地址0x0不stack'd，malloc分配或（最近）free'd

-1

我目前工作的一个程序来解决背包问题，我不得不使用指针的矩阵。我在学校被赋予使用伪后，我不断收到分段错误，并根据Valgrind的就是这个道理：地址0x0不stack'd，malloc分配或（最近）free'd

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==

我曾试图用过去的答案来解决这个问题，但我似乎无法图出来。在学校的其他人做了完全相同的事情，他们似乎并没有收到这个问题。是否有任何错误，我似乎没有看到或在代码中实现？

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]; 
}

来源

2017-03-18 Apostolescu Andrei

NULL指针解除引用。 –

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

在malloc电话提供的尺寸是没有意义的。而且你从未初始化过V[0]（或者任何V[i]）。它包含一个垃圾值。但是，您尝试访问V[0][w]（和稍后的V[i][w]）。这是未定义的行为。

如果你打算使用你的V作为一个二维数组，allocate it properly第一。

来源

2017-03-18 04:23:11 AnT

地址0x0不stack'd，malloc分配或（最近）free'd

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