Integer Representations(整数表示)

C语言整数数据类型的典型取值范围

C data type Minimum32bit Minimum64bit Maximum32bit Maximum64bit Bytes32bit Bytes64bit
[signed] char $-2^7$ $-2^7$ $2^{7}-1$ $2^{7}-1$ 1 1
unsigned char 0 0 $2^{8}-1$ $2^{8}-1$ 1 1
short $-2^{15}$ $-2^{15}$ $2^{15}-1$ $2^{15}-1$ 2 2
unsigned short 0 0 $2^{16}-1$ $2^{16}-1$ 2 2
int $-2^{31}$ $-2^{31}$ $2^{31}-1$ $2^{31}-1$ 4 4
unsigned 0 0 $2^{32}-1$ $2^{32}-1$ 4 4
long $-2^{31}$ $-2^{63}$ $2^{31}-1$ $2^{63}-1$ 4 8
unsigned long 0 0 $2^{31}-1$ $2^{64}-1$ 4 8
int32_t $-2^{31}$ $-2^{31}$ $2^{31}-1$ $2^{31}-1$ 4 4
uint32_t 0 0 $2^{32}-1$ $2^{32}-1$ 4 4
int64_t $-2^{63}$ $-2^{63}$ $2^{63}-1$ $2^{63}-1$ 8 8
uint64_t 0 0 $2^{64}-1$ $2^{64}-1$ 8 8

Unsigned Encodings

For vector $\vec{x} = [x_{w-1},x_{w-2},…,x_0]$

$$ B2U_{w}(\vec{x})\stackrel{.}{=}x_{w-1}{\cdot}2^{w-1}+x_{w-2}{\cdot}2^{w-2}+…+x_{0}{\cdot}2^{0}=\sum_{i=0}^{w-1}{x_{i}2^{i}} $$

for example:

$$ B2U_{4}([0101])=0{\cdot}2^3+1{\cdot}2^2+0{\cdot}2^1+1{\cdot}2^0=5 $$

$$ B2U_{4}([1011])=1{\cdot}2^3+0{\cdot}2^2+1{\cdot}2^1+1{\cdot}2^0=11 $$

十进制：

$$ B10U_{w}(\vec{y})\stackrel{.}{=}y_{w-1}{\cdot}10^{w-1}+y_{w-2}{\cdot}10^{w-2}+…+y_{0}{\cdot}10^{0}=\sum_{i=0}^{w-1}{y_{i}10^{i}} $$

2-2-1

Two’s Complement Encodeings

补码编码

For vector $\vec{x} = [x_{w-1},x_{w-2},…,x_0]$

$$ B2T_{w}(\vec{x})\stackrel{.}{=}x_{w-1}{\cdot}-2^{w-1}+x_{w-2}{\cdot}2^{w-2}+…+x_{0}{\cdot}2^{0}=-x_{w-1}2^{w-1}+\sum_{i=0}^{w-2}{x_{i}2^{i}} $$

for example:

$$ B2U_{4}([0101])=-0{\cdot}2^3+1{\cdot}2^2+0{\cdot}2^1+1{\cdot}2^0=5 $$

$$ B2U_{4}([1011])=-1{\cdot}2^3+0{\cdot}2^2+1{\cdot}2^1+1{\cdot}2^0=-11 $$

Unsigned Vs Signed

Unsigned Maximum

work size Binary Hexadecimal Decimal
8 1111 1111 0xFF $2^8-1$
16 $\underbrace{1111…1111}_{16}$ 0xFFF $2^{16}-1$
32 $\underbrace{1111…1111}_{32}$ 0xFFFFFFFF $2^{32}-1$
64 $\underbrace{1111…1111}_{64}$ $0x\underbrace{FFF…FFF}_{16}$ $2^{64}-1$

Signed Maximum

work size Binary Hexadecimal Decimal
8 0111 1111 0x7F $2^7-1$
16 $0\underbrace{111…1111}_{15}$ 0x7FF $2^{15}-1$
32 $0\underbrace{111…1111}_{31}$ 0x7FFFFFFF $2^{31}-1$
64 $0\underbrace{111…1111}_{63}$ $0x7\underbrace{FF…FFF}_{15}$ $2^{63}-1$

Signed Minimum

work size Binary Hexadecimal Decimal
8 1000 0000 0x80 $-2^7$
16 $1\underbrace{000…0000}_{15}$ 0x8000 $-2^{15}$
32 $1\underbrace{000…0000}_{31}$ 0x80000000 $-2^{31}$
64 $1\underbrace{111…1111}_{63}$ $0x8\underbrace{00…000}_{15}$ $-2^{63}$

