将多维数组传递给函数
将多维数组传递给函数时，多维数组遵循与单维数组相同的规则。然而，衰减到指针，运算符优先级和声明多维数组（数组数组与指针数组）的两种不同方式的组合可能使这些函数的声明不直观。以下示例显示了传递多维数组的正确方法。
#include <assert.h>
#include <stdlib.h>
/* When passing a multidimensional array (i.e. an array of arrays) to a
function, it decays into a pointer to the first element as usual. But only
the top level decays, so what is passed is a pointer to an array of some fixed
size (4 in this case). */
void f(int x[][4]) {
assert(sizeof(*x) == sizeof(int) * 4);
}
/* This prototype is equivalent to f(int x[][4]).
The parentheses around *x are required because [index] has a higher
precedence than *expr, thus int *x[4] would normally be equivalent to int
*(x[4]), i.e. an array of 4 pointers to int. But if it's declared as a
function parameter, it decays into a pointer and becomes int **x,
which is not compatable with x[2][4]. */
void g(int (*x)[4]) {
assert(sizeof(*x) == sizeof(int) * 4);
}
/* An array of pointers may be passed to this, since it'll decay into a pointer
to pointer, but an array of arrays may not. */
void h(int **x) {
assert(sizeof(*x) == sizeof(int*));
}
int main(void) {
int foo[2][4];
f(foo);
g(foo);
/* Here we're dynamically creating an array of pointers. Note that the
size of each dimension is not part of the datatype, and so the type
system just treats it as a pointer to pointer, not a pointer to array
or array of arrays. */
int **bar = malloc(sizeof(*bar) * 2);
assert(bar);
for (size_t i = 0; i < 2; i++) {
bar[i] = malloc(sizeof(*bar[i]) * 4);
assert(bar[i]);
}
h(bar);
for (size_t i = 0; i < 2; i++) {
free(bar[i]);
}
free(bar);
}