如何将void *转换为int [] [3]

Question

I need to pass a variable of type int[][3] to a callback function which only accepts void* as parameter. How can I do that?

The below code does not compile:

void myfunc(void *param) {
    int i[][3];
    i=*param;
    printf("%d\n",i[1][2]);
}

int main(int argc, char *argv[])
{
    int i[][3]={
        {1, 2, 3},
        {4, 5, 6},
        {7, 8, 9}};
    myfunc(i);
}

Answer 1

Use:

int (*i)[3] = param;

Then use it as before. That's a pointer to an array of integers.

Answer 2

Detailed answer:

• The type of i in main() is int[3][3] , an array of arrays.
• When you use that as part of an expression, it "decays" into a pointer to the first element.
• The first element of the array int[3][3] is of type int[3] .
• A pointer to that first element is an array pointer of type int(*)[3] , so that is what i ends up as when you type myfunc(i) .
• Therefore, int(*)[3] is the correct type to use inside the function.
• If you have int (*i)[3]; inside the function, then acessing i[x] will give you array number x in the array of arrays.
• i[x][y] will give you array number x , item number y . So the syntax ends up the very same as when accessing a 2D array directly.

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Similarly, you could have written the function as void myfunc(int param[3][3]) . Param would implicitly decay into a pointer int(*)[3] "between the lines", allowing you to use param[x][y] inside the function. It only looks like the array is passed by value, but it isn't - doing that isn't possible in C.