c Dynamic allocation of two-dimensional arrays of structural bodies

I have been struggling with this problem for a long time, because I have been busy (myself lazy), so I have not been able to study this well. I hope this article will help you.

1#include <stdio.h>2#include <stdlib.h>3#include <stddef.h>4 5typedefstructLnode {6     intF;7     structlnode*Next;8}lnode, *linklist;9 intMain ()Ten { Onelnode** map = (Lnode * *)malloc(5*sizeof(lnode*));//allocate 5 struct pointer space A      for(inti =0; I <5; ++i)//this loop allocates the corresponding space to 5 pointers. -     { -Map[i] = (Lnode *)malloc(Ten*sizeof(Lnode));//Allocate 10 node space the     } -  -      for(inti =0; I <5; ++i) -          for(intj =0; J <Ten; ++j) +         { -(Map[i] + j)->f =J; +         } A      for(inti =0; I <5; ++i) at     { -          for(intj =0; J <Ten; ++j) -         { -printf"%d\t", (Map[i] + j)F); -         } -printf"\ n"); in     } -     return 0; to}

Example One:

Example one thought: 1, the allocation of structural body pointer space;

2. Allocate the corresponding node space for the place where the pointer refers.

1#include <stdio.h>2#include <stdlib.h>3#include <stddef.h>4 5typedefstructlnode{6     intF;7     structlnode*Next;8}lnode,*Plnode;9 intMain ()Ten { Onelnode** map = (Lnode * *)malloc(5*sizeof(lnode*)); ALnode *tmap = (Lnode *)malloc(5*Ten*sizeof(Lnode)); -      for(inti =0; I <5; ++i) -     { theMap[i] = (Lnode *) (Char*) TMap + i *Ten*sizeof(Lnode));  -     } -      for(inti =0; I <5; ++i) -          for(intj =0; J <Ten; ++j) +(Map[i] + j)->f =J; -      for(inti =0; I <5; ++i) +     { A          for(intj =0; J <Ten; ++j) at         { -printf"%d\t", (Map[i] + j)F); -         } -printf"\ n"); -     } -      Free(TMAP); in      Free(map); -     return 0; to}

Example Two:

Example two ideas: 1, the allocation of structural body pointer space;

2, the allocation of the corresponding number of nodes space;

3. Use an array of pointers to split.

Note: I was just beginning to write "map[i" = TMap + i * * sizeof (LNODE); " So, since TMap here is the lnode struct pointer, he moves to "I * * * sizeof (Lnode)" *sizeof (Lnode). Give an example int *a; So a +3; Is the "A-point address" + 3*sizeof (int). So, if you want to add this, you can convert the TMAP to the address of the char* type in addition, and then force the type conversion, as shown in the code map[i] = (Lnode *) ((char *) TMAP + i * * sizeof (Lnode)). Of course, such as map[i] =tmap + i * 10.

1#include <stdio.h>2#include <stdlib.h>3#include <stddef.h>4 5typedefstructlnode{6     intF;7     structlnode*Next;8}lnode,*Plnode;9 intMain ()Ten { Onelnode** map = (Lnode * *)malloc(5*sizeof(lnode*) +5*Ten*sizeof(Lnode)); ALnode *head = (Lnode *) (map +5);//This is equivalent to the address indicated by map plus 5*sizeof (lnode*) -      for(inti =0; I <5; ++i) -     { theMap[i] = head + i *Ten;  -         //Here the principle is similar, must understand int *a, then A + 3 refers to the address is "a point to the address" + 3*sizeof (int).  -          -     } +      for(inti =0; I <5; ++i) -          for(intj =0; J <Ten; ++j) +(Map[i] + j)->f =J; A      for(inti =0; I <5; ++i) at     { -          for(intj =0; J <Ten; ++j) -         { -printf"%d\t", (Map[i] + j)F); -         } -printf"\ n"); in     } -      Free(map); to     return 0; +}

Example Three:

Here is a direct allocation of a large space, and then use the pointer to split. As long as we understand the previous two, it's not difficult.

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c Dynamic allocation of two-dimensional arrays of structural bodies