無法在共享 memory 中共享結構數組

Question

我創建了一個庫來初始化和填充 strutc 共享 2 進程數組。 問題是它不起作用，而是我創建了一個 int 數組並且它可以正常工作。 但是我需要共享 memory 中的這個數組，我想念什么？

EDIT: I have a theory but I need confirmation cause i'm studing now c and sh memory....is possible taht the problem are the transaction declared by pointer in the struct of block, and for that reason in the sh memory doesn '沒有任何結果？這個錯誤迫使進程因某種原因關閉？

編輯：所以我忘了張貼打印 function。 我將大小強制為 0，因為我只想插入 1 個元素並且至少只打印一個。 但是打印 function 在 if 之前的第一次打印之后卡住了，我不知道為什么。 我認為這是因為共享不起作用，但對於打印“Blocco n°0”，它表明Masterbook[i].id內部有一些價值。

編輯：更改了char*但也不起作用。

結構：

struct Transaction {
    char timestamp[24];
    int sender; /* pid user sent */
    int receiver;
    int reward;
    int money;
};

struct Block
{
    int id;
    struct Transaction* tr1;
    struct Transaction* tr2;
    struct Transaction* reward;
};

masterbook.c

    struct Block* MasterBook;
    int sizeMaster = -1;
        void initMasterBook(){
            int shmid,sizeid;
            shmid = shmget(SH_KEY_MASTERBOOK,MAX_MASTER*sizeof(struct Block),IPC_CREAT | 0666);
            if(shmid == -1) {
                perror("shmget error");
                exit(EXIT_FAILURE);
            }
            MasterBook = (struct Block*)shmat(shmid,NULL,0);
            if ( MasterBook == -1 ) {
                perror ( "Error in shmat: " );
                exit(EXIT_FAILURE);
            }
               
        }
        
        void freeMasterBook(){
            shmdt((void *) MasterBook); 
        }
        int addBlock(struct Block block){
            if(sizeMaster >= MAX_MASTER-1) return -1;
            ++sizeMaster;
            block.id = sizeMaster;
            MasterBook[sizeMaster] = block;
            return 0;
        }
int printMasterBook(){
    if(sizeMaster < 0) {
        printf("\nMasterbook vuoto");
        return -1;
    }
    int i = 0;
    for ( i; i <= sizeMaster; i++)
    {
        printf("\nBlocco n° : %d",MasterBook[i].id);
        
        if(MasterBook[i].tr1 != NULL){
            printf("\n\tTransazione n° 1 : ");
        struct Transaction *temp = MasterBook[i].tr1;
        printf("\n\t\tReceiver: %d",temp->receiver);
        printf("\n\t\tSender: %d",temp->sender);
        printf("\n\t\tReward: %d",temp->reward);
        printf("\n\t\tTimestamp: %s",temp->timestamp);
        }else printf("\nThere's no transaction 1 set!");

        if(MasterBook[i].tr2 != NULL){
            printf("\n\tTransazione n° 2 : ");
        struct Transaction *temp2 = MasterBook[i].tr2;
        printf("\n\t\tReceiver: %d",temp2->receiver);
        printf("\n\t\tSender: %d",temp2->sender);
        printf("\n\t\tReward: %d",temp2->reward);
        printf("\n\t\tTimestamp: %s",temp2->timestamp);
        }else printf("\nThere's no transaction 2 set!");

        if(MasterBook[i].reward != NULL){
            printf("\n\tTransazione Reward : ");
        struct Transaction *temp3 = MasterBook[i].reward;
        printf("\n\t\tReceiver: %d",temp3->receiver);
        printf("\n\t\tSender: %d",temp3->sender);
        printf("\n\t\tReward: %d",temp3->reward);
        printf("\n\t\tTimestamp: %s",temp3->timestamp);
        }else printf("\nThere's no transaction reward set!");
    
        printf("finish print");
    }
    return 0;
}

master.c：

initMasterBook();
printf("\nthe size id: %d",sizeMaster);
printMasterBook();
struct Block v;
struct Transaction t1;
t1.money = 4;
t1.receiver = 2;
t1.sender = 3;
strcpy(t1.timestamp,"ciao");
t1.reward = 4;
v.tr1 = &t1;
addBlock(v);
printMasterBook();
 printf("\nthe size id: %d",sizeMaster);

節點.c

int main(int argc, char const *argv[])
{
    initMasterBook();
    sizeMaster=0;
    printMasterBook();
    freeMasterBook();
    return 0;
}

Answer 1

我解決了 output 更改結構中的指針並重新排列代碼的問題。

struct Transaction {
    int empty;
    char timestamp[30];
    int sender; /* pid user sent */
    int receiver;
    int reward;
    int money;
};

struct Block
{
    int id;
    struct Transaction tr1;
    struct Transaction tr2;
    struct Transaction reward;
};

有問題的打印：

int printMasterBook(){
    if(sizeMaster < 0) {
        printf("\nMasterbook vuoto");
        return -1;
    }
    int i = 0;
    for ( i; i <= sizeMaster; i++)
    {
        printf("\nBlocco n° : %d",MasterBook[i].id);
        
        if(MasterBook[i].tr1.empty != -1){
            printf("\n\tTransazione n° 1 : ");
        printf("\n\t\tReceiver: %d",MasterBook[i].tr1.receiver);
        printf("\n\t\tSender: %d",MasterBook[i].tr1.sender);
        printf("\n\t\tReward: %d",MasterBook[i].tr1.reward);
        printf("\n\t\tTimestamp: %s",MasterBook[i].tr1.timestamp);
        }else printf("\nThere's no transaction 1 set!");

        if(MasterBook[i].tr2.empty!= -1){
            printf("\n\tTransazione n° 2 : ");
        printf("\n\t\tReceiver: %d",MasterBook[i].tr2.receiver);
        printf("\n\t\tSender: %d",MasterBook[i].tr2.sender);
        printf("\n\t\tReward: %d",MasterBook[i].tr2.reward);
        printf("\n\t\tTimestamp: %s",MasterBook[i].tr2.timestamp);
        }else printf("\nThere's no transaction 2 set!");

        if(MasterBook[i].reward.reward != -1){
            printf("\n\tTransazione Reward : ");
        printf("\n\t\tReceiver: %d",MasterBook[i].reward.receiver);
        printf("\n\t\tSender: %d",MasterBook[i].reward.sender);
        printf("\n\t\tReward: %d",MasterBook[i].reward.reward);
        printf("\n\t\tTimestamp: %s",MasterBook[i].reward.timestamp);
        }else printf("\nThere's no transaction reward set!");
    
        printf("finish print");
    }
    return 0;
}