我可以在没有 malloc 的情况下释放指针地址吗？

Question

Let's suppose this code:

int i,j=0;
char* block = (char*) (0x9000);
char table[4]= {0x01,0x02,0x03,0x04};
for (i=0; i< 45567; i++) {
    *(block +i)= table[j]; 
    j++; 
    if (j==4)
        j=0;
}

I would like to ask:

1. Is the memory for block allocated in the stack or in the heap?
2. What problems could happen with this code?
3. Can I use free(block) at the end of this code?

Answer 1

You didn't actually allocate any memory, neither on the stack nor on the heap. You're just pointing your variable at an address and then pretend that the memory there belongs to you.

This is not legal in either C or C++ and will generally not work . And “not work” can really mean anything here. And since the code is illegal, the question of whether the pointer can be free d is moot.

(In very specific settings, if the compiler and the hardware supports it, this is used to write to specific hardware addresses. But this isn't the case here.)

Answer 2

does the block memory allocated in the stack or in the heap?

Neither. It is not allocated at all: you assigned an arbitrary number to the pointer, so any attempt to read or write from block is undefined behavior.

Can I use free(block) at the end of this code?

That would be undefined behavior as well. Pointers that you are allowed to pass to free must come from malloc , calloc , or realloc .

Note: Undefined Behavior is the official name for what happens when your program does something that the standard does not allow, and the compiler does not catch. In many instances undefined behavior leads to a crash, but, unfortunately, in some cases your program appears to work.

Answer 3

you dont need to free anything you didnt ask to get space for , everything you declare without realloc malloc or calloc dont need free , and they store on static storage or the stack .

there are also strdup and some other function you need to free but simple declare dont require free

Answer 4

You should never call free on a pointer value that wasn't returned from one of malloc , calloc , or realloc .

Do you know if 0x9000 is a valid, writable address on your platform?

Answer 5

The array table is likely on the stack. The pointer block , it is to be hoped, is in some otherwise unoccupied memory, maybe it was allocated, and 0x9000 is some special address, there isn't enough code to determine context. Depending on the architecture, writing blindly to an arbitrary address may or may not be allowed.

So to answer your questions:

• you should not free without malloc/calloc/realloc, results are undefined; exceptions do exist, for example if a space is allocated by the system prior to your use, but then typically if the system allocates the memory, it will free it itself

• all kinds of problems can happen with this code, but depends on the architecture; I've written on platforms where this was perfectly legal, others will throw all kinds of exceptions; it might be that it fills 45567 bytes starting at 0x9000 with repeated sequences of 1 to 4, or it might crash your machine; C will happily let you shoot yourself in the foot

Answer 6

你所做的很危险。你不应该选择一个随机地址并将它分配给一个指针来将数据复制到内存中。这就像在地球上的一个随机地方倾倒核废料。相反，你应该使用 malloc 或 new 来分配内存安全。