有人可以解释一下我在使用malloc作为指针时发生了什么

Question

#include<stdio.h>
int main(){
    int * ptr=(int *)malloc(40);
    printf("%d",sizeof(ptr));
}

When I'm running this, output is coming as 8...what is happening here...why is output 8?

Answer 1

The sizeof operator is a compile time operator, and the size of any pointer is 8 bytes on your machine.

There is no way to retrieve the dynamic size of the malloc -ed memory zone. You have to know it otherwise, or to keep it somewhere.

You might consider using Boehm's conservative garbage collector then you'll call GC_malloc instead of malloc , you won't need to call free (or GC_free ), and you could use GC_size(p) to get the approximate size of the previously GC_malloc -ed memory zone starting at p (but I don't recommend using GC_size ).

If using malloc on Linux, learn how to use valgrind to hunt memory leak bugs, and compile with gcc -Wall -g

Answer 2

malloc() returns a pointer to 40 bytes of memory (probably 10 ints) and assigns it to ptr . The reason sizeof(ptr) is 8 is because you are using a 64 bit machine and pointers are 8 bytes in size.

You should use sizeof() inside the malloc() because it's good form and avoids problems if the type ever changes size (crossing platforms or whatever). If you really want space for 10 ints then use:

int *ptr = malloc(10 * sizeof *ptr);

This allocates 10 lots of the size of the type ptr points to, in this case int . The benefit of doing this is you can change the type without changing the malloc()

Answer 3

变量ptr是一个指向int的指针，在您的系统中，它恰好具有8个字节的大小。

Answer 4

ptr is a pointer.its printing the size fo the pointer on your system. as ptr holds an address, 8 bytes is the size required to hold an address on a 64 bit machine. This is compiler specific.You cannot retrieve the size allocated to a pointer.

if you getting this doubt,then you should also get a doubt of how does free() will free the memory without knowing the size that has been allocated to the pointer.

Answer 5

我已经在我的Windows 32位计算机上尝试过此操作，我得到的是4字节，我认为ptr是指向int的指针，其显示了正确的信息