How to get the sizeof data on the pointer, without using the link?

    4 answers 4

    If you write a type: int* ptr_int = ... int& link = *ptr_int; sizeof (link); int* ptr_int = ... int& link = *ptr_int; sizeof (link); it will give the size of the data, and how to do the same, but without resorting to the link?

    In your question, it seems the parser has eaten asterisks. If I put them right, the answer to the question is: sizeof(*ptr_int) .

    sizeof allows you to find out only the size known at compile time. If you have class B inherited from class A , and there will be such code:

     B b; A* ptrB=&b; cout << sizeof(*ptrB);

    then the size of the class A object will be displayed.

    If you need to know the size of the pointer, known in runtime, you can define these classes as follows:

     class A { virtual unsigned int GetMySize() {return sizeof(A);} int Avalue; }; class B: public A { virtual unsigned int GetMySize() {return sizeof(B);} double Bvalue; };

    You can get the real size of the object like this:

     cout << ptrB->GetMySize();

    It uses virtual functions that increase the size of objects by storing a pointer to a table of virtual functions, so sizes may vary.

      If the data type where the pointer points to is unknown, then there is no official method.

      • And if the data type is known, but is it a pointer? Simple dereferencing does not help - sizeof will display the size of the pointer. If you write a type: int * ptr_int = ... int & link = * ptr_int; sizeof (link); it will give the size of the data, and how to do the same, but without resorting to the link? - zzlavv
      • sizeof can take size from type. - KoVadim
      • In this case, sizeof will produce the size of the type, but the size of the data referenced by the pointer is needed. - zzlavv
      • If you write a type: int ptr_int = ... int & link = ptr_int; sizeof (link); This will give the size of the data. sizeof always gives the size of the type. In this case, sizeof will display the size of the TD, which is the link . That is why it is possible: int* a = NULL; sizeof(*a) int* a = NULL; sizeof(*a) ; although a does not point to any data, sizeof in compile-time looks at how TD is *a and returns this value. - fogbit
      • If in this example, instead of int, we substitute a structure, then sizeof will display the size of the data in the structure. Question: Is it possible to do the same through sizeof without using the link? - zzlavv

      If the memory was allocated dynamically, then you can use the malloc_usable_size function. It can return more than the value of * alloc was transferred during allocation, but this is the actual amount of memory that was allocated by this pointer. I only have a bad idea of ​​what situation such a situation might suddenly need ... How did it happen that you don’t know how much you allocate?)

      • I have 2 programs exchanging data through a file projected into memory. The size of the data naturally changes, the memory is allocated by the program and I cannot control it at all, but in order to project the data into a file you need to know the size of this data. I can only access them through pointers. So I'm trying to use sizeof to get the size of the data. So far it turns out only when using links. - zzlavv
      • 2
        The problem is not clear. There is a pointer, but it is not clear what? And if it is clear, why do you need exactly sizeof? And then as in win95, "Unknown device detected, install driver" ... - alexlz
      • Does the program that writes the data know the size of the data? If yes, then no problem. It can be done in such a way that not the data would lie on the pointer, but the structure of a known structure. And inside this structure, and the size can be put, and a pointer to the data. But if programs exchange pointers, who prevents another 4 bytes of size from being exchanged? - KoVadim
      • The question is not how to refine the program, but is it possible with the help of sizeof to get a pointer to a structure or class to find out the size of the data without resorting to a link? - zzlavv
      • one
        > Who has any suggestions?) I was funny to find out) Some left value. Most likely what is contained in the eax register, since the returned values ​​are passed through it. > The lecturer forced us to write the function of overloading the operator [] with the return of the link and the condition inside, that is, it turned out something like this: Are you sure you had to do this? It was necessary to crash the program or throw an exception. Even in this code it is not provided that i may be <0. - gammaker
       const char* p1="text"; int size1=sizeof(*p1); // == 1 (аналогично sizeof(char) int size2=sizeof(p1); // == 4 (аналогично sizeof(char*) const char p2[]="text"; int size3=sizeof(p2); // == 5

      References are not used anywhere.

      If the array is dimacic, then sizeof cannot be obtained from it in any way - sizeof calculates the size of the occupied memory area during compilation.