将表示二维数组的指针传递给 C++ 中的函数
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Passing a pointer representing a 2D array to a function in C++
提问by Derek
According to this site, I should be able to use the following code:
根据此站点，我应该可以使用以下代码：
double stuff[3][3];
double **p_stuff;
p_stuff = stuff;
But I get a complaint that the conversion is not allowed by assignment.
但是我收到投诉，说分配不允许转换。
Am I doing something wrong?
难道我做错了什么？
I have an extern "C" type function that I want to pass this double stuff[3][3] to. So I think i need to make it a pointer, right?
我有一个 extern "C" 类型的函数，我想将这个 double stuff[3][3] 传递给。所以我想我需要让它成为一个指针，对吗？
回答by Cubbi
Regarding the edit: to pass this double stuff[3][3]to a C function, you could
关于编辑：要将其传递double stuff[3][3]给 C 函数，您可以
1) pass a pointer to the whole 2D array:
1) 传递一个指向整个二维数组的指针：
void dostuff(double (*a)[3][3])
{
// access them as (*a)[0][0] .. (*a)[2][2]
}
int main()
{
double stuff[3][3];
double (*p_stuff)[3][3] = &stuff;
dostuff(p_stuff);
}
2) pass a pointer to the first 1D array (first row) and the number of rows
2）传递指向第一个一维数组（第一行）和行数的指针
void dostuff(double a[][3], int rows)
{
// access them as a[0][0] .. a[2][2]
}
int main()
{
double stuff[3][3];
double (*p_stuff)[3] = stuff;
dostuff(p_stuff, 3);
}
3) pass a pointer to the first value in the first row and the number of both columns and rows
3) 传递指向第一行第一个值的指针以及列数和行数
void dostuff(double a[], int rows, int cols)
{
// access them as a[0] .. a[8];
}
int main()
{
double stuff[3][3];
double *p_stuff = stuff[0];
dostuff(p_stuff, 3, 3);
}
(that this last option is not strictly standards-compliant since it advances a pointer to an element of a 1D array (the first row) past the end of that array)
（这最后一个选项并不严格符合标准，因为它将指向一维数组（第一行）元素的指针提前到该数组的末尾）
If that wasn't a C function, there'd be a few more options!
如果那不是 C 函数，那么还有更多选择！
回答by Prasoon Saurav
Your assigned is flawed. p_stuff;is pointer to pointer to doublewhereas stuffis two dimensional array( array of arrays)
你的分配有缺陷。p_stuff;是指向指针的指针，double而是stuff二维数组（数组数组）
A single dimension array decays to the pointer to its first element. A 2 dimensional array decays to pointer to a single dimension array.
一维数组衰减到指向其第一个元素的指针。二维数组衰减为指向一维数组的指针。
Try this
尝试这个
double stuff[3][3];
double (*p_stuff)[3]; // pointer to array of 3 int
p_stuff = stuff;
回答by Erik
double ** p_stuff;corresponds to an array of pointer to double. double stuff[3][3]doesn't have any pointers - it's a 2D array of double.
double ** p_stuff;对应于指向 double的指针数组。double stuff[3][3]没有任何指针 - 它是一个二维数组。
回答by Th?o Hoàng Minh
Below is maybe your answer:
下面也许是你的答案：
#include <stdio.h>
#include <stdlib.h>
typedef unsigned int uint;
uint m[10][20];
uint **ppm;
int main() {
int i;
ppm = (uint **)m;
for (i =0; i<10; ++i)ppm[i] = (uint *)&m[i];
m[1][1] = 10;
printf("0x%x vs 0x%x: %d vs %d\n", ppm,m, m[1][1], *(*(ppm+1)+1));
return 0;
}
The result's on console screen:
结果在控制台屏幕上：
0x6010a0 vs 0x6010a0: 10 vs 10
回答by Simon Woo
In addition to Cubbi's detailed answer, the following way is just natural to pass a 2-dim array to a function:
除了 Cubbi 的详细答案之外，以下方式很自然地将 2-dim 数组传递给函数：
void dostuff(void *a1, int rows, int cols)
{
double (*a)[cols] = (double (*)[cols]) a1;
// access them as a[0][0] .. a[rows-1][cols-1];
}
int main()
{
double stuff[3][3];
dostuff(stuff, 3, 3);
}
You don't have to cast them in advance since the newest C++ (C++14) supports run-time sized arrays.
由于最新的 C++ (C++14) 支持运行时大小的数组，因此您不必提前强制转换它们。
