如果没有创建该结构的对象，sizeof 如何确定内存中结构的确切大小？

Question

I came across an issue by determining the size of a structure. I´ve found out that it is possible to determine the size of a structure with the preceded keyword struct inside of the sizeof-operation, although no object has been created for the respective structure:

Example:

#include <stdio.h>

int main()
{
   struct struct1
   {
        char a[20];
        int v,i;
        double grw;
   };

   printf("Size of struct1 in Byte: %lu",sizeof(struct struct1));

   return 0;
}

Output:

Size of struct1 in Byte: 40

---

How is that possible?

How can sizeof determine the size of a structure with the help of the struct keyword inside of the sizeof-operation, if no object of this structure has been created?

Or has been an struct1 object created, I did not know about?

I´ve thought a structure is only a datatype, but not an object of its own type.

Answer 1

Other than variable length arrays, sizeof does not need an instance of the type to figure out its size, it can work it out just based on the type definition itself.

In the case you give (shown below), it actually knows, just from that type definition, how big the object will be (the size of each individual field plus whatever padding is needed for alignment between each field and after the final field) - the comments give one possibility:

struct struct1 {
    char a[20];  // 20 bytes @ 0.
    int v,i;     // Two 4-byte values @ 20 (a multiple of 4, so already aligned).
                 // 4 bytes padding to align next 8-byte double.
    double grw;  // 8 bytes @ 32 (a multiple of 8, aligned due to padding above).
};

Answer 2

sizeof needs to be able to determine the size of different types otherwise it wouldn't know how much space to allocate when you do need to.

The struct definition indicates how much space is going to be used.

One allocation might be like this:

• char a[20] : 20 bytes.
• int c, i : 16bit each, so 4 bytes.
• double grw : 128bit, so 16 bytes.

Total: 40 bytes