如何使用指针表达式访问 C 中二维数组的元素? [英] How to use pointer expressions to access elements of a two-dimensional array in C?
问题描述
我知道对于一维数组 x=a[i] 等价于 x=*(a+i)，但是我如何访问使用指针的二维数组?
I know that for single-dimensional arrays x=a[i] is equivalent to x=*(a+i), but how can I access elements of a two-dimensional arrays using pointers?
推荐答案
总结:如果你有一个多维数组定义为int [][]，那么x = y[a][b] 等价于 x = *((int *)y + a * NUMBER_OF_COLUMNS + b);
Summary: If you have a multidimensional array defined as int [][], then x = y[a][b] is equivalent to x = *((int *)y + a * NUMBER_OF_COLUMNS + b);
无聊的细节:
上述 y 的 (int *) 类型转换值得一些解释，因为它的必要性起初可能并不直观.要了解为什么它必须存在，请考虑以下几点:
The (int *) cast of y above deserves some explanation, as its necessity may not be at-first intuitive. To understand why it must be there consider the following:
C/C++ 中的类型指针算术在按标量加/减/递增/递减时，总是根据类型的大小(以字节为单位)调整类型指针值(即地址).
Typed pointer arithmetic in C/C++ always adjusts the typed pointer value (which is an address) by the size of the type in bytes when adding/subtracting/incrementing/decrementing by scalar.
多维数组声明的基本类型(不是元素类型；变量类型)是一维数组类型比最终尺寸.
The fundamental type of a multi-dimensional array declaration (not the element type; the variable type) is an array-type of one-less dimension than the final dimension.
后者(#2)确实需要一个例子来巩固.下面的变量ar1和ar2是等价的声明.
The latter (#2) of these really needs an example to solidify. In the following, variables ar1 and ar2 are equivalent declarations.
int ar1[5][5]; // an array of 5 rows of 5 ints.
typedef int Int5Array[5]; // type is an array of 5 ints
Int5Array ar2[5]; // an array of 5 Int5Arrays.
现在是指针算术部分.正如类型化结构指针可以按结构的大小(以字节为单位)前进一样，数组的整个维度也可以跳过.如果您考虑我上面声明的 ar2 的多维数组，这会更容易理解:
Now the pointer arithmetic part. Just as a typed structure pointer can be advanced by the size of the structure in bytes, so can a full dimension of an array be hopped over. This is easier to understand if you think of the multi-dimensioned array as I declared ar2 above:
int (*arptr)[5] = ar1; // first row, address of ar1[0][0].
++arptr; // second row, address of ar[1][0].
所有这一切都用一个裸指针消失了:
All of this goes away with a bare pointer:
int *ptr = ar1; // first row, address of ar1[0][0].
++ptr; // first row, address of ar1[0][1].
因此，在对二维数组进行指针运算时，以下方法无法获取多维数组的 [2][2] 处的元素:
Therefore, when doing the pointer arithmetic for two-dimensional array, the following would NOT work in getting the element at [2][2] of a multi-dimensioned array:
#define NUMBER_OF_COLUMNS 5
int y[5][NUMBER_OF_COLUMNS];
int x = *(y + 2 * NUMBER_OF_COLUMNS + 2); // WRONG
当您记得 y 是一个数组数组(声明性地说)时，原因很明显.将(2*5 + 2)加到y上的指针算法将添加12个行，从而计算和地址等价于&(y[12])，这显然是不对的，事实上，要么在编译时抛出一个严重的警告，要么完全无法编译.使用 (int*)y 的强制转换和基于裸指针到 int 的表达式的结果类型可以避免这种情况:
The reason is hopefully obvious when you remember that y is an array of arrays (declaratively speaking). The pointer arithmetic of adding the scaler (2*5 + 2) to y will add 12 rows, thereby computing and address equivalent to &(y[12]), which is clearly not right, and in fact, will either throw a fat warning at compile time or outright fail to compile altogether. This is avoided with the cast of (int*)y and the resulting type of the expression being based on an bare pointer-to-int:
#define NUMBER_OF_COLUMNS 5
int y[5][NUMBER_OF_COLUMNS];
int x = *((int *)y + 2 * NUMBER_OF_COLUMNS + 2); // Right!
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