结构尺寸（单位：C）

Question

      1 #include <stdio.h>
      2 
      3 
      4 struct test {
      5     char c;
      6     int i;
      7     double d;
      8     void *p;
      9     int a[0];
     10 };  
     11     
     12 int main (void) {
     13     struct test t;
     14     printf("size of struct is: %d\n", sizeof(t));
     15     return 0;
     16 }

Output:

size of struct is: 20

Why int a[0] is not considered?

I tried:

  1 #include <stdio.h>
  2 
  3 
  4 struct test {
  5     int a[0];
  6 };  
  7     
  8 int main (void) {
  9     struct test t;
 10     printf("size of struct is: %d\n", sizeof(t));
 11     return 0;
 12 }

And output:

size of struct is: 0

a[0] is a member of structure. Then how come it is not considered in the size of structure?

Answer 1

This is actually more complicated that it may look in the first place.

First of all, a member int a[0] is a constraint violation ("syntax error") in standard C. You must encounter an extension that is provided by your compiler.

Arrays of zero size had often be used by pre-C99 compilers to emulate the effect of flexible array members , that have the syntax int a[] without bounds as the last member of a struct .

For the size of the whole struct , such an array doesn't count by itself, but it may impose alignment constraints. In particular, it might add padding to the end of the other part of the structure. If you'd do the same test for

struct toto {
   char name[3];
   double a[];
};

(with [0] if your compiler needs it), you most probable see a size of 4 or 8 for it, since these are the usually alignment requirements for double .

Answer 2

Now try this:

 1 #include <stdio.h>
  2 
  3 
  4 struct test {
  5     int a[0];
  6 };  
  7     
  8 int main (void) {
  9     struct test t;
 10     printf("size of struct is: %d\n", sizeof(t)==sizeof(t.a));
 11     return 0;
 12 }

If you have no money in your bank account you don't have money at all :)

Answer 3

元素为零的array的大小为0可以使用sizeof(a[0])进行检查。因此不考虑结构的大小。