我有這段代碼,當我的函數loadRow(tmpPop,bestOf4Route,k,n)給出了我的SEGMENTATION FAULT;被稱爲第五次。特別是,函數在第一個循環(在p = 3時)被正確調用。對於每個k值都可以。我不明白爲什麼,當我第二次執行循環(p = 7)時,第一次被稱爲(k = 0),當它嘗試訪問基類tmpPop時返回SEGMENTATION F.C矩陣函數分段錯誤
randperm(popSize,randomOrder);
for(p = 3;p<popSize;p+=4)
{
load_rtes(rtes,pop,randomOrder,n,p);
load_dists(dists,totalDist,randomOrder,p);
find_min(dists, 4,&m,&idx);
for(j=0;j<n;j++) bestOf4Route[j]=rtes[j][idx];
X = rand_range(1,5);
Y = rand_range(1,5);
for(k =0;k<4;k++) //Mutate the Best to get Three New Routes
{
loadRow(tmpPop,bestOf4Route,k,n);
switch(k)
{
case 1: //Flip
flipMutation(tmpPop,k,X,Y);
break;
case 2: //Swap
swapMutation(tmpPop,k,X,Y);
break;
case 3: //Slide
slideMutation(tmpPop,k,X,Y);
break;
}
}
loadnewPop(newPop,tmpPop,p,n);
}
功能是:
void loadRow(int **mat,int *array,int k,int size)
{
int j;
for(j=0;j<size;j++)
{
mat[j][k] = array[j];
}
}
的參數是:
popSize = 16
n= 8
// create 4 x N matrix
tmpPop = (int**)malloc(n * sizeof(int*));
if(tmpPop==NULL) return 1;
for (i = 0; i < n; i++) {
tmpPop[i] = (int*)malloc(4 * sizeof(int));
if(tmpPop[i]==NULL) return 1;
}
// Creates an array of n
bestOf4Route = (int*)malloc(n * sizeof(int));
if(bestOf4Route==NULL) return 1;
clear_array(bestOf4Route,n);
而且她這是調試結果:
00401865 loadRow(墊= 0x3e1438,陣列= 0x3e1698中,k = 0,大小= 8)
void load_rtes(int **rtes,int **pop,int *randomOrder, int n,int p)
{
int i,j,r;
for(i=p-3;i<=p;i++)
{
//thakes the i element of randomOrder and use it as index for the pop row
r=randomOrder[i];
// copy the pop row in rtes
for(j=0;j<n;j++)
{
rtes[j][i]=pop[j][r];
}
}
}
void randperm(int n,int *perm)
{
int i, j, t;
for(i=0; i<n; i++)
perm[i] = i;
for(i=0; i<n; i++) {
j = rand()%(n-i)+i;
t = perm[j];
perm[j] = perm[i];
perm[i] = t;
}
}
你也可以發佈'load_rtes()'函數嗎?您間接使用它來填充'tmpPop'。 – 2013-02-13 12:06:54
好吧,我已將它添加到問題中! – Malo 2013-02-13 13:06:54
現在你初始化'randomOrder'的代碼:)你需要向我們展示**所有**代碼,這些代碼以某種方式與這個'loadRow'函數交互。 – 2013-02-13 13:46:46