Question

我有一个用C编写的函数，我将其转换为matlab可执行文件（mex文件）。当从matlab的命令行调用一次时，C函数工作正常，但是当在for循环中调用1000次以上时，它会自发崩溃。即使我在for循环的每次迭代中向它提供相同的输入，也会发生这种情况。

恐怕我有一个潜伏的c-bug。重复内存分配的一些问题，但我不知道足够的c来修复它:(

我已经将问题缩小到下面代码中的WHILE循环（如果在编译问题之前整个while循环被注释掉了）。请帮忙！

    #include "mex.h"
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>

    int sum_array(double a[], int from ,int to);
    // to compile this code in matlab do: mex -v ranksort.c

   void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray  *prhs[]){
           #define B_OUT       plhs[0]
           #define A_IN        prhs[0]

           int i,j,count,M,N; //declare some integers
           double *list,*sort_list,*rank_idx,*sort_idx,*ranksOrigOrder;//declare some pointers to doubles which will be allocated dynamically
           int *rank_list,*tielocs,*orig_indices;

           //set values based on input:
           M=mxGetM(A_IN);//get input row num
           N=mxGetN(A_IN);//get input col num
           if (M>N)
                 count=M; //get dimensions of A (columns...data must always be in columns)
           else
                 count=N;

           list =mxGetPr(A_IN); //grab data from pointer to inputted data
           sort_list = calloc(count,sizeof(double)); //allocate size and fill w/zeros all my 'double' arrays
           rank_idx =calloc(count,sizeof(double));
           sort_idx=calloc(count,sizeof(double));
           ranksOrigOrder=calloc(count,sizeof(double));
           tielocs =calloc(count+2,sizeof(double));
           orig_indices=calloc(count,sizeof(int)); //allocate size and fill w/ zeros all my 'int' arrays
           rank_list =calloc(count,sizeof(int));

            if (sort_list==NULL||tielocs==NULL||rank_list==NULL||orig_indices==NULL||ranksOrigOrder==NULL||rank_idx==NULL||list==NULL){ puts ("Error (re)allocating memory"); exit (1); }

           B_OUT = mxCreateDoubleMatrix(M, N, mxREAL); //create a matlab-style struct for output...
           ranksOrigOrder = mxGetPr(B_OUT); // set in-code variable to its pointer

           /* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CODE BODY STARTS HERE ~~~~~~~~~~~~~~~~~~~~*/
           /*Calculate the rank for each element in the arr_list*/
           for(i=0; i<count; i++){
                 for(j=0; j<i; j++){
                     if(list[i] >= list[j])
                           rank_list[i]++;
                     else
                           rank_list[j]++;
                  }
            }

           for(i=0; i<count; i++){
               sort_list[rank_list[i]] = list[i];
               orig_indices[rank_list[i]] =i;
               sort_idx[i]=i+1;
           }
           int tiesIdx=0; int *ties = NULL;
           for (i=0; i<count-1;i++){
                 if (sort_list[i]>= sort_list[i+1]){
                 ties = (int*) realloc (ties, (tiesIdx) *sizeof(int));       //reallocate size of array
                 if (ties==NULL){ puts ("Error (re)allocating memory"); exit (1); }
                  ties[tiesIdx]=i; //add location of tie to newly grown array  
                   tiesIdx++; //found a tie
                  }
           }
           // // append 2 after last element to ties
           ties = (int*) realloc (ties, (tiesIdx) * sizeof(int));         

           //reallocate size of array
           if (ties==NULL){ puts ("Error (re)allocating memory"); exit (1);           }
           ties[tiesIdx]=count+1;
           tiesIdx++;

           int tiecount=0; //step thru all the found ties

           // NO IN-LOOP CRASHING if this while loop is commented out....
           while (tiecount<tiesIdx){
                 int tiestart =ties[tiecount]; //grab this tie to check if it is one of a pair or a member of an island of consecutives
                 int ntied=2;

                 while (ties[tiecount+1] == ties[tiecount] + 1){                         //while it's a consecutive one...
                        tiecount++;
                        ntied++;
                 }
                 double mysum = (double)sum_array(sort_idx,tiestart,tiestart+ntied)/(double)ntied;        

                 for (int t=tiestart; t<tiestart+ntied;t++){
                       sort_idx[t]=mysum;
                  }
                 tiecount++;
           }

           for (i=0; i<count;i++){
                ranksOrigOrder[orig_indices[i]]=sort_idx[i];
           }

           free(sort_list);
           free(tielocs);
           free(rank_list);
           free(orig_indices);
           free(rank_idx);
           free(sort_idx);
           return;
   }
    int sum_array(double a[], int from ,int to){
        int i, sum=0;
        for (i=from; i<to; i++){
               sum = sum + (int)a[i];
         }
        return(sum);
     }

Answer 1

看一下这段代码：

 // // append 2 after last element to ties
 ties = (int*) realloc (ties, (tiesIdx) * sizeof(int));

 //reallocate size of array
 if (ties==NULL){ puts ("Error (re)allocating memory"); exit (1);           }
 ties[tiesIdx]=count+1;
 tiesIdx++;

 int tiecount=0; //step thru all the found ties

 // NO IN-LOOP CRASHING if this while loop is commented out....
 while (tiecount<tiesIdx){

您正在为tiesIdx元素重新分配空间，稍后您想要访问tiesIdx数组的ties位置。这是Undefined Behaviour，因为您的数组必须从0索引到tiesIdx-1

之后你将tiesIdx加1，然后对该递增的var执行循环，这显然是Undefined Behaviour，因为你将索引数组超出范围。

此外，内循环：

       while (ties[tiecount+1] == ties[tiecount] + 1){                         //while it's a consecutive one...
              tiecount++;
              ntied++;
       }

不检查数组边界：如果所有关系元素都是连续的，那么你将把数组索引越界。

注意：ranksOrigOrder上的内存泄漏不是free()。

由c代码制作的mex在循环中崩溃，但在运行一次时不会崩溃

1 个答案: