[postgraduate entrance examination planning group] conversion between signed and unsigned numbers

The landlord has something to say ：

Reviewing 《 The principle of computer organization 》 when , Encountered the problem of conversion between signed and unsigned numbers , To deal with this kind of problem , The landlord looked up the information carefully and summed up some experience in combination with the title , The contents are as follows （ If there is a new summary , It will be revised over and over again ）：

Be careful ：

1、 The word length is 16 position （ It's more convenient ）

2、 I suggest you go to have a general understanding of C++ Byte length and indication range of basic data type in ; Such as ：

char	1 byte	int	4 byte	short [int]	2 byte	float	4 byte	long [int]	4 byte
signed char	1 byte	signed int	4 byte	signed short [int]	2 byte	double	8 byte	signed long [int]	4 byte
unsigned char	1 byte	unsigned int	4 byte	unsigned short [int]	2 byte	long double	8 byte	unsigned long [int]	4 byte

3、（1） Range of fixed-point integers ( 16 bit ) ： A signed ：-32767 <= x <= 32767

（2） Range of fixed-point integer complement ( 16 bit ) ： A signed ：-32768 <= x <= 32767

4、 In the computer , Negative numbers are usually represented by complements . Be careful ： All the examples in this article , It is to convert the number to be assigned into the form of complement first .

5、 clarify 0 and 1 The original code of 、 Inverse code 、 Complement code .（0 There is only one form of complement and shift ; The original code of a positive number 、 Inverse code 、 The complement is the same ;)

	Original code	Inverse code	Complement code
[ +0 ]	0000 0000	0000 0000	0000 0000
[ -0 ]	1000 0000	1111 1111	0000 0000
[ +1 ]	0000 0001	0000 0001	0000 0001
[ -1 ]	1000 0001	1111 1110	1111 1111

One 、 Same type （ Like a signed number , Or both are unsigned numbers ） transformation

（ One ）1. Both are unsigned numbers （ From short to long ： Add... At the top “ 0 ”.）

unsigned short x = 255;  //short by 2 byte 

unsigned int y = x;  //int by 4 byte 

x = 255 D = 1111 1111 B

y = 0000 0000 1111 1111 B = 255 D

summary ： When both are unsigned numbers , From a shorter number to a longer number , That is, add... To the highest position “ 0 ”, be called “ Zero expansion ”.

（ One ）2. Both are unsigned numbers （ From long to short ： Add... At the top “ 0 ”.）

unsigned int x = 255;  //short by 2 byte 

unsigned short y = x;  //int by 4 byte 

x = 255 D = 0000 0000 1111 1111 B

y = 1111 1111 B = 255 D

summary ： When both are unsigned numbers , Change from a longer number to a shorter number , That is, remove the high order “ 0 ” byte , The lower byte remains unchanged .

（ Two ） Both are signed numbers （ Changing from short to long is the same as changing from long to short “（ One ） Both are unsigned numbers ” equally ）

short x = -120;  //short by 2 byte 

int y = x;  //int by 4 byte 

x = -120 D , Its original code is ：1111 1000 B, Its complement is ：1000 1000 B （= 88 H）

y = 1111 1111 1000 1000 B = -120 D （= FF 88 H）

summary ： When both are signed numbers , Use symbols to extend , When it is a negative number , High byte plus “ 1 ”; Positive number , High byte plus “ 0 ”.

Two 、 The conversion between signed and unsigned numbers

（ One ） Signed number to unsigned number

short n = -1;

unsigned short m = n;

n = -1 D, Its complement is ：1111 1111 B 

m = 1111 1111 B = 65535 D

notes ：n The highest position of “ 1 ” It represents the sign bit , because m Is an unsigned number , therefore m  The highest position of “ 1 ” It means “ 2 Of 7 Power ”.

summary ： When a signed number is changed to an unsigned number , When the signed number is negative , The key is to put the highest “ 1 ” become “ 2 To the power of ”.

（ Two ） Unsigned number to signed number

unsigned short a = 65535;

short b = -1;

a = 65535 = FF FF FF FF H

b The complement of is ：FF FF FF FF H = -1 D

notes ： because b Is a signed number , The highest position is “ 1 ”, therefore b It's a negative number , It can be seen that its value is -1.

summary ： When an unsigned number is converted to a signed number , First get the complement of the assigned number , It is the complement of the assigned number ; Then the complement of the assigned number is inversely added to get the original code , Calculate again to get the decimal value .

Another example ：

unsigned char i = 130;

char k = i;

i = 130 D = 1000 0010 B （ Be careful ： The original code of a positive number = A complement to a positive number = The inverse of a positive number ）

k The complement of is ：1000 0010 B, The original code is obtained by adding one to the reverse ：1111 1110 B, because k Is a signed number , The highest position is “ 1 ”, therefore k It's a negative number , That is to say -126 D.

practice ：（ Enclosed 2011 True questions of the annual unified examination , Easy to understand and consolidate ）

Suppose in a 8 Bit word long computers run as follows C Procedures section ：

unsigned int x = 134;
unsigned int y = 246;
int m = x;
int n = y;
unsigned int z1 = x - y;
unsigned int z2 = x + y;
int k1 = m - n;
int k2 = m + n;

If the compiler compiles 8 individual 8 Bit register R1 - R8 Assign to variables separately x、y、m、n、z1、z2、k1、k2, Please answer the following questions （ Tips ： Signed integers are represented by complements ）

（ 1 ） After executing the above program segment , register R1、 R5 and R6 What are the contents of ？（ In hexadecimal notation ）

（ 2 ） After executing the above program segment , Variable m and k1 What are the values of ？（ In decimal ）

（ 3 ） The above program segment involves the addition of signed integers / reduce 、 Unsigned integer plus / Subtraction operation , Can these four operations be realized by the same adder auxiliary circuit ？ Brief reasons .

