Question

The code states:

void (* log_msg)(char *msg)
    =printf;

void change_and_log(int *buffer, int offset, int value){
    buffer[offset] = value;
    log_msg("changed");
}

I’m most concerned with the first part:

Firstly, what does the signature void (* log_msg)(char *msg) mean? Is this code simply mapping the function log_msg to printf? In that case, why is the function name (* log_msg) and not simply log_msg?

Answer 1

It’s a function pointer.

The type of a function pointer is R (*)(Args...), where R and Args... are replaced with the return type and arguments, if any. It is read as “a pointer to a function that takes arguments Args... and returns R.”

Your code would read easier as:

// print_function is a type that is a function pointer
typedef void (*print_function)(char *msg); 

// log_msg is a variable of the type print_function: it points to a function
print_function log_msg = printf; // point to printf

And later, it’s just calling that function via a function pointer.