我可以假设调用realloc的非常小的尺寸将释放其余的？ [英] Can I assume that calling realloc with a smaller size will free the remainder?

问题描述

让我们考虑一下这个很短的code的片段：

 的#include＆LT;＆stdlib.h中GT;诠释的main（）
{
    字符* A =的malloc（20000）;
    字符* B = realloc的（一，5）;    免费（B）;
    返回0;
}
 

阅读手册页的realloc后，我不能完全肯定，第二行会造成19995额外的字节被释放。引述手册页：的realloc（）的函数改变内存块的大小由ptr大小字节，但是从这个定义，我可以肯定的。其余部分将被释放？

我的意思是，由指出b 块肯定包含5个免费字节，所以它会为一个懒惰相符分配器只是没有做的realloc的线什么就够了吗？

请注意：我使用的分配似乎释放19 995额外的字节，作为注释掉时，所示的valgrind的免费（B）行：

  == == 4457 HEAP摘要：
== == 4457使用在出口处：在1块5个字节
== == 4457总堆的使用情况：2 allocs，1的FreeS，2万5字节分配
 

解决方案

是由C标准保证，如果新的对象可以进行分配。

  

（C99，7.20.3.4p2）的realloc函数将释放旧的对象ptr指向并返回一个指针，指向具有由size指定一个新的对象。

Let’s consider this very short snippet of code:

#include <stdlib.h>

int main()
{
    char* a = malloc(20000);
    char* b = realloc(a, 5);

    free(b);
    return 0;
}

After reading the man page for realloc, I was not entirely sure that the second line would cause the 19995 extra bytes to be freed. To quote the man page: The realloc() function changes the size of the memory block pointed to by ptr to size bytes., but from that definition, can I be sure the rest will be freed?

I mean, the block pointed by b certainly contains 5 free bytes, so would it be enough for a lazy complying allocator to just not do anything for the realloc line?

Note: The allocator I use seems to free the 19 995 extra bytes, as shown by valgrind when commenting out the free(b) line :

==4457== HEAP SUMMARY:
==4457==     in use at exit: 5 bytes in 1 blocks
==4457==   total heap usage: 2 allocs, 1 frees, 20,005 bytes allocated

解决方案

Yes, guaranteed by the C Standard if the new object can be allocated.

(C99, 7.20.3.4p2) "The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size."

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