Section 13 Storage of Numerical Data

1. Basic data types

Test on your own computer!

Types of	Keyword	Number of bytes	Number of digits	Value
Integer	[singned] short	2	16	-32768~32767
	[singned] int / long	4	32	-214783648~2147483647
	[singned] long long	8	32	-9223372036854775808―9223372036854775807
	unsingned short	2	16	0~65535
	unsingned int / long	4	32	0~4294967295
	unsingned long long	8	32	0―18446744073709551615
Floating point	float	4	32	± (3.4×10-38~3.4×1038)
	double / long double	8	64	± (1.7×10-308~1.7×10308 )
Character type	[singned] char	1	8	-128~127
	[unsingned] char	1	8	0~256

2. Be sensitive to 2

Decimal	Binary	Decimal	Relationship with 2	Decimal	Relationship with 2
0	0	64	26	1	21-1
1	1	128	27	3	22-1
2	10	256	28	7	23-1
3	11	512	29	15	24-1
4	100	1024	210	31	25-1
5	101	2048	211	63	26-1
6	110	4096	212	127	27-1
7	111	8192	213	255	28-1
8	1000	16384	214	32767	215-1
16	10000	32768	215	65535	216-1
32	100000	65536	216	4294967295	232-1

3. Integer data storage method

1. Stored in binary form;
2. Stored in complement form, the highest bit represents the sign of the value (0 is positive, 1 is negative);

0	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1

The first digit of the short type is 0, which means the maximum value in a positive number = 2 ¹⁵ -1 = 32767;

1	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1

The first digit of the short type is 1, which means the minimum value in negative numbers =-2 ¹⁵ =-32768;

1	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1

unsigned short type is unsigned value itself represents all bits: max = - 2 ¹⁶ - 1 = - 65535;

4. Preliminary exploration of integer data overflow

 #include <stdio.h>

int main()
{
    
    
	short a = 32767;
	printf("%hd\n", a + 1);
}
输出结果:
-32768

Data overflow law:

① Maximum data <=> Minimum data -1;

② Maximum data + overflow value <=> minimum data + (overflow value -1);

③ The beginning and end of the data type form a closed loop;

5. The storage principle of floating-point numbers

Floating point number X is expressed as: X=M×r ^E

Codename	name	Description
M	mantissa	Significant digits of the number X, its digits reflect the accuracy of the data
r	Base,	The base of decimal is 10, the base of binary is 2
E	Order code	Determine the true position of the decimal point of the number X

Types of	Number of bytes	Number of digits	Number sign S	Order code E	Mantissa M	Value	Value
float	4	32	First place	8th place	23rd place	± (3.4×10-38~3.4×1038)	7-bit precision
double	8	64	First place	11th place	52nd place	± (1.7×10-308~1.7×10308 )	Accuracy 15 bits

Calculate the value range of float type data:

①The float type has 1 sign bit, 8 exponent bits, and 23 mantissa bits;

②The index range is -128~127, and the maximum index is 127;

③The maximum mantissa is +1.11...(23 ones after the point, the decimal number is 1.8388607);

④1.8388607×2127 ≈ 2×2127 = 3.4×1038, also for negative numbers.

Floating point precision problem

#include <stdio.h>
int main()
{
    
    
	double a, b, c;
	a = 1234567890.123456789;
	b = 987654321.987654321;
	c = a + b;
	printf("  %22.9lf\n", a);
	printf("+ %22.9lf\n", b);
	printf("------------------------\n");
	printf("= %22.9lf\n", c);
	return 0;
}
    输出结果:
    1234567890.123456717
+    987654321.987654328
------------------------
=   2222222212.111111164

#include <stdio.h>
int main()
{
    
    
	double a = 0.65f;
	double b = 0.6f;
	double c = a - b;
	if (c == 0.05)
		printf("对了！\n");
	else
		printf("错了！\n");
	printf("%10.8lf\n%10.8lf\n", a, b);
	return 0;
}
输出结果:
错了！
0.64999998
0.60000002