Question

我目前正在研究解决背包问题的程序，我不得不使用指向矩阵的指针。在使用伪代码后我在学校给出了我一直收到分段错误，根据valgrind，这就是原因：

1 errors in context 1 of 2:
==10545== Invalid write of size 4
==10545==    at 0x4009C3: DP (dp.c:70)
==10545==    by 0x400914: main (dp.c:39)
==10545==  Address 0x0 is not stack'd, malloc'd or (recently) free'd
==10545==
==10545==
=
=10545== 1 errors in context 2 of 2:
==10545== Use of uninitialised value of size 8
==10545==    at 0x4009C3: DP (dp.c:70)
==10545==    by 0x400914: main (dp.c:39)
==10545==

我试图用过去的答案解决问题，但我似乎无法弄明白。在学校的其他人做了完全相同的事情，他们似乎没有收到这个问题。在代码中我似乎没有看到或意识到的任何错误？

 int **V, **keep;  // 2d arrays for use in the dynamic programming solution
 // keep[][] and V[][] are both of size (n+1)*(W+1)

 int i, w, K;

 // Dynamically allocate memory for variables V and keep
 V = (int**)malloc((n+1) * (W+1) * sizeof(int*));
 keep = (int**)malloc((n+1) * (W+1) * sizeof(int*));

 //  set the values of the zeroth row of the partial solutions table to zero
 for (w = 0; w <= W; w++)
 {
    V[0][w] = 0;
 }

 // main dynamic programming loops , adding one item at a time and looping          through weights from 0 to W
 for (i = 1; i <= n; i++)
   for (w = 0; w <= W; w++)
     if ((wv[i] <= w) && (v[i] + V[i-1][w - wv[i]] > V[i - 1][w]))
     {
       V[i][w] = v[i] + V[i-1][w-wv[i]];
       keep[i][w] = 1;
     }

     else
     {
       V[i][w] = V[i-1][w];
       keep[i][w] = 0;
     } 

 K = W;
 for (i = n; i >= 1; i--);
 {
   if (keep[i][K] == 1)
   {
     printf("%d\n", i);
     K = K - wv[i];
   }
 } 
 return V[n][W];
 }

Answer 1

V = (int**)malloc((n+1) * (W+1) * sizeof(int*));
...
for (w = 0; w <= W; w++)
{
  V[0][w] = 0;
}

malloc来电中提供的尺寸没有任何意义。并且您从未初始化V[0]（或任何V[i]。它包含垃圾值。但您尝试访问V[0][w]（以及V[i][w]之后）。这是未定义的行为。

如果您打算将V用作2D数组，请先allocate it properly。

地址0x0不是堆栈＆＃39; d，malloc＆d; dd或（最近）空闲＆＃d; d

1 个答案: