CODEWITHSUNDEEP
carraysstructfunction-pointers
umphish
umphish

Reputation: 75

initialize an array of structs with function pointers in them C

I am trying to initialize an array of structs. the structs have function pointers in them and are defined as such:

typedef struct{
      char str[512];
      char *fptr;
} infosection;

then I try to make an array:

infosection section[] = {
      ("Software", *SoftwarePtr),
      ("Hardware", *HardwarePtr),
 }

I defined the function pointers using a simple tutorial. Function Software returns an int

int (*SoftwarePtr)()=NULL;
SoftwarePtr = &Software;

My question is about the warnings I get upon compiling.

Initialization makes integer from pointer without a cast

The warning references the lines in the section array.

So I have two doubts:

  1. Am I misusing the */&/neither for the pointer?I have tried combinations of each and i still seem to get the same warning.I am aware of the meaning of each, just unsure how they apply in this particular instance.

  2. Can I declare an instance of the infosection struct in an array as i do here?I have seen many examples where people declare their array of struct in a for loop, However my final product requires a long list of structs to be contained in that array.I would like to make my code portable such that people can add structs ( strings which correspond with functions to point to) in an easy list-like fashion as seen about. Is this the only way to declare an instance of the array

      infosection section[];
  section[0].str="software";
  section[0].fptr=SoftwarePtr;

Just to clarify, i have done quite a bit of research on structs, array of structs, and function pointers. It just seems that the combination of the 3 is causing trouble.

Upvotes: 5

Views: 5681

Answers (3)

Grijesh Chauhan
Grijesh Chauhan

Reputation: 58271

One

One mistake I an find, in structure declaration of function pointer is incorrect: 

typedef struct{
      char str[512];
      int (*fptr)();  // not char* fptr
} infosection;

Two

declaration: 

infosection section[] = {
      ("Software", *SoftwarePtr),
      ("Hardware", *HardwarePtr),
 }

Should be:

infosection section[] = {
      {"Software", SoftwarePtr},
      {"Hardware", HardwarePtr}
 }

Remove *. and replace inner ( ) with { }.

three: 

infosection section[];
section[0].str="software"; // compiler error
section[0].fptr=SoftwarePtr;

Is wrong you can't assign string in this way. you need to use strcpy() as follows;

strcpy(section[0].str, "software");

Upvotes: 7

Mr.Monshaw
Mr.Monshaw

Reputation: 450

been a while since i was playin around with pointers but maybe a constructor would help?

struct infosection{
      char str[512];
      char *fptr;
      infosection(char* strN, char* fptrN): str(strN), fptr(fptrN) {}
};

infosection[] isxn = { new infosection(&str1,fptr1), 
               new infosection(&str2,fptr2)},
               ... };

my apologies...where c stops and c++ begins has become blurred in my mind, i have seen a workaround for c that looks like this (but not exactly, im sure some of the pointer stuff is mixed up but this was the idea)

struct infosection{
  char str[512];
  char *fptr;
};

infosection* myIS (char *strN, char *fptrN) {
  infosection i = new infosection();
  i.str = &strN;
  i.fptr = fptrN;
  return *i;
}

//infosection* newInfoSection = myIS(myStrPtr,myFptr);

Upvotes: 1

krs
krs

Reputation: 4154

The struct member is currently a char*, change to the correct pointer type for your function signature.

typedef struct{
  char str[512];
  int (*fptr)();
} infosection;

* dereference the pointer, read *foo as "what foo points at". You want to set the struct member to the actual pointer, also use bracets instead of paranthesis.

infosection section[] = {
  {"Software", SoftwarePtr},
  {"Hardware", HardwarePtr}
}

Upvotes: 2

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