I am not 100% sure whether I correctly allocated an array with 10 elements
char *str;
str = (int*)malloc(10 * sizeof(int));
The task was to allocate memory of string with 10 elements.
Have I done it correctly or do I have to add "+ 1" because of the '\0' at the end.
Thanks!
You have allocated many times more than you need. (sizeof(int) times to be precise). The correct would be
#define MAXLEN 10
...
str = malloc(sizeof *str*(MAXLEN+1));
Note that this will be sizeof(char) which is 1. So you can do this also
str = malloc(MAXLEN+1);
Check the return value of malloc as shown below: (malloc might not be able to service the request, it might return a null pointer. It is important to
check for this to prevent later attempts to dereference the null pointer).
str = malloc(MAXLEN+1);
if(!str){
perror("malloc");
exit(EXIT_FAILURE);
}
Also void* to char* conversion is implicit - you don't need to cast the return value of malloc.
sizeof(*str) is a cleaner way to get the size of the type of the element for which we are allocating memory in str. The benefit is when you are later changing the code and making str point to an allocated memory which will contain int-s you don't need to look for sizeof(char) and then replace it with sizeof(int) it is done here by using sizeof(*str) automatically.
malloc(11). When you read the code later - whats 11? (why not 12 or 42?)
#define MAXLEN 10 for the number of characters, and MAXLEN+1 for the size of the needed buffer. Then if it has to change, you only have to change the value in #define MAXLEN 10.
char *str, why what where?!str = malloc(11);\0terminator only applies to strings. If you need a string of 10 characters, then you need 11 bytes. Nowhere in your question did you mention a "string" or "character string", so it's unclear what you want.