Variadic Functions

#include <stdarg.h>
#include <stdio.h>

void varfun(int n,...){
va_list ptr;
int num;
va_start(ptr,n);
num=v开发者_JAVA百科a_arg(ptr,int);
printf("\n%d",num);

}


int main(int argc, char **argv){

varfun(3,7.5,-11.2,0.66);

return 0;
}

Look at the above code, i expect the output to be the first variable parameter value casted to int i.e 7.5 casted to int, i.e. 7. But the output is 0. What is wrong in this?

---

The va_arg does not convert the argument. It interprets it as the indicated type. And if the types don't match, you invoke Undefined Behaviour.

va_arg(ptr, int); /* take the next 4 bytes from the stack and interpret them as an `int` */
va_arg(ptr, double); /* take the next 8(?) bytes ... and interpret as double */
(int)va_arg(ptr, double); /* ... convert to int */

Also note the cast isn't really needed in your snippet. The compiler will convert automatically

void varfun(int n, ...) {
    va_list ptr;
    int num;
    va_start(ptr, n);
    num = va_arg(ptr, double); /* interpret as double, then convert to int */
    printf("%d\n",num);
}

---

Your second argument, 7.5 is of type double, but your call va_arg(ptr, int) is treating it as an int. This means undefined behavior.

A variadic function has to be able to figure out the number and type(s) of the arguments -- and if the caller lies, you're out of luck. printf-like functions do this via the format string; other functions might treat all their arguments as pointers and use a trailing null pointer as a sentinel.

---

This is not really casting but merely "reading the first few bytes" (*) of a floating point variable and considering it an integer.

You need to read this argument as its actual type, a double, and then casting as an int.

num = (int) va_arg(ptr, double);

(*) Technically, the behavior associated with va_arg(ptr, wrong_type) is undefined. "reading the first few bytes" is a figure of speech, only describing what typically may happen. No serious program should rely on such implementation details.

---

It won't automatically cast the value 7.5 into an int like that. You'll need to use va_arg to retrieve the value as a double and then do the cast afterwards.

va_arg(ptr,int) is actually expecting the value stored at ptr to be a bitwise representation of an int, and it's not.