how to print big amount of number?

Hi im new to the forum and a newbie in C programming too

I need to ask how can i print large amount of number for example,

Code:

#include<stdlib.h>
#include<stdio.h>
#include<conio.h>d
#include<math.h>
#include<limits.h>  /* lol i tried so many header files already. believe me :S */

int main(void)

{
int apple,mango,result;

apple = 36000;
mango = 2;

result = apple*mango;

printf ("result : %d\n",result);

return 0;
}

the result should be printed out as '72000'.. but what I'd been getting is '6464'.. I tried googling but i couldn't find the solution. which is strange. I heard about integer could only bear numbers from -36000+ up to 36000+ (sorry if i'm wrong).

anyone can help me fix my code? i really need your kind help Thanks

Are you using the Turbo-C compiler?

On your machine, an int may be 16 bits, limiting it to -32768 to 32767. Try it with longs instead of ints:

long apple = 36000, mango = 2;
printf("%l\n", apple * mango);

Are you working with a 16 bit processor? Integers on most machines nowadays are between +-2 billion or so. I think specifying you data type to long should force it to use a 4 bytes.

IE: long apple,mango,result;

See This

Well, 1 byte variable offers 2^8 numbers. 2 bytes 2^16 and so on. Or less for signed variables

Try doing this and tell us the results. It will so you how many bytes an int has

Code:

printf("Int size = %d\n", sizeof(int)");

as far as the headers go, I think you only need stdio.h

how thanks for the response.

bithub: yea i'm using TurboCPP program.

Epy: i tried using your suggestion

Code:

long apple = 36000, mango = 2;
printf("%l\n", apple * mango);

it just prints out %l

Code:

%l

C_ntua: i believe you mean printf("Int size = %d\n", sizeof(int)); without the extra ". I run it and it says Int size = 2. does that mean int can only bear up to 65536? (2^16 = 65536). so how can i reach 72000 or higher?


I really appreciate the help but I still can't get it right. maybe I misunderstood some part of your feedback. Please let me know where I done wrong

here it comes again..

Code:

int X = 36000;
long L;

L = X*2;

printf ("L = %lu\n",L);

like i said before, it gives me L = 6464 rather than the correct one, L = 72000 :'( where did i do wrong?

signed values have a limit of one less than their bits, so 2^15. Stop using int for god's sake.

long X = 36000, L = X * 2;
printf("L = %ld\n", L);

%ld specifies long, %lu specifies unsigned long.

sorry guys my bad for being such a n00b :/

I GOT IT SOLVED NOW! i replaced all '%d' with '%ld'.. & replace all 'int' to 'long' thank you so much for your help