1 / 23

Representing Numbers Using Bases

Representing Numbers Using Bases. Numbers in base 10 are called decimal numbers, they are composed of 10 numerals ( ספרות ). 9786 = 9*1000 + 7*100 + 8*10 + 6*1 = 9*10 3 + 7*10 2 + 8*10 1 + 6*10 0

katima
Download Presentation

Representing Numbers Using Bases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Representing Numbers Using Bases • Numbers in base 10 are called decimal numbers,they are composed of 10 numerals ( ספרות ).9786 = 9*1000 + 7*100 + 8*10 + 6*1 = 9*103 + 7*102 + 8*101 + 6*100 • Numbers in base 2 are called binary numbers, they are composed of the numerals 0 and 1.110101 = 1*32 + 1*16 + 0*8 + 1*4 + 0*2 + 1*1 = 1*25 + 1*24 + 0*23 + 1*22 + 0*21 + 1*20 • 1111 1111 1111 1111 = 32768 + 16384 + 8192 + 4096 + 2048 + 1024 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 215 + 214 + … + 21 + 20 = S2n = 216 -1

  2. Converting Between Bases • From base 2 to base 10 is easy: just multiply each numeral by its exponent. 1001b = 1*8 + 1*1 = 9d • From base 10 to base 2 is slightly more complicated: In a loop from 31 to 0 (in the case of a 32 bit word) perform the following: • Divide the decimal number by 2n and write the result in the nth position. • Replace the number with the number minus the result multiplied by 2n • Decrement n by 1.

  3. A Base Conversion Example In C++ it looks like this: for(n=31;n>=0;n--){ res = decimal/(int)pow(2,n); decimal = decimal - res*(int)pow(2,n); cout << res; } 500/29 = 0 ; 500 - 0 = 5004/23 = 0 ; 4 - 0 = 4 500/28 = 1 ; 500 - 256 = 244 4/22 = 1 ; 4 - 4 = 0 244/27 = 1 ; 244 - 128 = 116 0/21 = 0 ; 0 - 0 = 0 116/26 = 1 ; 116 - 64 = 52 0/20 = 0 ; 0 - 0 = 0 52/25 = 1 ; 52 - 32 = 20 20/24 = 1 ; 20 - 16 = 4 500d = 111110100b

  4. Representation (יצוג) of Numbers • int, long, short, char - positive numbers are stored as a binary number where the MSB (Most Significant Bit) is 0A short is represented by 16 bits 100d = 26 + 25 + 22 =0000000001100100An int is represented by 32 bits 65545d = 216 + 23 + 20 = 00000000000000010000000000001001A char is represented by 8 bits‘a’ = 48d = 25 + 24 = 00110000

  5. Signed and Unsigned Numbers • In order to calculate both positive and negative numbers we must be able to distinguish (להבדיל) between them. • The most obvious solution is to add a bit that represents the sign. The name of this scheme is called: sign and magnitude (גודל). • There were several problems with this scheme: • where do we put the sign? • an extra step is needed to calculate the sign bit • there is both a positive and negative zero Thus a new representation was chosen

  6. Two's Complement • In the representation chosen leading 0s means positive and leading 1s means negative. • The positive half of the numbers from 0 to 2,147,483,647d (231 -1) uses the same representation as before. • The smallest negative number is -2,147,483,648 (-231) and is represented by 10000…00000b • It is followed by -2,147,483,647 (1000…0001b) up to -1 (1111…1111b). • There is a negative number (-231) with no positive partner, beware when programming. • The sum of a number and its inverse (נגדי) is -1.

  7. Negative Integers • Negative integers are represented using the two's-complement (משלים לשניים) method. The number is represented in binary, the bits are flipped and 1 is added. Thus the MSB is always 1 • Thus in a char:-100d = -01100100 = 10011011 + 1 = 1001110011111111 = 00000000 + 1 = -1d-128d = -10000000 = 01111111 + 1= 1000000011000101 = 00111010 + 1 = 00111011 = -59d • In a short:-25,000d = -0110000110101000 = 1001111001010111 + 1 = 10011110011000

  8. Unsigned Integers • Are stored as binary numbers. The MSB (sign bit) can be 1 or 0, the number is still positive.Thus in a char:10011100 = 27 + 24 + 23 + 22 = 156 11111111 = 27 + … + 20 = 28 - 1 = 255 • A problem arises when we have to compare numbers. How do we know if the number is an unsigned large number or a negative number. The solution is to add instructions which handle unsigned numbers. sltu $t0,$s0,$s1sltiu $t0,$s0,100

  9. Hexadecimal Numbers • Number in base 16 are called hexadecimalnumbers,they are composed of 16 numerals (0-9,a-f).9786hex = 9*163 + 7*162 + 8*161 + 6*160 = = 9*4096 + 7*256 + 8*16 + 6*1 = 38790d0xabcdef = 10*165 + 11*164 + 12*163 + 13*162 + 14*161 + 15*160 =11259375d • The conversion from binary to hex is very easy, each hex digit is 4 binary digits:0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 00110x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 01110x8 = 1000 0x9 = 1001 0xa = 1010 0xb = 10110xc = 1100 0xd = 1101 0xe = 1110 0xf = 1111

