The following code yields this compiler warning and it will print, what I think, is a memory address location. I can pass a struct with the same syntax and it works fine. I'm not very good with C, but I enjoy it. Any suggestions as to how I might resolve this?
w
arning: format '%d' expects type 'int', but argument 2 has type 'int *'|
Code:#include <stdio.h> void lax(int *fs); int main() { int fs = 0; lax(&fs); return 0; } void lax(int *fs) { printf("fs: %d\n", fs); return; }
Use %p to print a pointer.
Code:while(!asleep) { sheep++; }
This is the output wuth %d: fs: 2686748
This is the output wuth %p: fs: 0028FF1C
Last edited by juuuugroid; 03-27-2012 at 03:42 PM.
0028FF1C is hex for 2686748, so you're getting the address either way.
You passed a pointer to the function. The name of the pointer is fs.
If you want to print the value that fs points to, then print *fs, instead of fs.
Yes, that's right. The first is the address represented as a decimal integer, the second is the same thing in hexadecimal.
If you want the decimal version but without the compiler warnings you can cast the pointer to an int
Code:printf("fs: %d\n", (int) fs);
Code:while(!asleep) { sheep++; }
Ahh perfect! Thanks much guys. I appreciate. This is an awesome board.
