user-0 提问: 请问C语言怎么用realloc函数分配一个二维数组这样的内存?

回复:
user: user-1
content: int **a; a = (int **)realloc(NULL,10 * sizeof(int *)); for (int i = 0; i < 10; i ++)     a[i] = (int *)realloc(NULL,200 * sizeof(int));

user: user-1
content: realloc Reallocate memory blocks.  void *realloc( void *memblock, size_t size );  Routine Required Header Compatibility  realloc <stdlib.h> and <malloc.h> ANSI, Win 95, Win NT    For additional compatibility information, see Compatibility in the Introduction.  Libraries  LIBC.LIB Single thread static library, retail version  LIBCMT.LIB Multithread static library, retail version  MSVCRT.LIB Import library for MSVCRT.DLL, retail version    Return Value  realloc returns a void pointer to the reallocated (and possibly moved) memory block. The return value is NULL if the size is zero and the buffer argument is not NULL, or if there is not enough available memory to expand the block to the given size. In the first case, the original block is freed. In the second, the original block is unchanged. The return value points to a storage space that is guaranteed to be suitably aligned for storage of any type of object. To get a pointer to a type other than void, use a type cast on the return value.  Parameters  memblock  Pointer to previously allocated memory block  size  New size in bytes  Remarks  The realloc function changes the size of an allocated memory block. The memblock argument points to the beginning of the memory block. If memblock is NULL, realloc behaves the same way as malloc and allocates a new block of size bytes. If memblock is not NULL, it should be a pointer returned by a previous call to calloc, malloc, or realloc.  The size argument gives the new size of the block, in bytes. The contents of the block are unchanged up to the shorter of the new and old sizes, although the new block can be in a different location. Because the new block can be in a new memory location, the pointer returned by realloc is not guaranteed to be the pointer passed through the memblock argument.  realloc calls malloc in order to use the C++ _set_new_mode function to set the new handler mode. The new handler mode indicates whether, on failure, malloc is to call the new handler routine as set by _set_new_handler. By default, malloc does not call the new handler routine on failure to allocate memory. You can override this default behavior so that, when realloc fails to allocate memory, malloc calls the new handler routine in the same way that the new operator does when it fails for the same reason. To override the default, call   _set_new_mode(1)  early in your program, or link with NEWMODE.OBJ.  When the application is linked with a debug version of the C run-time libraries, realloc resolves to _realloc_dbg. For more information about how the heap is managed during the debugging process, see Using C Run-Time Library Debugging Support.  Example  /* REALLOC.C: This program allocates a block of memory for  * buffer and then uses _msize to display the size of that  * block. Next, it uses realloc to expand the amount of  * memory used by buffer and then calls _msize again to  * display the new amount of memory allocated to buffer.  */  #include <stdio.h> #include <malloc.h> #include <stdlib.h>  void main( void ) {    long *buffer;    size_t size;     if( (buffer = (long *)malloc( 1000 * sizeof( long ) )) == NULL )       exit( 1 );     size = _msize( buffer );    printf( "Size of block after malloc of 1000 longs: %u\n", size );     /* Reallocate and show new size: */    if( (buffer = realloc( buffer, size + (1000 * sizeof( long )) ))          ==  NULL )       exit( 1 );    size = _msize( buffer );    printf( "Size of block after realloc of 1000 more longs: %u\n",              size );     free( buffer );    exit( 0 ); }   Output  Size of block after malloc of 1000 longs: 4000 Size of block after realloc of 1000 more longs: 8000   Memory Allocation Routines  See Also   calloc, free, malloc

user: user-2
content: 再提供一个分配连续内存的方法：   int **a; a = (int **)malloc(10 * sizeof(int *)); a[0] = (int*)malloc(10*200*sizeof(int)); for (int i = 1; i < 10; i ++)     a[i] = a[0] + i * 200;

user: user-3
content: int **a; a = (int **)malloc(10 * sizeof(int *)); for (int i = 0; i < 10; i ++) 	a[i] = (int *)malloc(200 * sizeof(int));  realloc在a已经分配好的情况下，用于扩展内存。