是二维数组一个双指针？ [英] Is 2d array a double pointer?

问题描述

  INT的main（）
{
    矩阵[2] [4] = {{11,22,33,99}，{44,55,66,110}};
    INT ** PTR =（INT **）矩阵;
    的printf（％D，**矩阵，* PTR）;
}
 

但是，当一个2-D阵列，因为它是类型强制转换成（*矩阵）一个参数传递[2] ..
什么类型的呢编译器存储这个数组...是存储为一个2-D数组或一个双指针或指向数组的指针。如果它被存储为一个数组是怎么做的间$ P $点在不同不同的情况下像上面。请帮我明白了。

解决方案

  

是2D的阵列中的双指针？

没有。你的程序的这一行不正确：

  INT ** PTR =（INT **）矩阵;
 

<一个href=\"http://stackoverflow.com/questions/7351331/please-explain-the-ambiguity-in-pointers-in-c/7351390#7351390\">This与同一主题答案便宜的价格

如果你想具体的形象如何多维数组来实现：

有关多维数组的规则不是从那些普通阵列不同，刚刚替补内部数组类型的元素类型。该阵列项目后直接相互存储在内存中：

 矩阵：11 22 33 99 44 55 66 110
        -----------矩阵的第一元素
                    ------------矩阵的第二元件
 

因此​​，要解决的元素矩阵[X] [Y] ，你把矩阵的基地址+ X * 4 + Y （四是内部数组的大小）。

当数组传递给函数，它们衰变为指针，以他们的第一个元素。当你注意到了，这将是 INT（*）[4] 。那么 4 中的类型会告诉编译器内部类型的大小，这就是为什么它的工作原理。当类似的指针做指针算法，编译器将元素大小的倍数，所以对 matrix_ptr [X] [Y] ，你得到 matrix_ptr + X * 4 + Y ，这是完全一样的上方。

演员 PTR =（INT **）矩阵因此，不正确。这一次， * PTR 将意味着储存在矩阵的地址的指针值，但没有任何。其次，没有一个指针矩阵[1] 在程序的存储器中的任何

注：在这篇文章的计算假设的sizeof（int）的== 1 ，以避免不必要的复杂性。

int main()
{
    matrix[2][4] = {{11,22,33,99},{44,55,66,110}};
    int **ptr = (int**)matrix;
    printf("%d%d",**matrix,*ptr);
}

But when a 2-d array is passed as a parameter it is typecasted into (*matrix)[2] .. what type does the compiler store this array as... is it storing as a 2-d array or a double pointer or an pointer to an array .. If it is storing as an array how does it interprets differently at different situations like above. Please help me understand.

解决方案

Is 2d array a double pointer?

No. This line of your program is incorrect:

int **ptr = (int**)matrix;

This answer deals with the same topic

If you want concrete image how multidimensional arrays are implemented:

The rules for multidimensional arrays are not different from those for ordinary arrays, just substitute the "inner" array type as element type. The array items are stored in memory directly after each other:

matrix: 11 22 33 99 44 55 66 110
        -----------               the first element of matrix
                    ------------  the second element of matrix

Therefore, to address element matrix[x][y], you take the base address of matrix + x*4 + y (4 is the inner array size).

When arrays are passed to functions, they decay to pointers to their first element. As you noticed, this would be int (*)[4]. The 4 in the type would then tell the compiler the size of the inner type, which is why it works. When doing pointer arithmetic on a similar pointer, the compiler adds multiples of the element size, so for matrix_ptr[x][y], you get matrix_ptr + x*4 + y, which is exactly the same as above.

The cast ptr=(int**)matrix is therefore incorrect. For once, *ptr would mean a pointer value stored at address of matrix, but there isn't any. Secondly, There isn't a pointer to matrix[1] anywhere in the memory of the program.

Note: the calculations in this post assume sizeof(int)==1, to avoid unnecessary complexity.

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