Special Numeric

Value 8 16 32 64
-1 1111 1111 $\underbrace{1111…1111}_{16}$ $\underbrace{1111…1111}_{32}$ $\underbrace{1111…1111}_{64}$
UMax 1111 1111 $\underbrace{1111…1111}_{16}$ $\underbrace{1111…1111}_{32}$ $\underbrace{1111…1111}_{64}$

Error : 1000 0001

weight $\pm2^{15}$ $2^{14}$ $2^{13}$ $2^{12}$ $2^{11}$ $2^{10}$ $2^{9}$ $2^{8}$ $2^{7}$ $2^{6}$ $2^{5}$ $2^{4}$ $2^{3}$ $2^{2}$ $2^{1}$ $2^{0}$
$12345_2$ 0 0 1 1 0 0 0 0 0 0 1 1 1 0 0 1
$12345_{10}$ 0 0 8192 4096 0 0 0 0 0 0 32 16 8 0 0 1
$-12345_2$ 1 1 0 0 1 1 1 1 1 1 0 0 0 1 1 1
$-12345_{10}$ -32768 16384 0 0 2048 1024 512 256 128 64 0 0 0 4 2 1
$53191_2$ 1 1 0 0 1 1 1 1 1 1 0 0 0 1 1 1
$53191_{10}$ 32768 16384 0 0 2048 1024 512 256 128 64 0 0 0 4 2 1

Conversion between Signed and Unsigned

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#include <stdio.h>

int main() {
    short int a = -12345;
    unsigned short b = (unsigned short) a;
    printf("a = %d, b = %u", a, b);
    return 0;
}

运行结果

a = -12345, b = 53191

-12345: 1100 1111 1100 0111

53191: 1100 1111 1100 0111

Two’s Complement to Unsigned

$$ B2U_{w}(\vec{x})\stackrel{.}{=}x_{w-1}{\cdot}2^{w-1}+x_{w-2}{\cdot}2^{w-2}+…+x_{0}{\cdot}2^{0} $$

$$ B2T_{w}(\vec{x})\stackrel{.}{=}x_{w-1}{\cdot}-2^{w-1}+x_{w-2}{\cdot}2^{w-2}+…+x_{0}{\cdot}2^{0} $$

$$ B2U_{w}-B2T_{w}=x_{w-1}{\cdot}2^{w-1}-x_{w-1}{\cdot}-2^{w-1}=x_{w-1}{\cdot}2^{w} $$

$T2U_{w}(x)$

$$ B2U_{w}(\vec{x})=B2T_{w}(\vec{x})+x_{w-1}{\cdot}2^{w} $$

$$ T2U_{w}(x)= \begin{cases} x+2^w, & \text {x < 0 } \\ x, & \text{ x $\geq$ 0} \end{cases} $$

$U2T_{w}(u)$

$$ B2T_w=B2U_w-x_{w-1}{\cdot}2^w $$

$$ U2T^w(u)= \begin{cases} u, & \text {u $\leq$ $TMax_w$ } \\ u-2^w, & \text{ u > $TMax_w$} \end{cases} $$

Signed versus Unsigned in C

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#include <stdio.h>

int main() {
    int a = -1;
    unsigned int b = 0;
    if (a < b) { // 隐式转换， 无符号数a会转换为无符号数，即-1 = 2^{32} -1 = 4294967295, 所以if (4294967295 < 0) 不成立输出 -1 > 0
        printf("-1 < 0");
    } else {
        printf("-1 > 0");
    }
    return 0;
}

Expanding the bit Representation of a Number

unsigned char a;

$x_7$ $x_6$ $x_5$ $x_4$ $x_3$ $x_2$ $x_1$ $x_0$

unsigned short b;

0 0 0 0 0 0 0 0 $x_7$ $x_6$ $x_5$ $x_4$ $x_3$ $x_2$ $x_1$ $x_0$

Signed Extension

有符号数 符号扩展

char b;

$x_7$ $x_6$ $x_5$ $x_4$ $x_3$ $x_2$ $x_1$ $x_0$

short b;

$$ \underbrace{x_{7}\text{|}x_{7}\text{|}x_{7}\text{|}x_{7}\text{|}x_{7}\text{|}x_{7}\text{|}x_{7}\text{|}x_{7}}_{sign\ extension} $$

$$ x_{7}\text{|}x_{6}\text{|}x_{5}\text{|}x_{4}\text{|}x_{3}x_{2}\text{|}x_{1} $$

$x_7 = 0$

0 0 0 0 0 0 0 0 $x_7$ $x_6$ $x_5$ $x_4$ $x_3$ $x_2$ $x_1$ $x_0$

$x_7=1$

1 1 1 1 1 1 1 1 $x_7$ $x_6$ $x_5$ $x_4$ $x_3$ $x_2$ $x_1$ $x_0$

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