function pointer array with different number of arguments

Question

I wanted to replace below switch case statement with following code

switch (n)
  {
      case 0:
            i2c_scan();
            break;
        case 1:
            i2c_read(45);
            break;
        case 2:
            i2c_write(45,5);
            break;
        default:
            printf("\nwrong input\n");
            break;
  }

Please check following code

#include<stdio.h>

void i2c_scan()
{
    printf("\ni2c scan\n");
}
void i2c_read(int x)
{
    x= x+1;
   printf("\ni2c read: %d\n",x); 
}
void i2c_write(int x , int y)
{
    x=x+y;
    printf("\ni2c write:%d\n",x);
}

int main() {
   int n;
   
void (*fun_ptr_arr[])() = {i2c_scan,i2c_read,i2c_write}; 
    (*fun_ptr_arr[0])(45,5);  //have put array index to show results ,i could have took input from user and put it here like switch case: case number) 
    (*fun_ptr_arr[1])(45,5);
    (*fun_ptr_arr[2])(45,5);
    
}

Output:

i2c scan

i2c read: 46

i2c write:50

How above code is compiling and running without any error when i am passing more arguments then required to the function ? And how is it working correctly like i2c_read takes first argument and gives result ?

Passing more or less arguments to a C function is considered undefined behaviour. If you really want to pass a variable number of arguments to a function you should have a look at variadic functions — DanBrezeanu
– DanBrezeanu, Commented Aug 16, 2020 at 18:38
Several people mentioned it in this thread. You have links there to the official C documentation. — DanBrezeanu
– DanBrezeanu, Commented Aug 16, 2020 at 19:34

chux · Accepted Answer · 2020-08-16 20:58:21Z

0

How above code is compiling and running without any error when i am passing more arguments then required to the function ?

As @DanBrezeanu wells points out, the behavior is UB.
Anything may happen including apparently "working correctly".

answered Aug 16, 2020 at 20:58

community wiki

chux

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Comments

root · Accepted Answer · 2020-08-17 03:45:06Z

0

The declaration

void (*fun_ptr_arr[])()

Means that fun_ptr_arr is an array of pointers to functions of the form:

void f() { ... }
void g() { ... }

and that is different from:

void f(void) { ... }
void g(void) { ... }

This answer elaborates more, but the gist is: the second form is a function that accepts no arguments, whereas the first form accepts any number of arguments.

answered Aug 17, 2020 at 3:45

root

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