1. Structured Program Modular structure Program grouped into blocks, each performing one task Clear interface between blocks, Easy to write, read, debug and maintain Top-down design Write outline, then refine Start from basic case, then add

2. Mathematics Input (–1) S B e–X 0.M Conversion to IEEE format Sign: OK Base: B e–X = 2 (log 2 B)(e–X) E.g. 4 10 =2 2*10, 8 10 =2 3*10, 16 10 =2 4*10 Mantissa: 0.M = 2 –i 1.m

3. Main Structure of Lab3 Read input B, E, X Read H and convert (code given) Extract exp from H, convert to IEEE Extract M from H, convert to IEEE Extract S from H Put everything together Print loop

4. Read Input B, E and X... msg1:.asciiz "Base (2, 4, 8 or 16): " msg2:.asciiz "Number of bits for exponent (5 to 10): " msg3:.asciiz "Excess of exponent: “... main:la $a0, msg1# print msg1 li $v0, 4 syscall li $v0, 5# read base syscall or $s1, $v0, $zero# $s1 stores B (base)... or $s2, $v0, $zero# $s2 stores E (no bits for e)... or $s3, $v0, $zero# $s3 stores X (excess)

5. Main Structure Read input B, E, X Read H and convert (code given) Extract exp from H, convert to IEEE Extract M from H, convert to IEEE Extract S from H Put everything together Print loop

6. Read H and Convert (given) loopjal read# read hex word into $s0... j loop read:# read 8 hex and store binary in $s0 li $t0, 0x00000030# t0 stores '0' li $t1, 0x00000061# t1 stores 'a‘... rend:jr $ra

7. Main Structure Read input B, E, X Read H and convert (code given) Extract exp from H, convert to IEEE Extract M from H, convert to IEEE Extract S from H Put everything together Print loop

8. Extract exp and Convert # H is in $s0 # assume E = 5 for now, extend later sll $t0, $s0, 1# get rid of sign srl $t0, $t0, 27# extract exp, 27=32-5 sub $t0, $t0, $s3# $t0 = exp–X # convert from base B to base 2 or $t1, $zero, $zero# $t1 = 0 bloop:srl $s1, $s1, 1# divide base by 2 beq $s1, $zero, bend add $t1, $t1, $t0 j bloop bend:addi $s1, $t1, 127# $s1 in excess 127 code

9. Main Structure Read input B, E, X Read H and convert (code given) Extract exp from H, convert to IEEE Extract M from H, convert to IEEE Extract S from H Put everything together Print loop

10. Extract M and Convert # H is in $s0, exp is in $s1 # assume E = 5 for now, extend later # extract mantissa in $t0 sll $t0, $s0, 6# get rid of exp, 6=5+1 srl $t0, $t0, 9# extracted mantissa # normalize li $t1, 0x00800000# $t1 = mask for hidden bit norm:and $t2, $t0, $t1 bne $t2, $zero, sign sll $t0, $t0, 1# shift left 1 bit addi $s1, $s1, -1# subtract exp by 1 j norm normd:li $t1, 0x007fffff and $t0, $t0, $t1# get rid of hidden bit

11. Main Structure Read input B, E, X Read H and convert (code given) Extract exp from H, convert to IEEE Extract M from H, convert to IEEE Extract S from H Put everything together Print loop

12. The Rest # H is in $s0, exp is in $s1, mantissa is in $t0 sign:li $t1, 0x80000000# $t1 = mask for sign and $t1, $s0, $t1# $t1 = sign # merge to for IEEE format sll $s0, $s1, 23# position exponent or $s0, $s0, $t1# merge sign and exponent or $s0, $s0, $t0# merge mantissa # print floating point sw $s0, 12($a1) lwc1 $f12, 12($a1) li $v0, 2# print single precision syscall

13. Extend to Arbitrary E # extract mantissa in $t0 srl $t0, $t0, 27# extract exp, 27=32-5... # extract mantissa in $t1 sll $t0, $s0, 6# get rid of exp, 6=5+1 addi $t1, $zero, 32 sub $t1, $t1, $s2# $s2 contains E jal shift... addi $t1, $s2, 1 sub $t1, $zero, $t1 jal shift

14. Shift Subroutine shift:# shift $t0 to right by $t1 bits slt $t2, $zero, $t1 beq $t2, $zero, left right:beq $t1, $zero, finish srl $t0, $t0, 1 addi $t1, $t1, -1 j right left:beq $t1, $zero, finish sll $t0, $t0, 1 addi $t1, $t1, 1 j left finish:jr $ra

15. High Level Language C++ void main() { int b, e, x; unsigned int h; char buf[100]; cout << “Input Base :”; cin >> b; cout << “Input No of bits for exp:”; cin >> e; cout << “Input excess:”; cin >> x; cout << “Input hexdecimal:” h = Char2Hex(buf); cout << Value(h, b, e, x); }

16. Convert String to HEX unsigned int Char2Hex(char a[]) { int i; unsigned int c, result; for (i = 0; i < 8; i ++) { if (isalpha(a[i])) c = a[i]-’a’+10; else c = a[i]-’0’; result = (result << 4) | c; } return result; }

17. Compute Value float value( unsigned int s0, // hex int s1,// base int s2,// no of bits for exp int s3)// excess { unsigned int m, e; m = ((s0 > 2; e = ((s0 > (32-s2)) – s3; return m*power(2, -30)*power(s1, e); }