How is this memory released?

meisetaisi

I wrote a function a few days ago, and to adapt to the situation with one parameter and two parameters, I thought of writing overload.
Later, I thought, do I have to write an overload? I tried the default parameter method. This is the following piece of code.
The question now is, how to judge whether to release the memory and how to judge whether that memory has been generated?

#include <iostream.h>

int f (int i, int&j = * new int (1))
{

j = i;
return i;
}

void main ()
{
int a, b, c, d = 0;

a = f (d);
b = f (d, c);
cout << "a =" << a << endl;
cout << "b =" << b << endl;
cout << "c =" << c << endl;
cout << "d =" << d << endl;
}

superandy

int f (int i, int j = 1)
{

j = i;
return i;
}

jotoos

No mistake,
1) The function always has a convention, you have to agree that the incoming parameters must be dynamically allocated.
2) Try to use overload instead of default parameters.

qingclie

no method

you should record the pointer

julines

I don't understand why this place is new.
You lost the pointer, how to delete?

fatman

What you create should be regarded as j itself, and can only be processed inside the function; assuming that the function is processed inside, what about j?

baiyunyy

int f (int i, int j = 1) It seems that this is also ok.

ap2000ap

int f (int i, int&j = * new int (1))
What is the definition, not to mention how strange this definition is, j is a reference, is j assigned to the address of new? If the address is changed, you can make it clear to everyone, otherwise you have a problem with the definition.

reekee_max

// crt_va.c
/ * The program below illustrates passing a variable
 * number of arguments using the following macros:
 * va_start va_arg va_end
 * va_list va_dcl (UNIX only)
 * /

#include <stdio.h>
#define ANSI / * Comment out for UNIX version * /
#ifdef ANSI / * ANSI compatible version * /
#include <stdarg.h>
int average (int first, ...);
#else / * UNIX compatible version * /
#include <varargs.h>
int average (va_list);
#endif

int main (void)
{
   / * Call with 3 integers (-1 is used as terminator). * /
   printf ("Average is:% d\n", average (2, 3, 4, -1));

   / * Call with 4 integers. * /
   printf ("Average is:% d\n", average (5, 7, 9, 11, -1));

   / * Call with just -1 terminator. * /
   printf ("Average is:% d\n", average (-1));
}

/ * Returns the average of a variable list of integers. * /
#ifdef ANSI / * ANSI compatible version * /
int average (int first, ...)
{
   int count = 0, sum = 0, i = first;
   va_list marker;

   va_start (marker, first); / * Initialize variable arguments. * /
   while (i! = -1)
   {
      sum + = i;
      count ++;
      i = va_arg (marker, int);
   }
   va_end (marker); / * Reset variable arguments. * /
   return (sum? (sum / count): 0);
}
#else / * UNIX compatible version must use old-style definition. * /
int average (va_alist)
va_dcl
{
   int i, count, sum;
   va_list marker;

   va_start (marker); / * Initialize variable arguments. * /
   for (sum = count = 0; (i = va_arg (marker, int))! = -1; count ++)
      sum + = i;
   va_end (marker); / * Reset variable arguments. * /
   return (sum? (sum / count): 0);
}
#endif

xpyuanfly

The method on the first floor is very good, you can directly plug a default value to it