c - char** 和 struct 错误(空)

Question

我尝试将 char** 保存在结构中，我的代码是:

job * proceso = new_job(123,args[0],args,BACKGROUND);
printf("%s",(proceso->commando)[0]);

其中 args 是字符**；

并出现(空)； 结构是:

job * new_job(pid_t pid, const char * command,char ** commando, enum job_state state){
    job * aux;
    aux=(job *) malloc(sizeof(job));
    aux->pgid=pid;
    aux->state=state;
    aux->command=strdup(command);
    aux->commando=malloc(1000);
    aux->next=NULL;
    return aux;}


typedef struct job_{
   pid_t pgid; /* group id = process lider id */
   char * command; /* program name */
   char ** commando;
   enum job_state state;
   struct job_ *next; /* next job in the list */} job;


job * new_job(pid_t pid, const char * command,char ** commando, enum job_state state);

char** com;
#define new_list(name)           new_job(0,name,com,FOREGROUND)  // name must be const char *

我不知道错误；有人可以帮助我吗？

Answer 1

要复制char **变量，您需要知道数组中有多少个指针。就像在 main 中一样，您会得到一个 argc 来告诉您 argv 中有多少个指针。添加该数据后，您可以:

// Added num_commando parameter
job * new_job(pid_t pid, const char * command, char ** commando, size_t num_commando, enum job_state state)
{
    job * aux;
    aux=(job *) malloc(sizeof(job));
    aux->pgid=pid;
    aux->state=state;
    aux->command=strdup(command);
    // Create array of pointers
    aux->commando = malloc(sizeof(char *) * num_commando); // or sizeof(*aux->commando)
    // Duplicate every string
    for (size_t i = 0; i < num_commando; i++) {
        aux->commando[i] = strdup(commando[i]);
    }
    aux->next=NULL;
    return aux;
}

您还应该将 num_commando 参数添加到结构中，以便知道要 free() 有多少个字符串。