C ++中的内存分配和回收

Question

I am new to c++ and just read some online tutorial about it. I am pretty curious about my experiment result in pointer. Here is my code:

int *p_value;
sizeof(*p_value) <--- I expected 0 but it shows 4
p_value = new int;
sizeof(*p_value) <--- it shows 8, make sense
delete(p_value)
sizeof(*p_value) <--- I expect 0, but it shows 4 again...

More importantly, I don't understand why does p_value point to the same memory address after delete operation. Is it necessary to be true?

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Thanks for pointing out my misunderstanding over the operator "sizeof". But here is something I didn't state it too clear in my question.

here is another code: int *p_value = new int; cout << p_value <-- it prints the address of the new created int, say 0x000A; delete p_value; cout << p_value <-- it shows the same address before the "delete" called

Can i consider the "delete" just mark the 0x000A is not occupied in Map, I guess. But it won't change the content of the p_value?

Answer 1

Your results cannot be true. All your cases output size of type int in bytes (and sizeof(int) cannot be changed due program run).

Answer 2

I think the key thing your missing is that sizeof is not a function. It doesn't look at the value of its argument at run time like a function does, just its type at compile time.

int *p_value;   
sizeof(*p_value) <--- I expected 0 but it shows 4

Since p_value is a pointer to an integer, *p_value is an integer. Integer are four bytes on your platform.

p_value = new int;   
sizeof(*p_value) <--- it shows 8, make sense

That's odd. Since p_value is a pointer to an integer, *p_value is an integer. So apparently, integers are 8 bytes on your platform.

delete(p_value)  
sizeof(*p_value) <--- I expect 0, but it shows 4 again...

Why do you expect 0? Since p_value is a pointer to an integer, p_value is an integer. So its size is 4 (or however many bytes an integer occupies on your platform). How else would p_value = malloc (sizeof (*p_value)); work?

more importantly, I don't understand why p_value point to the same memory address after delete operation. Is it necessary to be true?

Passing the value of a variable to a function doesn't change the value of that variable. So delete (p_value) doesn't change the value of p_value , except it now points to garbage.

Answer 3

All of the results returned by the above sizeof(* ptr) give you information about the size of the type it points to, specific to your platform , specifically the size of int , ie 4 means size: 4 bytes or 32 bits, thus the int can hold up to 2 ^{32 - 1-sign-bit} different numbers.

You could think of the returned results as results from: sizeof(int) , which won't change regardless of the location* of your int of whether you free dynamically allocated memory, with delete or not, fact that, as @ForEveR, already said, makes your results dubious**.

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*Static, stack or heap memory.

**Checked on two different platforms and the result is 4 , not 8 .

Answer 4

int *p_value;
sizeof(*p_value) <--- I expected 0 but it shows 4

*p_value is an integer, the size is typically 4 bytes, thus it shows 4. This does depends on the compiler too (C++ standard only mentioned that it should be at least 4 bytes). My test returns 8 though.

p_value = new int;
sizeof(*p_value) <--- it shows 8, make sense

If the above shows 4, then this one should shows 4 as well (unless you're using different compiler/target arch). The sizeof operator shows the number of bytes taken by the data type. The value itself is irrevalent.

delete(p_value)
sizeof(*p_value) <--- I expect 0, but it shows 4 again...

The same as above.