Handling an array of pointers

Question

I'm attempting to create a struct that has an array of char pointers as one of its members and am having trouble attempting to set/access elements of the array. Each char pointer is going to point to a malloc'd buffer. This is the struct currently.

struct rt_args {
    long threadId;
    char (*buffers)[];
    FILE* threadFP;
};

And when I attempt to access an element of buffers via

char *buffer = malloc(100);

struct rt_args (*rThreadArgs) =  malloc( sizeof(long) + 
                                         (sizeof(char *) * (numThreads)) +
                                         sizeof(FILE*)
                                 );
rThreadArgs->buffers[0] = buffer;

I get the error "invalid use of array with unspecified bounds". I don't know what the size of the array is going to be beforehand, so I can't hardcode its size. (I've tried de-referencing buffers[0] and and adding a second index? I feel as though its a syntactical error I'm making)

Some programmer dude · Accepted Answer · 2014-03-05 18:01:53Z

2

You can't have arrays without a size, just like the error message says, at least not n the middle of structures. In your case you might think about pointers to pointers to char? Then you can use e.g. malloc for the initial array and realloc when needed.

Something like

char **buffers;

Then do

buffers = malloc(sizeof(buffers[0]) * number_of_pointers_needed);

Then you can use buffers like a "normal" array of pointers:

buffers[0] = malloc(length_of_string + 1);
strcpy(buffers[0], some_string);

answered Mar 5, 2014 at 18:01

Some programmer dude

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4 Comments

alk Over a year ago

Please excuse, the declaration of struct rt_args is perfectly valid.

Some programmer dude Over a year ago

@alk Not with a flexible array (array without a size). Those has to be placed last in the structure, so when not being last the structure is not valid.

alk Over a year ago

char (*buffers)[]; isn't a "flexible array" but a pointer to one.

Some programmer dude Over a year ago

@alk Yes, but it's not what the OP wants, if we read the question and its title.

haccks · Accepted Answer · 2014-03-05 18:04:33Z

0

char (*buffers)[SIZE];

declares buffers as a pointer to char array not the array of pointers. I think you need this

char *buffers[SIZE];

NOTE: Flexible array member can be used only when it is the last member of the structure.

edited Mar 5, 2014 at 18:04

answered Mar 5, 2014 at 17:58

haccks

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4 Comments

Danzo Over a year ago

What if I don't know the size of the array? It will depend on input.

haccks Over a year ago

If you are using flexible array member then it must be a last member and you need to allocate space, depends on your input, for it before using it.

Utkan Gezer Over a year ago

Why exactly would you need it to be the last member? I don't think that that would be required... It's just a pointer.

haccks Over a year ago

@ThoAppelsin; Flexible array members are must be a last member of a structure.

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