将三重指针传递给函数，然后分配一个二维数组

Question

遇到分段错误并且不明白为什么，必须传入三重指针进行赋值，因此无法更改...

这是功能

    void alloc2d(double*** a, int m, int n) {
     int i, j;

     **a  = malloc(sizeof(double *) * m);
     a[0] = malloc(sizeof(double) * n * m);

     for(i = 0; i < m; i++)
      a[i] = (*a + n * i);
    }

这是函数的调用...

double** m;
alloc2d(&m, 5, 10);

double count = 0.0;

for (int i = 0; i < 5; i++)
    for (int j = 0; j < 10; j++)
        m[i][j] = ++count;

for (int i = 0; i <  5; i++)
    for (int j = 0; j < 10; j++)
        printf("%f ", m[i][j]);

Answer 1

您想为 wherea所指的指针分配内存。更改**a = malloc(sizeof(double *) * m);为*a = malloc(sizeof(double *) * m);。你已经为 分配了内存a[0]，所以用 index 开始你的循环1。该类型a[0]是类似*a的double**，但你要设置在指针a[0]指。更改a[0]到(*a)[0]和a[i]到(*a)[i]。像这样调整你的代码：

void alloc2d(double*** a, int m, int n) {

     *a = malloc( sizeof( double* ) * m );       // allocate the memory for (m) double*

     (*a)[0] = malloc( sizeof(double) * n * m ); // allocate the linearized memory for (n*m) double

     for( int i=1; i<m; i++ )
         (*a)[i] = (*a)[0] + n*i;                // initialize the pointers for rows [1, m[
}

我认为这会更容易理解：

void alloc2d(double*** a, int m, int n) {

     double **temp = malloc( sizeof( double* ) * m ); // allocate the memory for (m) double*

     temp[0] = malloc( sizeof(double) * n * m );      // allocate the linearized memory for (n*m) double
     for ( int i=1; i<m; i++ )
         temp[i] = temp[0] + n*i;                     // initialize the pointers for rows [1, m[

     *a = temp;                                       // assign dynamic memory pointer to output parameter
}