  10. Binary to Hexadecimal • Using the previous table it is easy to represent 32 bit binary numbers in a short form:0100 1100 0010 1101 1000 1100 0111 1001 = 4 a 2 d 8 c 7 9 =0x4a2d8c79. • Conversion from decimal to hexadecimal is similar to converting decimal to binary. for(n=7;n>=0;n--){ res = decimal/(int)pow(16,n); decimal = decimal - res*(int)pow(16,n); if(res>9) cout << (char)((res -10) + 'a'); else cout << res; }

  11. Addition and Subtraction • Lets add 6 to 7:00…00111 = 7d00…00110 = 6d00…01101 = 13d • Subtraction uses addition, the operand is simply negated before being added:7 - 6 = 7 + (-6) =00…00111 = 7d11…11010 = -6d00…00001 = 1d • The pseudoinstruction neg $t0,$t1 #$t0=-$t1is implemented by: sub $t0,$zero,$t1

  12. Overflow (גלישה) • Overflow occurs when the result of a operation can't be represented by the hardware. This can happen when we add two numbers of the same sign or subtract two large numbers of opposing signs. • Overflow is detected by the following table:Operation A B Result A+B >=0 >=0 <0 A+B <0 <0 >=0 A- B >=0 <0 <0 A- B <0 >=0 >=0 • How is overflow detected in unsigned numbers? It isn't. The instructions add, addi, sub cause exceptions when overflow occurs. The instructions addu, addiu, subu ignore overflow. Guess what instructions MIPS C compilers use?

  13. Logical Operations • All processors provide logical instructions that operate on the bits of the operands. • and, andi, or, ori, xor and xori perform the AND, OR and XOR (eXlusive OR) operations. • If $t0 contains 10 and $t1 contains 9 then:and $t2,$t0,$t1 # $t2=10&9=8or $t2,$t0,$t1 # $t2=10|9=11xor $t2,$t0,$t1 # $t2=10&9=3 • MIPS provides a NOR (Not OR) instruction as well:nor $t2,$t0,$t1 # $t2=~(10|9)=41100 NOR 1001 = 0100 • The logical NOT is a pseduinstruction in MIPS.

  14. Shift Instructions • In order to access bits or fields of bits the logical operations and the shift operations are used. • They move all bits in a word to the left or right, filling the emptied bits with 0s. The instructions are shift left logical (sll) , and shift right logical(srl).These 2 instructions use the shamt field of the R-type instruction. • sll $t0,$t1,6 # $t0 = $t1<<8;0000 0000 0000 1101 -> 0000 1101 0000 0000srl$t2,$t0,3 # $t2 = $t0>>3;0000 1101 0000 0000 -> 0000 0001 1010 0000 • The instructions sllv and slrv contain the shift amount in a register.sllv $t0,$t1,$t2 # $t0=$t1>>$t2

  15. The Basic Building Blocks

  16. Building an ALU • The arithmetic logic unit or ALU is the device that performs arithmetic and logical operations in the computer. We will build it out of 4 basic building blocks. The AND gate, Or gate, Inverter and Multiplexor. • We have now a 1-bit logical unit for AND and OR:

  17. 1-bit Full Adder • 1-bit addition has 3 inputs (a, b, CarryIn) and 2 outputs (Sum, CarryOut). • The CarryOut is set if at least 2 out of a, b, and CarryIn are 1. • The Sum is set when exactly one input is 1 or all three are 1.

  18. 1-bit ALU • We now have a 1-bit ALU that can perform AND, OR and addition. • By connecting 32 of these 1-bit ALUs we can build a 32-bit ALU. The CarryOut of the N 1-bit ALU is connected to the CarryIn of the N+11-bit ALU. This adder is called a ripple carry or ripple adder. A single carry out in the least significant bit (LSB) can cause a carry out of the MSB.

  19. 32-bit ALU • Subtraction is performed by adding the negative value. Thus we have to add an inverter to our ALU.

  20. Adding the slt instruction • slt produces 1 if rs < rt. Thus slt will set all but the LSB to 0 with the LSB being dependent on the result. We add a new input to the multiplexor called Less, this will be the output of slt. To set the LSB we perform rs-rt. If the result is negative we must set the output. Thus the sign bit is the correct output. But it isn't the output of slt, so a new output is added to the 1-bit ALU of the MSB, Set.

  21. The New ALU • As we can see all the Less inputs are hardwired to zero except the LSB Less input which is wired to the Set output of the MSB 1-bit ALU.

  22. Comparison in the ALU • In order to implement the bne and beq instructions we must compare between two numbers. a is equal to b if a-b=0The output line Zero is set to 1 if all the Results are 0, which happens if a == b.

  23. ALU Summary • Our ALU can perform AND, OR, Addition, Subtraction, Set on less then (slt), and Equality tests. The universal symbol of the ALU is shown here:

More Related