I understand that & is used to reference the address of object so &char* = char**. Is there anyway to reverse this so that I can get char* from char**?
So I have:
char** str; //assigned to and memory allocated somewhere printf ("%s", str); //here I want to print the string. How would I go about doing this?
You can use the dereference operator.
The dereference operator operates on a pointer variable, and returns an l-value equivalent to the value at the pointer address. This is called "dereferencing" the pointer.
char** str; //assigned to and memory allocated somewhere printf ("%s", *str); //here I want to print the string. Dereference str:
print ("%s", *str); /* assuming *str is null-terminated */ If you have a T* (called a pointer to an object of type T) and want to get a T (an object of type T), you can use the operator *. It returns the object pointed by your pointer.
In this case you've got a pointer to an object of type char* (that's it: (char*)*) so you can use *.
Another way could be that of using operator [], the one you use to access the arrays. *s is equal to s[0], while s[n] equals *(s+n).
If your char** s is an array of char*, by using printf( "%s", *str ) you'll print the first one only. In this case it's probably easier to read if you use []:
for( i = 0; i < N; ++ i ) print( "%s\n", str[i] ); Althought it's semantically equivalent to:
for( i = 0; i < N; ++ i ) print( "%s\n", *(str+i) ); 