C语言double 型数组转short

[sinat_38089000]

举个例子如下，为什么程序无法运行。
#include "stdio.h"
#include "stdlib.h"

int main()
{
int i,j;
short *T;
double **X;
X = (double **)malloc( sizeof(double*) * 10000);
for(i=0;i<10000;i++)
{
for(j=0;j<2;j++)
{
X[i][j] = -0.003+0.2*(j+1);
}
}
T = (short *)malloc( sizeof(short) * 10000);
for(i=0;i<10000;i++)
{
T[i] = (short)(X[i][0]*3000);
}

for(i=0;i<10000;i++)
{
printf("%d", T[i]);
printf("\n");
}

free(X);
X=NULL;
free(T);
T=NULL;

return 0;

}

[自信男孩]

#include "stdio.h"
#include "stdlib.h"

int main()
{
    int i,j;
    short *T;
    double **X;
    X = (double **)malloc( sizeof(double*) * 10000); 
    for(i=0;i<10000;i++)
    {
        X[i] = (double *)malloc(sizeof(double) * 2);    /*申请第二维的空间*/
        for(j=0;j<2;j++)
        {
            X[i][j] = -0.003+0.2*(j+1);
        }
    }
    T = (short *)malloc( sizeof(short) * 10000); 
    for(i=0;i<10000;i++)
    {
        T[i] = (short)(X[i][0]*3000);
    }

    for(i=0;i<10000;i++)
    {
        printf("%d", T[i]);
        printf("\n");
    }

    for(i=0;i<10000;i++)    /*释放注意循环释放，因为是循环申请的*/
        free(X[i]);
    free(X);
    X = NULL;
    free(T);
    T = NULL;

    return 0;
}

详见上面的代码

[sinat_38089000]

已经解决了，举的这个例子有点问题，哈哈，不过还是谢谢你