无法从int [] []转换为int*

Question

我有一个3x3数组,我正在尝试创建一个指针,我不断得到这个数组,是什么给出的？

我如何定义指针？我已经尝试了[]和*的每个组合.

是否有可能做到这一点？

int tempSec[3][3];

int* pTemp = tempSec;

Answer 1

你可以做 int *pTemp = &tempSec[0][0];

如果要将3x3数组视为int*,则应该将其声明为a int[9],并使用tempSec[3*x+y]而不是tempSec[x][y].

或者,也许你想要的是int (*pTemp)[3] = tempSec什么？那将是指向tempSec的第一个元素的指针,第一个元素本身就是一个数组.

实际上,您可以使用指向2D数组的指针:

int (*pTemp)[3][3] = &tempSex;

然后你会像这样使用它:

(*pTemp)[1][2] = 12;

这几乎肯定不是你想要的,但在你的评论中,你确实要求它......

Answer 2

它更容易使用typedef

typedef int  ThreeArray[3];
typedef int  ThreeByThree[3][3];

int main(int argc, char* argv[])
{
    int         data[3][3];

    ThreeArray* dPoint    = data;
    dPoint[0][2]          = 5;
    dPoint[2][1]          = 6;   

    // Doing it without the typedef makes the syntax very hard to read.
    //
    int(*xxPointer)[3]    = data;
    xxPointer[0][1]       = 7;

    // Building a pointer to a three by Three array directly.
    //
    ThreeByThree*  p1     = &data;
    (*p1)[1][2]           = 10;

    // Building a pointer to a three by Three array directly (without typedef)
    //
    int(*p2)[3][3]        = &data;
    (*p2)[1][2]           = 11;

    // Building a reference to a 3 by 3 array.
    //
    ThreeByThree&  ref1   = data;
    ref1[0][0]            = 8;

    // Building a reference to a 3 by 3 array (Without the typedef)
    //
    int(&ref2)[3][3]      = data;
    ref2[1][1]            = 9;


    return 0;
}