Question

#include <stdio.h>

int main()
{
    long double n = (long double)1208925819614629174706175; // (1<<80)-1 (generated in python)
    printf("%LG\n",n);
}

On compiling this with gcc, I get a warning

test.c:5:31: warning: integer constant is too large for its type [enabled by default]

On running it prints -1

On doing a sizeof(long double), it prints 12.
So a signed long double should accomodate (1<<83)-1 right?

How do I get it working correctly?

PS: The original problem I was working on, dealt with storing signed integers as large as 12*(10¹⁸). I don’t want to write multiplication, addition, and other calculation routines. So how do I do it?

Answer 1

Long double will handle 10**300 and above, that’s not what the compiler is complaining about. It says the integer that you are trying to cast into a long is too big for an integer. ( hint – add a decimal point ) – starbolin