Hi all, this could be a really stupid question here... but here it goes.
I am trying to pass the address of the variable dNum to a function that takes a parameter of type void *. But nothing prints.. Any ideas?
Code:#include <stdio.h> void output(void *); int main(void) { double dNum; dNum = 23.2; output(&dNum); return 0; } void output(void *pData) { printf("%f", pData); }
When you pass a variable as a void * don't forget to convert it back before using it. And to retrieve the data that a pointer points to you need to dereference it.
-PreludeCode:#include <stdio.h> void output(void *pData) { printf("%.1f", *(double *)pData); } int main(void) { double dNum = 23.2; output(&dNum); return 0; }
My best code is written with the delete key.
Prelude - Your a legend!
It was so straight forward, but it solved a really big problem I was having with quite a large program. I was overcomplicating the issue essentially.
Thanks!
>I was overcomplicating the issue essentially
You'd be surprised how many people do this.
-Prelude
My best code is written with the delete key.
You weren't over complicating the issue, you simply forgot to type cast your void pointer when you dereferenced it. Actually, you forgot to dereference it when you went to use it with printf. That was your only mistake.
Quzah.
Hope is the first step on the road to disappointment.
No, I mean I was overcomplicating the issue in the REAL program. What I have here is just a cut down version of what I thought the problem was.. which it isn't. The problem here was syntax related but was giving similar results.
It just (through some rather bizzare lateral thinking) highlighted the issue I was having with the rest of the program.
But yeah - there were some pretty fundamental errors in this code snippet!
