malloc_in_function.c 

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 
 5 void malloc_in_function(char **myArray, int size)
 6 {
 7   int i = 0;
 8 
 9   *myArray = (char *) malloc(size * sizeof(char));
10   if (*myArray == NULL)
11   {
12     fprintf(stderr, "Error allocating memory for myArray!\n");
13     exit(0);
14   }
15 
16   /* this is how to access array members in the function */
17   for (i = 10; i < 20; i++)
18   {
19     /* This line will cause a segmentation fault. Why? */
20       /* *myArray[i] = 'a' + (i - 10); */
21 
22     /* This line works much better. */
23     (*myArray)[i] = 'a' + (i - 10);
24   }
25 }
26 
27 int main(int argc, char **argv)
28 {
29   /* char myArray[20]; */
30   char *myArray;
31   int size = 20;
32   int i = 0;
33 
34   malloc_in_function(&myArray, size);
35 
36   for (i = 0; i < 10; i++)
37   {
38     myArray[i] = 'A' + i;
39   }
40     
41   for (i = 0; i < 20; i++)
42   {
43     printf("myArray[%d] = %c\n", i, myArray[i]);
44   }
45 
46   free(myArray);
47 }
48 
49 /* in main(), what does (myArray + 10) and &myArray[10] have in common?
50  * What about *(myArray + 10) and myArray[10]?
51  * What is *myArray + 10? Is it the same as myArray[0] + 10?
52  * If the malloc was for an int, would (myArray + 10) be the same as &myArray[10]?
53  */

 

posted @ 2020-04-29 23:07  Jasper2003  阅读(164)  评论(0)    收藏  举报