聲明C函數以返回數組如何創建返回數組的函數？我試過這個const int WIDTH=11;const int HEIGHT=11;int main() {
char A[WIDTH][HEIGHT];
A=rand_grid(WIDTH,HEIGHT);
return 0;}// Initializes a random board.char[][] rand_grid(int i, int k) {
char* A[i][k];
for(j=0;j<i;++j) {
for(l=0;l<k;++l) {
A[j][l]=ran(10);
}
}
return A;}// Returns a random number from the set {0,...,9}.int ran(int i) {
srand((unsigned int) time(0));
return(rand()%10);}
3 回答
喵喵時光機
TA貢獻1846條經驗 獲得超7個贊
char A[WIDTH][HEIGHT]; A=rand_grid(WIDTH,HEIGHT);
char (*foo(int width))[HEIGHT]{
/**
* dynamically allocate memory for a widthxHEIGHT array of char
*/
char (*newArr)[HEIGHT] = malloc(sizeof *newArr * width);
/**
* initialize array contents here
*/
return newArr;}foo -- foo foo(int width) -- is a function -- taking an int parameter *foo(int width) -- returning a pointer (*foo(int width))[HEIGHT] -- to a HEIGHT-element arraychar (*foo(int width))[HEIGHT] -- of char
void foo (T *p) {...}...T arr[N];foo(arr);void foo (T (*p)[M]) {...}...T arr[N][M];foo(arr);void foo(T *base, size_t rows, size_t cols) {...}...T arr[N][M];foo (&arr[0][0], N, M);void rand_grid(char *base, size_t rows, size_t cols){
size_t i, j;
for (i = 0; i < rows; i++)
{
for (j = 0; j < cols; j++)
{
/**
* Since base is a simple char *, we must index it
* as though it points to a 1-d array. This works if
* base points to the first element of a 2-d array,
* since multi-dimensional arrays are contiguous.
*/
base[i*cols+j] = initial_value();
}
}}int main(void){
char A[WIDTH][HEIGHT];
rand_grid(&A[0][0], WIDTH, HEIGHT);
...}即使這些表情 &A[0][0]和 A產生相同的值(A的基址)，這兩個表達式的類型是不同的。第一個表達式計算為指向char的簡單指針( char *)，而第二個值計算為指向char的二維數組的指針( char (*)[HEIGHT]).
守著星空守著你
TA貢獻1799條經驗 獲得超8個贊
autostructmalloc()struct.
struct
#define WIDTH 11#define HEIGHT 11typedef struct {
unsigned char cell[WIDTH * HEIGHT];} Board;Board board_new(void){
Board b;
size_t i;
for(i = 0; i < sizeof b.cell / sizeof *b.cell; i++)
b.cell[i] = rand() & 255;
return b;}void board_init(Board *b);
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