malloc(n*sizeof(int));
    l=1;
    for (k=0;k<=n-2;k++)
      { d=0.0;
        for (i=k;i<=n-1;i++)
          for (j=k;j<=n-1;j++)
            { t=fabs(a[i*n+j]);
              if (t>d) { d=t; js[k]=j; is=i;}
            }
        if (d+1.0==1.0) l=0;
        else
          { if (js[k]!=k)
              for (i=0;i<=n-1;i++)
                { p=i*n+k; q=i*n+js[k];
                  t=a[p]; a[p]=a[q]; a[q]=t;
                }
            if (is!=k)
              { for (j=k;j<=n-1;j++)
                  { p=k*n+j; q=is*n+j;
                    t=a[p]; a[p]=a[q]; a[q]=t;
                  }
                t=b[k]; b[k]=b[is]; b[is]=t;
              }
          }
        if (l==0)
          { free(js); printf("fail\n");
            return(0);
          }
        d=a[k*n+k];
        for (j=k+1;j<=n-1;j++)
          { p=k*n+j; a[p]=a[p]/d;}
        b[k]=b[k]/d;
        for (i=k+1;i<=n-1;i++)
          { for (j=k+1;j<=n-1;j++)
              { p=i*n+j;
                a[p]=a[p]-a[i*n+k]*a[k*n+j];
              }
            b[i]=b[i]-a[i*n+k]*b[k];
          }
      }
    d=a[(n-1)*n+n-1];
    if (fabs(d)+1.0==1.0)
      { free(js); printf("fail\n");
        return(0);
      }
    b[n-1]=b[n-1]/d;
    for (i=n-2;i>=0;i--)
      { t=0.0;
        for (j=i+1;j<=n-1;j++)
          t=t+a[i*n+j]*b[j];
        b[i]=b[i]-t;
      }
    js[n-1]=n-1;
    for (k=n-1;k>=0;k--)
      if (js[k]!=k)
        { t=b[k]; b[k]=b[js[k]]; b[js[k]]=t;}
    free(js);
    return(1);
  }

//&Ouml;÷&ordm;&macr;&Ecirc;&yacute;
void main()
{
    int i;
    static double a[2][2]={{1,1},{1,-1}};
    static double b[2]={3,1};
     if (agaus(a,b,2)!=0)
     {
         for (i=0;i<=1;i++)
             printf("x(%d)=%.2f\n",i,b[i]);
     }
}
请问agaus(a.b,n)函数下面的double a[],b[];是什么意思？这行代码的位置似乎还不能动。。。

为什么主函数定义二维double a[][],而agaus中的数组却定义为double a[];