Truncation after 17. Digit on C

有些话、适合烂在心里 提交于 2019-12-02 22:47:01

问题


I was trying to write a simple program on C about the classical history between the king, chessboard and rice grains. The total amount of rice grains that is needed to fill the chessboard is : 18 446 744 073 709 551 615 .

But i'm getting : 18 446 744 073 709 552 000. on C.

Aren't there any solution to increase 17 digits resolution?

Here is my code.

#include <stdio.h>
#include <math.h>

int main( void )
{
  double i=1.0;
  while(i<=64)  
  {
    printf("%2.0lf.Casilla = %.0lf\n", i, pow(2.0,(i-1.0)));
    i++;
  }
  printf("\n\n***En Total = %.0lf Granos.\n\n",pow(2.0,64.0)-1);
  return 0;
}

Thanks in advance!


回答1:


double of accuracy is approximately 17 decimal digits(52bit+1). so You'll use the type that has a 64-bit or higher accuracy.

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

uint64_t pow_2_i(int n){//2^n
    if(0 > n || n >= 64)
        return 0;
    uint64_t x = 1;
    return x << n;
}

int main(void){
    int i;
    uint64_t x, sum = 0;
    for(i = 1; i <= 64; ++i){
        sum += (x = pow_2_i(i-1));
        printf("%2d.Casilla = %" PRIu64 "\n", i, x);
    }
    printf("\n\n***En Total = %" PRIu64 " Granos.\n\n", sum);//No Cheat^-^
    return 0;
}

If long double has large accuracy than double

#include <stdio.h>

int main( void ){
    long double x=1.0;//1.0L
    int i;
    for(i=1; i<=64;++i){
        printf("%2d.Casilla = %.0Lf\n", i, x);
        x *= 2.0L;
    }
    printf("\n\n***En Total = %.0Lf Granos.\n\n", x-1.0L);//2^64-1:Σar^k (k=0->n) =a(r^(n+1)-1)/(r-1):1(2^(63+1)-1)/(2-1)
    return 0;
}


来源:https://stackoverflow.com/questions/29388592/truncation-after-17-digit-on-c

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