Copy char array to char pointer

Hi ,

char date[2] = "02";
char *temp_date;

how can i copy date to temp_date.
I tried strcpy and it gives a memory exception violation;

Please advice,

thanks

That depends on what you mean by "copy date to temp_date". If you just want temp_date to point to the date you've got, then do so, with "temp_date = date". If you want a separate copy, then you will first need to acquire enough memory for the separate copy to live in (using malloc and friends).

Not only that, but "02" takes up 3 bytes (don't forget the null terminator), yet you have only allocated 2 for your date array.

Quote:

---

Originally Posted by tabstop

If you want a separate copy, then you will first need to acquire enough memory for the separate copy to live in (using malloc and friends).

---

Not sure how many friends malloc() would need here. Anyway, another option:

Code:

---

char date[3] = "02";
char temp_date[3];

---

Quote:

---

Originally Posted by MK27

Not sure how many friends malloc() would need here. Anyway, another option:

Code:

---

char date[3] = "02";
char temp_date[3];

---

---

strdup() anyone?

Quote:

---

Originally Posted by MK27

Not sure how many friends malloc() would need here.

---

In this case, just one (free).

Quote:

---

Originally Posted by Kennedy

strdup() anyone?

---

Possibly, except that it is non-standard.

Quote:

---

Originally Posted by laserlight

Possibly, except that it is non-standard.

---

Really? According to who's standards?

Quote:

---

Originally Posted by Kennedy

According to who's standards?

---

The 1999 edition of the C Standard, and without checking I daresay the previous edition as well.

Who cares how standard it is when implementing it yourself is like, a two line function?

Code:

---

char *MyStrdup( const char *str )
{
    char *dup = malloc( strlen( str ) + 1 );
    return dup ? strcpy( dup, str ) : dup;
}

---