In C, I have an array of structs defined like:
struct D { char *a; char *b; char *c; }; static struct D a[] = { { "1a", "1b", "1c" }, { "2a", "2b", "2c" } }; I would like to determine the number of elements in the array, but sizeof(a) returns an incorrect result: 48, not 2. Am I doing something wrong, or is sizeof simply unreliable here? If it matters I'm compiling with GCC 4.4.
sizeof a / sizeof a[0]; This is a compile-time constant, so you can use it to, for example, create another array:
#define N sizeof a / sizeof a[0] int n_a[N]; 
struct.
extern (although it is not the case in the original question). The compiler would not able to determine the sizeof(a) at compile time if it is declared in another source file.
sizeof gives you the size in bytes, not the number of elements. As Alok says, to get the number of elements, divide the size in bytes of the array by the size in bytes of one element. The correct C idiom is:
sizeof a / sizeof a[0] 
sizeof a / sizeof *asizeof returns the size in memory of the passed element. By dividing the size of an array by a single element size, you get the elements count.
Note that the element size may include some padding bytes as well. For this reason, a padded struct (e.g. when a char member is followed by a pointer) will have a sizeof value greater than it members size sum.
On the other hand, don't let it bother you when counting elements in an array: sizeof(a) / sizeof(a[0]) will still work as smooth as expected.
ssize_t portfoySayisi = sizeof(*portfoyler);
THIS ONE WORKS