（ 4 ） How to judge the addition of signed integers in the computer / Whether the result of subtraction overflows ？ In the above program segment , The execution results of signed integer operation statements will overflow ？

Explain ：（ 1 ） register R1、 R5 and R6 The content of , They correspond to each other x, z1（x - y) , z2（x + y）.

x = 134 D, The original code is equal to the complement code 1000 0110 B, That is to say 86 H, That is to say  R1

y = 246 D, The original code is equal to the complement code 1111 0110 B, That is to say F6 H

z1 = x - y = x Complement + [-y] Complement = 1000 0110 B + 0000 1010 B = 1001 0000 B = 90 H, That is to say  R5

z2 = x + y = x Complement + y Complement = 1000 0110 B + 1111 0110 B = (1) 0111 1100 B = 7C H, That is to say  R6

（ Be careful ：z2 Overflow at , But the addition and subtraction of unsigned integers , Generally, overflow is not considered , Only when output, if the highest bit of the signed number is the sign bit , intend z2 If it is converted to a signed number , The highest bit is 1.）

therefore R1 = 134 = 86 H;
        R5 = 90 H;
        R6 = 7C H;

（ 2 ）m Is an unsigned integer x To signed integer ,n Is an unsigned integer y To signed integer , from （1） know

x The complement of is 1000 0110 B,y The complement of is 1111 0110 B, therefore

m The complement of is 1000 0110 B, Its original code is 1111 1010 B, Its value is -122 D

n  The complement of is 1111 0110 B, Its original code is 1000 1010 B, Its value is -10 D

namely k1 = m - n = m  Complement + -n  Complement = 1000 0110 B + 0000 1010 B = 1001 0000 B = 90 H, Get the original code as 1111 0000 B = -112 D

By the way k2 = m + n = m  Complement + n  Complement = 1000 0110 B + 1111 0110 B = (1) 0111 1100 B = 7C H（ Be careful ： There is overflow here ）

therefore ,m = -122;
           k1 = -112;

Digression （ Yes OF 、SF、CF Also pay attention to distinguish ）：

k2 = m + n, From the known  m = -122,n = -10, The sum of the two is still negative , But the result is positive . therefore , overflow flag OF = 1, It means overflow , Description register R6 The content in is not the real result ; Symbol sign bit SF = 0, Indicates that the result is a positive number （ The overflow flag is 1, There is an error in the symbol ）, Carry flag bit CF = 1, It only indicates that the adder has the highest carry , It doesn't mean anything to the result of the operation .

( 3 ) can .n The bit adder implements modulo 2 Of n Power unsigned integer addition operation . Because in a computer , Both positive and negative numbers are represented by complements , The addition of signed integers and unsigned integers , It can be realized directly by adder . For subtraction , As long as the complement of the subtracted number plus the complement of the negative number of the subtracted number , But consider the overflow situation . So signed integers plus / reduce 、 Unsigned integer plus / The subtraction operation can be realized by the same adder auxiliary circuit （ The overflow judgment circuit is different ）.

（ 4 ） yes . The computer internally judges that a signed integer plus / Whether the result of subtraction overflows , Yes 3 Methods . One is if two inputs of the adder （ Add ） The same symbol as , And different from the output （ and ） The symbol of , The result overflows . Second, when the adder completes the addition operation , If the second highest （ Highest digit ） Carry and highest bit of （ Sign bit ） The carry of is different , The result overflows . Third, both signed integers are negative numbers , When the two are added together , The result is less than 8 The smallest negative number that a binary can represent , The result overflows .

In the implementation of k2 Overflow occurs when the statement of .k2 = 1000 0110 B + 1111 0110 B = (1) 0111 1100 B, In parentheses is the carry of the adder , because 2 Signed integers are all negative numbers , After addition, the highest bit of the result is 0 , So overflow .

（ Overflow judgment method simple version ： Double sign decision 、 Highest carry 、 The symbol of the operand with the same symbol is different from that of the original operand , The result overflows .）

notes ：

Double sign decision ：

Double sign bit operation with complement （ The positive sign is 00, The negative sign is 11）, The first sign bit represents the symbol of the final result , The second sign bit indicates whether the operation result overflows . The first sign bit is the same as the second sign bit , There is no overflow ,; If different , Then overflow . As follows ：

If the sign bit of the operation result is 01, Is a positive overflow ;

If the double sign of the result is 10, Negative overflow ;

If the double sign bit of the result is 00, The result is a positive number , No overflow .

If the double sign bit of the result is 11, The result is a negative number , No overflow .