如何获得的大小动态（使用malloc或释放calloc）分配的内存？ [英] How to get the size of dynamically (using malloc or calloc )allocated memory?

问题描述

  

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/1281686/determine-size-of-dynamically-allocated-memory-in-c\">determine在C 结果动态分配的内存大小
  有关的malloc和sizeof 结果新手问题
  <一href=\"http://stackoverflow.com/questions/232691/how-can-i-get-the-size-of-an-array-from-a-pointer-in-c\">How我可以用C的指针获取数组的大小？

  
  

<一个href=\"http://stackoverflow.com/questions/5213356/malloc-how-much-memory-has-been-allocated?rq=1\">Malloc - ＆GT;多少内存已分配？

  INT ** arrofptr;
arrofptr =（INT **）的malloc（sizeof的为（int *）* 2）;
arrofptr [0] =（INT *）malloc的（的sizeof（int）的* 6144）;
arrofptr [1] =（INT *）malloc的（的sizeof（int）的* 4800）;
 

现在我必须知道多少字节arrofptr分配，arrofptr [0]，arrofptr [1]？有没有办法知道的大小？

如果我们将打印

 的sizeof（arrofptr）;
的sizeof（arrofptr [0]）;
的sizeof（arrofptr [1]）;
 

然后将打印4

解决方案

没有。如果不使用额外数据的地方，存储分配的大小。

Possible Duplicate:
determine size of dynamically allocated memory in c
newbie questions about malloc and sizeof
How can I get the size of an array from a pointer in C?

Malloc -> how much memory has been allocated?

int **arrofptr;
arrofptr = (int **)malloc(sizeof(int *) * 2);
arrofptr[0] = (int *)malloc(sizeof(int)*6144);
arrofptr[1] = (int *)malloc(sizeof(int)*4800);

Now i have to know that how many bytes are allocated in arrofptr,arrofptr[0],arrofptr[1]? is there any way to know the size?

if we will print

sizeof(arrofptr);
sizeof(arrofptr[0]);
sizeof(arrofptr[1]);

then it will print 4

解决方案

No. Not without using extra data somewhere that stores the allocated sizes.

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