Lab7_CS22B036 - HackMD

# Lab 7 Name: M V Sonith Roll No.: CS22B036 --- # Question 1 **Algorithm** Multiply given number by 2 , if it is greater than 1, mark 1 in binary format else 0. Do until you get zero **Steps** | Integer part | Decimal Part | | ------------- | ------------- | x | 0.1| 0 | 0.2| 0 | 0.4| 0 | 0.8| 1 | 0.6| 1 | 0.2| repetition continues after this **Answer** 0.1~decimal~ = 0.00011001~binary~ --- # Question 2 **Answer** From above question we know that 0.1~decimal~ = 0.00011001~binary~ **Scientific notation is** 0.00011001~binary~ = 1.1001 * 2^-4^ signed = 0 exponential = 01111011 mantissa = 10011001100110011001100 **IEEE 754 Format is** 00111101110011001100110011001100 --- # Question 3 **Steps** | Integer part | Decimal Part | | ------------- | ------------- | x | 0.2| 0 | 0.4| 0 | 0.8| 1 | 0.6| 1 | 0.2| So 0.2~decimal~ = 0.0011001~binary~ **Scientific notation is** 0.0011001~binary~ = 1.1001 * 2^-3^ signed = 0 exponential = 01111100 mantissa = 10011001100110011001100 **IEEE 754 Format is** 00111110010011001100110011001100 **Converting Back** 1.10011001100110011001100 x 2^-3^ which is ***Conversion:*** (0.00110011001100110011001100)₂ = (0 × 2⁰) + (0 × 2⁻¹) + (0 × 2⁻²) + (1 × 2⁻³) + (1 × 2⁻⁴) + (0 × 2⁻⁵) + (0 × 2⁻⁶) + (1 × 2⁻⁷) + (1 × 2⁻⁸) + (0 × 2⁻⁹) + (0 × 2⁻¹⁰) + (1 × 2⁻¹¹) + (1 × 2⁻¹²) + (0 × 2⁻¹³) + (0 × 2⁻¹⁴) + (1 × 2⁻¹⁵) + (1 × 2⁻¹⁶) + (0 × 2⁻¹⁷) + (0 × 2⁻¹⁸) + (1 × 2⁻¹⁹) + (1 × 2⁻²⁰) + (0 × 2⁻²¹) + (0 × 2⁻²²) + (1 × 2⁻²³) + (1 × 2⁻²⁴) + (0 × 2⁻²⁵) + (0 × 2⁻²⁶) = (0.19999998807907104492)₁₀ --- # Question 4 **Algorithm** For a floating point number we can use a representations that is using two registers where one is to store fractional other is to store integer part of number As the precision is upto 3 fractional places after the point we can use 10 bits rightmost 10 bits in a register for storing. To detect overflow we add 24 to fractional part of the result obtained by adding fractional parts of two numbers, because 1000 is then 1024 which is easy to detect overflow as 11th bit in register becomes 1 ***So according to Representation*** * 5.391 is stored as 0000 0000 0000 0000 0000 0000 0000 0101 (Integer part) 0000 0000 0000 0000 0000 0001 1001 0111 (Fractional Part) * 3.012 is stored as 0000 0000 0000 0000 0000 0000 0000 0011 (Integer part) 0000 0000 0000 0000 0000 0000 0000 1100 (Fractional Part) ***adding them*** * Integer part is 0000 0000 0000 0000 0000 0000 0000 1000 which is 8 * Fractional part is 0000 0000 0000 0000 0000 0001 1001 0011 which is 403 adding 24 to fractional part for checking overflow we get 0000 0000 0000 0000 0000 0001 1010 1011 in which 11th bit is zero so there is no overflow **So, the number is 8.403 accoridin to representation** --- # Question 5 **Code** ```assembly= .data sig: .word 0x80000000 pow: .word 0x7F800000 man: .word 0x007FFFFF num1: .word 0xBF400000 #-0.75 num2: .word 0x3FC00000 #1.5 n1: .word 0x00800000 .text lw x4 sig lw x5 pow lw x6 man lw x7 num1 lw x8 num2 lw x15 n1 and x9 x7 x4 #s and x10 x7 x5 #p and x11 x7 x6 #m and x12 x8 x4 #s and x13 x8 x5 #p and x14 x8 x6 #m srli x10 x10 23 #getting pow srli x13 x13 23 add x16 x11 x15 #make number from mantisa add x17 x14 x15 beq x10 x13 equal #checking powers for equalizing blt x10 x13 lesser blt x13 x10 greater lesser: srli x16 x16 1 addi x10 x10 1 bne x10 x13 lesser j equal greater: srli x17 x17 1 addi x13 x13 1 bne x10 x13 greater j equal j exit equal: xor x18 x12 x9 #checking whether numbers are of same sign beq x18 x0 same_sign diff_sign: beq x9 x0 num2neg #determining which number is negative num1neg: sub x20 x17 x16 #subtracting beq x20 x0 zeroc1 # if both are equal answer is zero and x21 x20 x4 #checking resultant is negative or positive beq x21 x0 next1 #if negative store only absolute value sub x20 x16 x17 next1: lw x22 sig srli x22 x22 8 underflow1: and x23 x22 x20 #checking underflow beq x23 x0 zerox1 #updating mantisa if underflow j nozerox1 zerox1: addi x10 x10 -1 #updating mantisa and power slli x20 x20 1 j underflow1 nozerox1: slli x10 x10 23 # shifting back power to its place or x31 x21 x10 sub x20 x20 x15 or x31 x31 x20 #storing result in x31 j exit zeroc1: addi x31 x0 0 j exit num2neg: sub x20 x16 x17 beq x20 x0 zeroc and x21 x20 x4 beq x21 x0 next sub x20 x17 x16 next: lw x22 sig srli x22 x22 8 underflow: and x23 x22 x20 beq x23 x0 zerox j nozerox zerox: addi x10 x10 -1 slli x20 x20 1 j underflow nozerox: slli x10 x10 23 or x31 x21 x10 sub x20 x20 x15 or x31 x31 x20 j exit zeroc: addi x31 x0 0 j exit same_sign: add x20 x16 x17 addi x1 x1 1 srli x21 x20 24 beq x1 x21 overflow nooverFlow: sub x20 x20 x15 slli x10 x10 23 or x31 x20 x10 or x31 x31 x9 j exit overflow: addi x10 x10 1 srli x20 x20 1 sub x20 x20 x15 slli x10 x10 23 or x31 x20 x10 or x31 x31 x9 j exit exit: add x0 x0 x0 ``` **Output** we observe hexadecimal value 0x3f400000 in x31 register which is 0.75 in decimal ---