[英]Getting one-dimensional arrays from a two-dimensional array
I have an array like 我有一个类似的数组
int outer[4][3] = {
{ 1, 2, 3 },
{ 2, 3, 5 },
{ 1, 4, 9 },
{ 10, 20, 30 }
};
and I would like to get a pointer/array for the n-th one-dimensional array inside outer
, something like 我想获得用于内的第n个一维数组的指针/阵列
outer
,像
void foo () {
printf("%d %d %d\n", outer[1][0], outer[1][1], outer[1][2]);
int inner[3] = outer[1]; /* Is there some way to do this assignment? */
printf("%d %d %d\n", inner[0], inner[1], inner[2]);
/* so that this line gives the same output as the first */
}
Of course this is possible with pointer math, but I feel like there is some syntax for this that I've forgotten. 当然这可以通过指针数学来实现,但我觉得有一些我已经忘记的语法。
For pointer to array, declare
inner
as a pointer to an array of 3
int
对于指向数组的指针,将
inner
声明为指向3
int
数组的指针
int (*inner)[3] = &outer[1];
If you want a pointer to first element of array outer[1]
then 如果你想要一个指向数组
outer[1]
第一个元素的指针
int *inner = outer[1];
will do the job. 会做的。 You can also do
你也可以
int *inner = &outer[1][0];
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