C - struct变量是指针吗？

Question

I have searched a lot but i could't find any topic about this. My question is: Is struct variable a pointer? And if every input are stored in struct members' addresses, so what are struct variable used for?

This is the code i used for checking struct variable, when i printed what is stored in the variable, it gave me another address but i couldn't figure what that address is used for.

Here is my code and its output:

 struct test
 {
 int a;
 int b;
 char c;
 }; 

 int main()
 {

 struct test f;

 printf("%u", &f);
 printf("\n%d", f);
 printf("\n%u", &f.a);
 printf("\n%u", &f.b);
 }

Output:

6487616

6487600

6487616

6487620

Answer 1

Short answer: No .

Long version: All your print statements in the aforesaid program invoke undefined behaviour, as the supplied argument does not match the conversion specifier.

To elaborate, to print a pointer, you need to use %p conversion specifier, with a cast to void * for the argument.

There is no generic ( one-for-all sort ) format specifier for a structure variable. You need to choose individual members (or member of member, thereof) which has a corresponding format specifier, and pass that as the argument.

That said, regarding the first member and the access,

[...] A pointer to a structure object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides), and vice versa. [...]

Answer 2

Is struct variable a pointer?

No. Given the struct definition:

struct test
{
  int a;
  int b;
  char c;
}; 

and the declaration:

struct test f;

you get the following in memory (assuming 4-byte int s):

   +---+ ---+
f: |   |    |
   +---+    |
   |   |    |
   +---+    +--- f.a
   |   |    |
   +---+    |
   |   |    |
   +---+ ---+
   |   |    |
   +---+    |
   |   |    |
   +---+    +--- f.b
   |   |    |
   +---+    |
   |   |    |
   +---+ ---+
   |   | f.c
   +---+

Members are laid out in the order declared - fa occupies the first four bytes of the object, fb occupies the next four bytes, and fc occupies the last byte. Additional "padding" bytes may be present to satisfy alignment requirements (in this case, the compiler will likely add 3 padding bytes after fc so that the next object starts on an address that's a multiple of 4).

As you can see, the address of the first member of the struct ( fa ) will be the same as the address of the entire struct object. There's no sort of padding or metadata before the first member.

To look at the various addresses and sizes, use the following:

printf( "address of f: %p\n", (void *) &f );
printf( "size of f: %zu\n", sizeof f );

printf( "address of f.a: %p\n", (void *) &f.a );
printf( "size of f.a: %zu\n", sizeof f.a );

printf( "address of f.b: %p\n", (void *) &f.b );
printf( "size of f.b: %zu\n", sizeof f.b );

printf( "address of f.c: %p\n", (void *) &f.c );
printf( "size of f.c: %zu\n", sizeof f.c );

Like I said above, the compiler will most likely add padding bytes after fc , so sizeof f will likely be larger than sizeof fa + sizeof fb + sizeof fc .

Answer 3

The & operator as you are using it is the address-of operator . It will return the address of any variable, heap allocated or not.