将参数传递给函数指针

Question

I just can't figure out how to pass an Argument like in the following scenario:

#include<stdio.h>

void quit(const char*);

int main(void){
    const char *exit = "GoodBye";
    void (*fptr)(const char*) = quit;
    (*fptr)(exit);

    return 0;
}

void quit(const char *s){
    printf("\n\t%s\n",s);
}

This is how my program should work and it does, but when I make a text menu i just can't figure out how to do it:

#include<stdio.h>
#include<stdlib.h>

int update(void);
int upgrade(void);
int quit(void);
void show(const char *question, const char **options, int (**actions)(void), int length);

int main(void){
    const char *question = "Choose Menu\n";
    const char *options[3] = {"Update", "Upgrade", "Quit"};
    int (*actions[3])(void) = {update,upgrade,quit};

    show(question,options,actions,3);
    return 0;
}

int update(void){
    printf("\n\tUpdating...\n");
    return 1;
}

int upgrade(void){
    printf("\n\tUpgrade...\n");
    return 1;
}

int quit(void){
    printf("\n\tQuit...\n");
    return 0;
}

void show(const char *question, const char **options, int (**actions)(void), int length){
    int choose = 0, repeat = 1;
    int (*act)(void);

    do{
        printf("\n\t %s \n",question);
        for(int i=0;i<length;i++){
            printf("%d. %s\n",(i+1),options[i]);
        }

        printf("\nPlease choose an Option:  ");
        if((scanf("%d",&choose)) != 1){
            printf("Error\n");
        }
        act = actions[choose-1];
        repeat = act();

        if(act==0){
            repeat = 0;
        }
    }while(repeat == 1);
}

Here I need to change the quit function ( int quit(void); to int quit(char *s){}; ) like in the First example and call it with an argument like const char *exit = "GoodBye"; ==>> (*fptr)(exit);

I know that at this point my program takes only void as argument, but I done it only to illustrate the problem.

I'm very confused about this.

EDIT:

this int (*actions[3])(void) I think is an Array of Function pointers and all 3 function pointers takes void as argument, but I need to know if i can use one pointer to take an argument or i have to re-code the whole program.

Answer 1

Since you have an array of function pointers, all the functions need to be of the same type. So at the very least each function should take a const char * (not all functions need to use it) and the array type should be changed to match.

If you want something more flexible, you can have the functions accept a single void * so each function can be passed a different parameter which it then casts to the appropriate type. This is how pthreads passes parameters to functions which start a new thread. You will lose some compile-time type checking with this, so be careful if you go this route.

EDIT:

An example of the latter:

#include<stdio.h>
#include<stdlib.h>

int update(void *);
int upgrade(void *);
int quit(void *);

int main(void){
    const char *question = "Choose Menu\n";
    const char *options[3] = {"Update", "Upgrade", "Quit"};
    int (*actions[3])(void *) = {update,upgrade,quit};

    show(question,options,actions,3);
    return 0;
}

int update(void *unused){
    printf("\n\tUpdating...\n");
    return 1;
}

int upgrade(void *unused){
    printf("\n\tUpgrade...\n");
    return 1;
}

int quit(void *message){
    printf("\n\tQuit...%s\n", (char *)message);
    return 0;
}

void show(const char *question, const char **options, int (**actions)(void *), int length){
    ...
    if (act == quit) {
        repeat = act("GoodBye");
    } else {
        repeat = act(NULL);
    }
    ...
}

Answer 2

Since you are using a an array of function pointers, you don't know which ones to take which arguments. But have You can avoid re-coding it by making the functions to take "unspecified number of arguments". ie Remove the void from as the parameter from function definitions and prototypes from of the function pointers and from the quit() function.

int quit(const char*);
void show(const char *question, const char **options, int (**actions)(), int length);

int main(void){
    const char *question = "Choose Menu\n";
    const char *options[3] = {"Update", "Upgrade", "Quit"};
    int (*actions[3])() = {update,upgrade,quit};
    ...
}

int quit(const char *msg){
    printf("\n\tQuit...%s\n", msg);
    return 0;
}

void show(const char *question, const char **options, int (**actions)(), int length){
    ....
    int (*act)();
   ....
}

This works because C allows a function with no explicit parameters to take "unspecified number of arguments". Otherwise, you need to make all functions have similar signatures.