Unable to pass array as parameter, using void pointer, s.t. the contents too are accessed in the called function.

dugan · 06-17-2023, 12:11 AM

Apart from the fact that COLOR isn't declared (as pointed out by the compiler errors), what is this even supposed to do?

Code:

[COLOR="YELLOW"]*((int *)(p)+i));

The part with the square brackets isn't C.

ajiten · 06-17-2023, 12:47 PM

Quote:

Originally Posted by dugan

Apart from the fact that COLOR isn't declared (as pointed out by the compiler errors), what is this even supposed to do?

Code:

[COLOR="YELLOW"]*((int *)(p)+i));

The part with the square brackets isn't C.

Sorry, for being too careless.
Have deleted that part, but the original question is still there.

NevemTeve · 06-17-2023, 01:58 PM

Here is the fixed version of the original program (make sure not to modify it before testing):

Code:

#include <stdio.h>

void foo(int type, unsigned n, void *p)
{
    if (type=='C') {
        for (unsigned i=0; i<n; ++i) {
            printf("%d: '%c'\n", i, ((char *)p)[i]);
        }
    } else if (type=='I') {
        for (unsigned i=0; i<n; ++i) {
            printf("%d: %d\n", i, ((int *)p)[i]);
        }
    }
}

int main()
{
    int x[10] = {1,2,3,4,5,6,7,8,9,10};
    char c[5] = "ABCDE";

    foo('I', 10, &x);
    foo('C',  5, &c);
    return 0;
}

GazL · 06-18-2023, 04:42 AM

You'd really be much better off just sticking with individual fooI() and fooC() functions. There's a lot of "Just because you can, doesn't mean you should" with C and abusing void pointers to make C work like a dynamically typed language is going to lead to tears somewhere down the line.

dugan · 06-18-2023, 07:52 PM

Quote:

Originally Posted by GazL

There's a lot of "Just because you can, doesn't mean you should" with C and abusing void pointers to make C work like a dynamically typed language is going to lead to tears somewhere down the line.

There's a famous rant about a library that took this too far:

https://what.thedailywtf.com/topic/15001/enlightened