How to display 23 bytes long double answer in C?

How to display 23 bytes long double answer in C?
long double can only show 16 bytes in windows......?
"15648934568488.45684 * 1561 +1265465.56
=24427986862675946.68724"
how to show?
i can only show
24427986862675944.00000

source code:
-----------------------------------------------------------
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

#define MAXOP 100
#define NUMBER '0'

#define MAXVAL 100

#define BUFSIZE 100

int getop(char []);
void push(long double);
long double pop(void);

int sp = 0;
long double val[MAXVAL];


char buf[BUFSIZE];
int bufp = 0;

int getch(void);
void ingetch(int);

main() {
int type;
long double op2;
char s[MAXOP];

while ((type = getop(s)) != EOF) {
switch (type) {
case NUMBER:
push(atof(s));
break;
case '+':
push(pop() + pop());
break;
case '*':
push(pop() * pop());
break;
case '-':
op2 = pop();
push(pop() - op2);
break;
case '/':
op2 = pop();
if (op2 != 0.0)
push(pop() / op2);
else
printf("error: zero divisor\n");
break;
case '\n':
printf("%.5lf\n", pop());
break;
default:
printf("error: unknown command %s\n", s);
break;
}
}
return 0;
}

void push(long double f) {
if (sp < MAXVAL)
val[sp++] = f;
else
printf("error: stack full, can't push %g\n", f);
}

long double pop(void) {
if (sp > 0)
return val[--sp];
else {
printf("error: stack empty\n");
return 0.0;
}
}

int getch(void) {
return (bufp >0) ? buf [--bufp] : getchar();
}

void ungetch(int c) {
if (bufp >= BUFSIZE)
printf("ungetch: too many characters\n");
else
buf[bufp++] = c;
}

int getop(char s[]) {
int i, c;

while ((s[0] = c = getch()) == ' ' || c == '\t')
;
s[1] = '\0';
if (!isdigit(c) && c != '.')
return c;
i = 0;
if (isdigit(c))
while (isdigit(s[++i] = c = getch()))
;
if (c == '.')
while (isdigit(s[++i] = c = getch()))
;
s[i] = '\0';
if (c != EOF)
ungetch(c);
return NUMBER;
}
-------------------------------------------------
input as follow:
15648934568488.45684 1561 * 1265465.56 + <enter>

Thanks a lot~~~

Does this help?

Code:

---

int main ( ) {
    long double foo = 12345.67890;
    printf( "%Lf\n", foo );
    printf( "%.0Lf\n", foo );
    return 0;
}

---

> How to display 23 bytes long double answer in C?
Technically, you can't. In my implementation at least, long double's have only 18 decimal digits of accuracy, so the last 5 digits of anything you write will be in effect noise.

Why not typedef your own 'long triple'? That's how the compiler maker would do it. That is what 'typedef' is for-- creating new types...

Neither the computer nor the compiler care one wit about the size of the number-- the control is on _your_ hands...

But typedef cannot create a type to show all the digits......maximum is still long double...
Am I right?