C Allocating memory for array of structures containing multiple char*

Question

I have problem allocating memory for array of structure containing char*.

I have one stracture "Person"

typedef struct
{
    char *name;
    char *surname;
    char *phonenumber;
} Person;

What I want to do is to read some data from file and fill the array ( Person *array ) of people where I have to dynamically allocate memory.

---

At the moment, I have something like that:

array = malloc(sizeof(Person) * arraysize);
Person *buff;
char   text[100];
char   *result;

for(i=0; i<arraysize; i++)
{
    buff = &array[i];
    fgets(text, 100, f );

    //Read first name
    result = strtok(text,":");
    buff->name= malloc(strlen(result));
    buff->name= result;

    //Read surname
    result  = strtok(0,":");
    buff->surname = malloc(strlen(result));
    buff->surname = result;

    //Read phone number
    result = strtok(0, ":");
    buff->phonenumber = malloc(strlen(result));
    buff->phonenumber = result;
}

When I print out the whole array I don't get any valid data. I'm wondering what am I doing wrong. I appreicate for your answers in advance!

Answer 1

buff->name= result

resets the pointer buff->name to point to the same location in memory as result . If you want to copy in the string contents of result , use strcpy :

strcpy(buff->name, result);

But note that you have to reserve space for the trailing NUL character as well in the call to malloc :

buff->name = malloc(strlen(result) + 1);

If you're on a POSIX system (Linux, Mac OS X, any Unix) you can replace malloc , strlen and strcpy with the much more convenient strdup :

buff->name = strdup(result);

Answer 2

The problem is:

result = strtok(text,":");
buff->name= malloc(strlen(result));
buff->name= result;

You need to use strcpy for copying and also the length of the malloced string should be one more than the string length to accommodate the NUL char.

buff->name= malloc(strlen(result) + 1);

Answer 3

This:

buff->name= malloc(strlen(result));
buff->name= result;

is broken. You can't assign "strings" like that in C, you need to copy characters. The pointer returned by malloc() is overwritten by result , leaking memory. Also, you need to include space for the terminator.

So, this should be:

buff->name= malloc(strlen(result) + 1);
strcpy(buff->name, result);

Answer 4

strtok overwrite the pointer everytime. So you lose the pointer to data as the iteration goes.

You can do something like this, if you wish to directly assign the pointer:

  result = strtok(text,":");
  buff->name= strdup(result);

strdup is POSIX function. So if it's not available you can easily implement it.

Answer 5

you need to copy the text in result in the newly allocated memory block. now you just allocate memory (assign the pointer to the pointer), then you discard that pointer and assign a new pointer result to it. you need to strncpy the result to the new memory allocation. Also strlen does not include the terminating zero byte, so you do not have enough space in the new memory block. There is a function strdup which does what you want.