用一個C語言小程序解釋爲什麼在java中2-0.1是1.9，而 2-1.1結果是0.8999999999999999

#include<stdio.h>

void showDoubleInMemory(double *d){
	int i,k = 0;
	char *p = (char*)d;
	printf("In Memory:\n");
	for(i=7;i>=0;i--)
	{
 		unsigned char j=0x80,tmp=*(p+i);
 		for(;j;j>>=1){
			if(j&tmp) printf("1");
			else printf("0");
			k++;
			if(k == 1 || k == 12) printf(" ");
		}	
	}
	printf("\n\n");
}

int main()
{
/*
	Pay attention to the accuracy of the date type double.
*/	

	double d1 = 2.0, d2 = 0.1, d3, d4 = 1.9;
	d3 = d1 - d2;
	printf("%1.15lf\n",d1);
	showDoubleInMemory(&d1);
	printf("%.16lf\n",d2);
	showDoubleInMemory(&d2);
	
	printf("The value from calculation(2-0.1):\n%1.15lf\n",d3);
	showDoubleInMemory(&d3);
	printf("The value from definition:\n%1.15lf\n",d4);
	showDoubleInMemory(&d4);
	
	printf("------------------------------------------------------------------\n");

	d2 = 1.1;
	d3 = d1 - d2;
	d4 = 0.9;
	
	printf("%1.15lf\n",d1);
	showDoubleInMemory(&d1);
	printf("%1.15lf\n",d2);
	showDoubleInMemory(&d2);
	printf("The value from calculation(2-1.1):\n%0.16lf\n",d3);
	showDoubleInMemory(&d3);
	printf("The value from definition:\n%0.16lf\n",d4);
	showDoubleInMemory(&d4);
	return 0;
}

運行結果（2-0.1）：

2.000000000000000
In Memory:
0 10000000000 0000000000000000000000000000000000000000000000000000

0.1000000000000000
In Memory:
0 01111111011 1001100110011001100110011001100110011001100110011010

The value from calculation(2-0.1):
1.900000000000000
In Memory:
0 01111111111 1110011001100110011001100110011001100110011001100110

The value from definition:
1.900000000000000
In Memory:
0 01111111111 1110011001100110011001100110011001100110011001100110

------------------------------------------------------------------

（2-1.1）

2.000000000000000
In Memory:
0 10000000000 0000000000000000000000000000000000000000000000000000


1.100000000000000
In Memory:
0 01111111111 0001100110011001100110011001100110011001100110011010


The value from calculation(2-1.1):
0.8999999999999999
In Memory:
0 01111111110 1100110011001100110011001100110011001100110011001100


The value from definition:
0.9000000000000000
In Memory:
0 01111111110 1100110011001100110011001100110011001100110011001101

這個程序展現了相減後的結果與定義時的浮點數在內存中的差別，因爲double精度的原因只能有16位有效數字，經過四捨五入，就形成了這樣的結果。