Passing a pointer representing a 2D array to a function in C++ (转载)
原文链接:http://stackoverflow.com/questions/5329107/passing-a-pointer-representing-a-2d-array-to-a-function-in-c
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According to this site, I should be able to use the following code:
double stuff[3][3];
double **p_stuff;
p_stuff = stuff;
But I get a complaint that the conversion is not allowed by assignment.
Am I doing something wrong?
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?
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Regarding the edit: to pass this double stuff[3][3] to a C function, you could
1) pass a pointer to the whole 2D array:
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
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
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)
If that wasn't a C function, there'd be a few more options!

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