Question

Suppose there is non-POD struct below, is the alignment taken effect? If not, what would be expected?

struct S1
{
    string s;
    int32_t i;
    double d;
} __attribute__ ((aligned (64)));

EDIT: the output of the sample code below is 64 even s is set to a long string.

int main(int argc,char *argv[])
{

        S1 s1;

        s1.s = "123451111111111111111111111111111111111111111111111111111111111111111111111111";
        s1.i = 100;
        s1.d = 20.123;
        printf("%ld\n", sizeof(s1));
        return 1;
}

Answer 1

Yes, the alignment takes effect in GCC.

Example code:

#include <string>
#include <iostream>
#include <cstddef>
using namespace std;


struct S1
{
    string s;
    int i;
    double d;
} __attribute__ ((aligned (64)));

struct S2
{
    string s;
    int i;
    double d;
} __attribute__ ((aligned)); // let the compiler decide

struct S3
{
    string s;
    int i;
    double d;
};

int main() {
    cout << "S1 " << sizeof(S1) << endl;
    cout << "S2 " << sizeof(S2) << endl;
    cout << "S3 " << sizeof(S3) << endl;
}

Output:

S1 64
S2 16
S3 16

---

Also: You have observed that S1 is non-POD. Note that std::string is allowed to store its data externally (it usually does so because the data can be of arbitrary length, so a memory buffer must be allocated dynamically).

Remember that sizeof is only calculated during compilation, it can’t depend on runtime values. Specifically, you can’t ever query a size of dynamically allocated memory this way.

Also remember that every element in an array is always of the same type, so of the same size too.