关于将C语言转换为C#中的指针问题

Question

将C语言的代码转换为C#的代码，C中的指针在C#中应该是用数组吧？
那位大神看看下面的转换
void __ogg_fdrffti(int n, double *wsave, int *ifac){  if (n == 1) return;
  drfti1(n, wsave+n, ifac);
}/*c代码*/
        void __ogg_fdrffti(int n,  double[] wsave, int[] ifac)
        {
            if (n == 1) return;
            drfti1(n,wsave+n,ifac);
        }/*转换后的C#代码*/
但是在wsave+n下提示有错误为“运算符“+”无法应用于double[]和int类型的操作数”
求错误的的解决方法，不胜感激

Answer 1

C# CODE:
 void __ogg_fdrffti(int n, double[] wsave, int[] ifac)
  {
  if (n == 1) return;
  drfti1(n,wsave[n],ifac);
  }给分

Answer 2

wsave+n,这样c#不支持，c#只能通过下标访问，wsave[n]

Answer 3

这就是数组首地址+偏移量
在C#里面就是array[offse]呗。。如上#4所述

Answer 4

打开“unsafe”编译开关，就支持了。

Answer 5

关键是看你的drfti1是如何改写的。

Answer 6

        static void drfti1(int n,double[] wa,double[] ifac){            
             int[] ntryh ={ 4,2,3,5 };
             double tpi = 6.28318530717958647692528676655900577;
             double arg,argh,argld,fi;
             int ntry=0,i,j=-1;
              int k1, l1, l2, ib;
             int ld, ii, ip;
             int ls, nq, nr;
             int ido, ipm, nfm1;
             int nl=n;
             int nf=0;             int temp;
            L101:
             j++;
             if (j < 4)
               ntry=ntryh[j];
             else
               ntry+=2;            L104:
             nq=nl/ntry;
             nr=nl-ntry*nq;
             if (nr!=0) goto L101;             nf++;
             ifac[nf+1]=ntry;
             nl=nq;
             if(ntry!=2)goto L107;
             if(nf==1)goto L107;             for (i=1;i<nf;i++){
              ib=nf-i+1;
             ifac[ib+1]=ifac[ib];
             }
             ifac[2] = 2;             L107:
             if(nl!=1)goto L104;
             ifac[0]=n;
             ifac[1]=nf;
             argh=tpi/n;
              ls=0;
             nfm1=nf-1;
             l1=1;             if(nfm1==0)return;             for (k1=0;k1<nfm1;k1++){
               ip=(int)ifac[k1+2];
                ld=0;
               l2=l1*ip;
               ido=n/l2;
               ipm=ip-1;               for (j=0;j<ipm;j++){
                 ld+=l1;
                 i=ls;
                 temp = i + n;
                 argld=(double)ld*argh;
                 fi=0;
                 for (ii=2;ii<ido;ii+=2){
             fi+=1;
             arg=fi*argld;
             wa[temp++]=Math.Cos(arg);
             wa[temp++]=Math.Sin(arg);
                 }
                 ls+=ido;
                }
               l1=l2;
              }
    }
这是经过修改的C#写的drftil

Answer 7

既然你的drfti1是改写得来的，那么顺便把参数列表也改写下，增加1个参数，用来传递那个wsave+n的n值，内部处理就是数组的偏移量了。