Question

I am studying functions that accept arguments of arbitrary datatypes using void pointers. Such a function is the following:

void funct(void *a) {
    float *p = a;
    printf("number = %f\n",*p);
}

Here is a successful invocation to funct:

float x = 1.0f;
funct(&x);

x is declared to be a float and then its pointer, namely &x (which is of type float*) is passed to funct; quite straightforward!

There is however yet another way to declare a variable in C and get its pointer. This is:

float *p;
*p = 1.0f;

But then the call funct(&x); returns a Segmentation fault: 11! How is that possible?

Additionally, assume that I want to create a method that accepts a “number” (i.e. float, integer, double, float or anything else (e.g. even u_short)) and adds 1 to it. What would the most versatile implementation possibly be? Should I consider the following prototype:

void add_one(void* x);

?

Answer 1

First off, I must make something clear. float *p and float* p are equivalent statements. If you’re a beginner in C, I recommend that you write the latter instead of the former, because it makes clearer what the type is and what the variable is – it will save you a lot of confusion in the future. With that in mind, please remember that in the following block p, not *p, is the variable.

float* p;
*p = 1.0f;

The first line allocates a float* on the stack. As it happens with all variables allocated on the stack, its initial value is undefined (most probably garbage). This is not exclusive to pointers: even if you allocate an int and you don’t initialize it, it will have an undefined value.

The same happens with p, so you should think of p as containing random data initially. If p contains random data and it’s a pointer, it means that it’s pointing to a random address. Attempting (in the second line) to write the value 1.0f (or anything else) to that random address will almost always cause a segmentation fault, because a random address at any given time has little chance of belonging to your program.