How to call pointer to function defined in typedef struct

Question

what is wrong with the following code?
parseCounter1() and parseCounter1() below are two functions.
I put their pointers in const OptionValueStruct so that
they can be called accordingly when each element of option_values[]
are gone through:

typedef struct OptionValueStruct{  
    char counter_name[OPTION_LINE_SIZE];  
    int* counter_func;  
} OptionValueStruct_t;  

const OptionValueStruct option_values[] = {    
    {"Counter1", (*parseCounter1)(char*, char**)},  
    {"Counter2", (*parseCounter2)(char*, char**)},  
   };  

const OptionValueStruct *option = NULL;

for(int i = 0; i< sizeof(option_values)/sizeof(OptionValueStruct_t); i++){
    option = option_values + i ;  
    result = option->counter_func(opt_name, opt_val);  
}

nos · Accepted Answer · 2017-03-10 09:28:41Z

10

You have declared your counter_func member to be a pointer to an int, not a function pointer , while you have something resembling a function pointer declaration in your option values. Here's what you want (assuming your return type is int )

typedef struct OptionValueStruct{
  char counter_name[OPTION_LINE_SIZE];
  int (*counter_func)(char*, char**);
} OptionValueStruct_t;

const OptionValueStruct_t option_values[] = {
  {"Counter1", parseCounter1},
  {"Counter2", parseCounter2},
};

for(int i = 0; i< sizeof(option_values)/sizeof(OptionValueStruct_t); i++){
  result = option_values[i]->counter_func(opt_name, opt_val); 
  // don't know what you relly want to do with result further on..
}

edited Mar 10, 2017 at 9:28

answered Aug 9, 2010 at 20:30

nos

231k60 gold badges436 silver badges516 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Nathan Over a year ago

Is the colon (instead of semicolon) at the end of int (counter_func)(char, char**) a typo? or some special syntax?

nos Over a year ago

No it was a typo

ysap · Accepted Answer · 2010-08-09 20:33:45Z

0

If you are compiling as C code (as your tag suggests), then you should change the type of option_values[] and option to OptionValueStruct_t. In C++, however, this is OK.

Alternatively, you can eliminate the trailing _t from the custom type name.

answered Aug 9, 2010 at 20:33

ysap

8,24110 gold badges68 silver badges136 bronze badges

3 Comments

user399517 Over a year ago

Thanks for the answer. the type of option_values[] is already OptionValueStruct_t and the type of option is OptionValueStruct_t *. as I show in the original post. do you mean I should use struct OptionValueStruct{...}, then use declaration struct OptionValueStruct option_value[]?

ysap Over a year ago

@lilili08 - at the time of writing this comment, the type of the two variables is const OptionValueStruct and not const OptionValueStruct_t. However, OptionValueStruct is the tag of the struct definition and not the name of the custom type (at the typedef statement). In C, a struct tag is NOT a new type. In C++, though, it is. Now, the tag and the custom type name can be the same, so you can remove the _t characters from the typedef statement and have the newly defined type to be OptionValueStruct so you can use it in further variable declarations.

ysap Over a year ago

@lilili08 - If you are using the Visual C++, most probably the compilation is done as C++ code, so the compiler does not warn you about this. You can change the Project Properties/Configuration Properties/C/C++/Advanced/Compile As to Compile as C code so you know for sure you are C compliant.

Collectives™ on Stack Overflow

How to call pointer to function defined in typedef struct

2 Answers 2

2 Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related