Question

For pointers, I’m getting confused with declarations and function parameters on when to use char ** or char * or *array[n], etc. Like if a function takes a (*array[n]) parameter, do I pass it a **type?

I try using the Right-Left rule and know that p would be a pointer to a pointer to a char (char **p), and p is an array of n pointers (*p[n]), but someone said that *p[n] and **p are essentially equivalent. Is that true?

Answer 1

Reading C declarators (that’s the part of the variable with the * and []) is fairly nuanced. There are some websites with tips:

• http://www.antlr.org/wiki/display/CS652/How+To+Read+C+Declarations
• http://www.ericgiguere.com/articles/reading-c-declarations.html

A char** is a pointer to (possible multiple) pointer(s) to (possibly multiple) char(s). For example, it might be a pointer to a string pointer, or a pointer to an array of string pointers.

A char*[] is an array of pointers to char. When you have a function that takes this as a parameter, the C compiler makes it “decay” into a char**. This only happens to the first layer… so, taking a complicated example, char*[4][] becomes char*(*)[4]. Read the links above so you can understand what the heck that means.

Or you can do a (very sensible) thing and make a bunch of typedefs. I don’t do this, but until you’re good at reading declarators, it’s a good idea.

typedef char * stringp;
void func(stringp array[]) { ... }
static stringp FOUR_STRINGS[4] = { ... };