Memory size of structure

Question

I have the following structure.

    typedef struct {
        int8_t     tmsi[4];        /**< TMSI value. */
        int8_t     ptmsi[4];       /**< PTMSI value. */
        int8_t     gprs_attach_status;
        int8_t     rplmn[3];               /**< PLMN info */
        uint32_t T3212_value;
        uint32_t T3312_value;
        uint8_t   cs_reject_cause;
        uint8_t   ps_reject_cause;
        int8_t     qos[28];  /** QoS  parameters for a PDP context. */
        int8_t     pdp_addr_len;
        int8_t     pdp_address[10];
        uint16_t apn_addr_len;   /**< APN address length. */
        int8_t     apn_address[20];   /**< APN address. */
    }nas_ftd_umts_nas_info_s_type_v01 ;

The size of this structure without padding should be 83.But when compiled on 64 bit processor the size is showing as 84. Compiler is allocation one extra byte for pdp_address[10].Not sure why this extra byte is allocated. Can anyone let me know the reason for this?

I checked the offset of each member with the following code:

#define OFFSETOF(TYPE, ELEMENT) ((size_t)&(((TYPE *)0)->ELEMENT)) 

    int main()
    {
        temp a; 
        test_ping_req_msg_v01 t;
        nas_ftd_umts_nas_info_s_type_v01 info;
        nas_ftd_umts_network_info_s_type_v01 lte;
        nas_umts_ftd_info_ind_msg_v01 ftd;
        cout << sizeof(info) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, ptmsi) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, gprs_attach_status) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, rplmn) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, T3212_value) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, T3312_value) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, cs_reject_cause) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, ps_reject_cause) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, qos) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, pdp_addr_len) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, pdp_address) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, apn_addr_len) << endl;
        cout << OFFSETOF(nas_ftd_umts_nas_info_s_type_v01, apn_address) << endl;
    }

output :

yamunarani@yamunarani:~$ ./a.out
84
4
8
9
12
16
20
21
22
50
51
62
64

An n-bit integer (typically) needs to be aligned on a n-bit boundary, where n is 8, 16, 32, etc. Roughly speaking, that means the address of a n-bit integer is a multiple of n bits (or n/CHAR_BIT bytes). The same goes for every element of an array. In your case, every element of pdp_address (and pdp_address itself) will be aligned on a 16-bit boundary, and every second element of pdp_address will be aligned on a 32-bit boundary (since 32 is twice 16). If pdp_address[9] is on a 32-bit boundary, the only way apn_addr_len can be on a 32-bit boundary is if there is padding between them — Peter
– Peter, Commented Oct 19, 2019 at 0:04
If you want to eliminate padding between fields, force the structure to use 1-byte alignment via #pragma pack(1) or equivalent, depending on the compiler you are using. — Remy Lebeau
– Remy Lebeau, Commented Oct 19, 2019 at 4:14

Daniel · Accepted Answer · 2019-10-19 00:00:41Z

6

The alignment requirement of uint16_t is that it starts on an even address. Without padding of pdp_address[10], you have 61 bytes (i.e. odd) until apn_addr_len, so the compiler pads pdp_address with one byte to get apn_addr_len onto an even address.

answered Oct 19, 2019 at 0:00

Daniel

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