Thread: Confusing array of char pointer type mismatch

The warning message is self-explanatory. If it is not self-explanatory to you, then you are making the mistake of believing a pointer and an array are interchangeable. They are not.

In the caller, input_args is an array of 20 pointers to char. &input_args is therefore of type "pointer to array of 20 pointers to char". The function, on the other hand, is expecting a "pointer to a pointer to a pointer to char".

The simplest way to shut the compiler up would be for the caller to define input_args as

Code:

    char **input_args[MAX_ARGS];

and to pass input_args rather than &input_args to the function.

That is only part of your problem though - depending on what parseArgs() is doing, you will need to initialise the elements to something useful (otherwise parseArgs() will simply read or write to lots of random areas of memory). You will need to read the documentation of parseArgs() to work out what initialisation is needed. Your compiler will not diagnose problems related to incorrect initialisation of input_args.

Yes, it's easy to get confused when mixing array notation and pointer notation.

Bascially you want the the function to taking a char ***, a pointer to a pointer which points to a list of pointers. Whilst you can also specify how big the lists are, this tends to cause trouble.