如何将char *分配给函数内的struct成员？

Question

My problem is that when I try and print (Room.dscrptn) outside of the function "createRoom", nothing shows up. Is this because because I declared the character array inside the function? what should i do?

struct roomInfo{
    int rmNm;
    char* dscrptn;
    int   nrth;
    int   sth;
    int   est;
    int   wst;
};

void createRoom(struct roomInfo* Room, char* line){
    int i = 0, tnum = 0;
    char tstr[LINE_LENGTH];
    tnum = getDigit(line, &tnum);       
    Room->rmNm = tnum;               //Room.rmNm prints correct outside the function
    getDescription(line, &i, tstr);
    Room->dscrptn = tstr;           //Room.dscrptn wont print outside createRoom

}

void  getDescription(char* line, int* i,char* tstr){
    //puts chars between [$,$] into tstr
    //tstr[0] == '0' if error
    int cash = 0, j = *i, t = 0;
    while (cash < 2 && line[j] != '\0'){
        if (line[j] == '$'){
            ++cash;
        }
        if (cash > 0){
            tstr[t] = line[j];
            ++t;
        }
        ++j;
    }
    tstr[t] = '\0';
    if (tstr[0] == '$' && tstr[t-1] == '$'){
        *i = j;
    }
    else{
        tstr[0] = '0';
    }

}

Answer 1

As suspected, the problem was the declaration of the array inside of createRoom(). To solve the problem, I declared an array in the main function

int main(){

    struct roomInfo R[TOTAL_ROOMS];
    char tstr[LINE_LENGTH];
    createRooms(R, tstr); //CreateRooms calls createRoom, therefore,
                            it was neccessary to declar tstr in the 
                            global namespace

   }

Answer 2

You need to understand one thing that all local variables goes to stack. You are out of function and the variable is gone. Two solution of this

1. Create a global variable char tstr[LINE_LENGTH];. The problem with this solution that you cannot contain more than one name. for a good solution go for second option

2. do a malloc.

    char *tstr = (char *) malloc(LINE_LENGTH); 

This will solve your problem