Question

here are part of my code.

float a = 12.5;
printf("%d\n", a);
printf("%d\n", (int)a);
printf("%d\n", *(int *)&a);

when I compile in windows, I got:

0
12
1094713344


and then, I compile in linux, I got:

-1437851864
12
1094713344


-1437851864 will be changed every time I excuted it.
my question is: in how does the “printf” function works in linux

Answer 1

It works very well, but why are you passing the wrong sort of data to it? The %d specifier expects and int, but you’re passing something else. Bad idea.

If float and int are differently sized across the varargs barrier, this is undefined behavior. And since float is typically promoted to double with varargs calls, if your int is smaller than your double this will break.

In short, this is really bad and broken code. Don’t do this.