c - 如何为矩阵中的特定元素动态分配内存？

Question

1 2 0 0 0
0 2 3 0 0
0 0 4 5 0
0 0 0 5 6
0 0 0 0 7

我正在尝试以这种格式分配一个矩阵，但不应动态分配等于 0 的元素。而且这种分配需要高效。

Answer 1

下面是一个可能的解决方案示例，可帮助您入门。其他方法也是可能的。此代码未经测试。

#include <stdlib.h>

typedef float T;

//  Attempt to allocate space for a bidiagonal matrix of size n-by-n.
T *MallocBidiagonalMatrix(size_t n) { return malloc((2*n-1) * sizeof(T)); }

//  Free a bidiagonal matrix.
void FreeBidiagonalMatrix(T *p) { free(p); }

//  Return a reference to element [r, c] of bidiagonal matrix p of size n-by-n.
T *BidiagonalMatrixElement(size_t n, T *p, size_t r, size_t c)
{
    static T Zero = 0;

    //  The diagonal elements are stored from p[0] to p[n-1].
    if (r == c)
        return p+r;

    //  The existing off-diagonal elements are stored from p[n] to p[2*n-2].
    if (r == c-1)
        return p+n+r;

    /*  For non-existing (zero) elements, return a reference to zero.  (Callers
        are expected not to write to this address.  If they do, the behavior is
        not defined.)
    */
    else
        return &Zero;
}

/*  Show how a bidiagonal matrix previously allocated with
    MallocBidiagonalMatrix might be initialized.
*/
void InitializeBidiagonalMatrix(size_t n, T *p)
{
    //  Initialize the diagonal elements.
    for (size_t i = 0; i < n; ++i)
        *BidiagonalMatrixElement(n, p, i, i) = rand();

    //  Initialize the off-diagonal elements.
    for (size_t i = 0; i < n-1; ++i)
        *BidiagonalMatrixElement(n, p, i, i+1) = rand();
}

/*  Show how a bidiagonal matrix might be used in arithmetic by demonstrating
    a simple matrix addition.  The n-by-n bidiagonal matrix p is added to
    n-by-n matrix q.
*/
void AddBidiagonalMatrixToDenseMatrix(size_t n, T *p, T (*q)[n])
{
    for (size_t r = 0; r < n; ++r)
        for (size_t c = 0; c < n; ++c)
            q[r][c] += *BidiagonalMatrixElement(n, p, r, c);
}

请注意，AddBidiagonalMatrixToDenseMatrix 可以优化为只处理存储的元素。但是，给定的实现演示了一种算法如何运行，而无需考虑实际存储的元素。