I want to allocate an array for 6 words of a maximum of 30 charachters for example.
My teacher says I should allocate 6 "+1" to count for a NULL that I have to set my self
after inserting my words.
How exactly is a char **array is allocated and why is my teacher insisting ?
It sounds like your teacher wants you do use a null pointer at the end of the array, as a sentinel value. This is similar to null termination of individual strings and can sometimes be useful when the amount of strings is unknown.
When you allocate a single string you do char* str = malloc(length + 1); where the +1 is for the null terminator '\0. The extra +1 ensures that given length = strlen("hello");, we can safely do strcpy(str, "hello") to fill this string.
You can do something similar for an array of strings by
char** arr = malloc(sizeof(char*[size+1]));
If you then assign arr[size] = NULL; as sentinel value, it will be possible to iterate over this array in this way:
for(const char** ptr=arr; ptr!=NULL; ptr++)
{
printf("%s\n", *ptr); // do something with each string
}
But if you know the size, then the preferred method is a normal loop:
for(size_t i=0; i<size; i++)
{
printf("%s\n", arr[i]); // do something with each string
}
malloc should have the +1 inside the square brackets, otherwise you're allocating only 1 byte for a pointer.
char **arr = calloc(7, sizeof(char*));
So let's break this down, calloc basically makes an array, first argument is the size of the array and the second is how much memory to allocate to every index of the array.
char**.
char** is supposed to point to an array of char*.char **arr = calloc(7, sizeof(char*));