pointer to char

[Maryan]

Hello,

i have the following code:

typedef struct
{
char *matrikelnr;
}student;


int main (void)
{

student *studall[2];
char tVal[10];

studall[0] = (student*)mallo c(sizeof(studen t));
strcpy(tVal,"12 345");
studall[0]->matrikelnr = tVal;
printf("studall[0]->matrikelnr: %s\n",studall[0]->matrikelnr);

studall[1] = (student*)mallo c(sizeof(studen t));
strcpy(tVal,"12 346");
studall[1]->matrikelnr = tVal;
printf("studall[1]->matrikelnr: %s\n",studall[1]->matrikelnr);

printf("studall[0]->matrikelnr: %s\n",studall[0]->matrikelnr);

return 0;
}

output:

studall[0]->matrikelnr: 12345
studall[1]->matrikelnr: 12346
studall[0]->matrikelnr: 12346

My Question: How can i get for stuall[0] the first value 12345? Could any one please help me.

thanks in advance.
Maryam

[weaknessforcats]

studall[0]->matrikelnr = tVal;

Stop changing matrikelnr.

matrikelnr is a char pointer. tVal is also a char pointer. (Remember, the name of an array is the address of element 0 and tVal[0] is a char). In effect all of your array elements point to tVal so they all report the final value placed in tVal.

What you need to to is copy the tVal array to matrikelnr. That means you a) need to allocate memory for the copy and b) actually copy the array instead of just assigning its address.

[Maryan]

Thank you for answering my Question. I have done it like this:

typedef struct
{
char *matrikelnr;
}student;


int main (void)
{

student *studall[2];

char *tVal;
studall[0] = (student*)mallo c(sizeof(studen t));

tVal=(char*)mal loc(10);
strcpy(tVal,"12 345");
studall[0]->matrikelnr = tVal;
printf("studall[0]->matrikelnr: %s\n",studall[0]->matrikelnr);

studall[1] = (student*)mallo c(sizeof(studen t));

tVal=(char*)mal loc(10);
strcpy(tVal,"12 346");
studall[1]->matrikelnr = tVal;
printf("studall[1]->matrikelnr: %s\n",studall[1]->matrikelnr);

printf("studall[0]->matrikelnr: %s\n",studall[0]->matrikelnr);
return 0;
}

and it works.

bye,
Maryan