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Homework

Homework. Turn in HW2 tonight HW3 is on-line already Questions?. Base for Integer Constants. Designating the base for an integer constant If constant begins with either: 0x It is Hex with a-f as Hex digits 0X It is Hex with A-F as Hex digits Otherwise, if constant begins with

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Homework

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  1. Homework • Turn in HW2 tonight • HW3 is on-line already • Questions?

  2. Base for Integer Constants • Designating the base for an integer constant • If constant begins with either: 0x It is Hex with a-f as Hex digits 0X It is Hex with A-F as Hex digits • Otherwise, if constant begins with 0 It is Octal • Otherwise, it is decimal 23 Decimal 23

  3. Base for Character Constants • Designating the base for a character constant • If constant begins with either: ‘\x It is Hex with a-f as Hex digits ‘\X It is Hex with A-F as Hex digits • Otherwise, if constant begins with ‘\0 It is Octal • Otherwise, it is the ASCII code for a character ‘a’ ASCII code for lower case character a

  4. Signed/Unsigned Constants • Constants are Signed by default! • Important when converted to 32 bits of significance • When sign bit (MSB) is equal to 0: 0x55 as a 32 bit quantity 0000 … 01010101 ‘\x55’ as a 32 bit quantity 0000 … 01010101 When sign bit (MSB) is equal to 1: 0xaa as a 32 bit quantity 0000 … 10101010 ‘\xaa’ as a 32 bit quantity 1111 … 10101010

  5. What we’d expect to see! • C Source Code int i = 0x55; /* int set = to int constant */ int j = '\x55'; /* int set = to char constant */ printf("i= %x, j= %x\n", i, j); • Executes and prints as i= 55, j= 55

  6. What Happened? • C Source code int i = 0xaa; /* int set = to int constant */ int j = '\xaa'; /* int set = to char constant */ printf("i= %x, j= %x\n", i, j); • Executes and prints: i= aa, j= ffffffaa

  7. Signed (Default) Behavior • Careful mixing modes when initializing variables! int i = 0xaa; (= 000000aa) probably as intended int j = ‘\xaa’; (= ff ff ff aa) sign bit extension! Sign Bit Extension char 0 int 00000000 00000000 00000000 0 Sign Bit Extension char 1 int 11111111 11111111 11111111 1

  8. Signed (Default) Behavior • Careful mixing modes when initializing variables! c = 0x23456789; (= ‘\x89’) truncates MSBs + int 00100011 01000101 01100111 10001001 - char 10001001 NOTE: char value is negative even though int was positive

  9. Unsigned Behavior • C Source Code unsigned int i = 0xaa; unsigned int j = '\xaa'; printf("i= %x, j= %x\n", i, j); • Executes and prints as: i= aa, j= aa

  10. One’s ComplementTwo’s Complement • One’s Complement Character Arithmetic ~ is the one’s complement operator ~ flips the value of each bit in the data All zeroes become one, All ones become zero ~’\xaa’ == ‘\x55’ ~10101010 == 01010101

  11. One’s ComplementTwo’s Complement • Two’s Complement Character Arithmetic - is the two’s complement operator • flips the value of each bit in the data and adds 1 It creates the negative of the data value - ‘\x55’ == ‘\xab’ - 01010101 == 10101011

  12. Two Special Case Values • char 0 (or zero of any length) -00000000 = 11111111 + 1 = 00000000 • char -27 (or -2n-1 for any length = n) -10000000 = 01111111 + 1 = 10000000

  13. Bit Manipulation • Bitwise Operators: ~ unary not & and | or ^ xor (exclusive or) << leftshift >> rightshift • Operate on pair of bits of the input data values!

  14. Binary Logic Tables AND OR & 0 1 | 0 1 NOT 0 0 0 0 0 1 ~ 1 0 1 1 1 1 0 1 XOR ADD 1 0 ^ 0 1 + 0 1 0 0 1 0 0 1 0 Carry 1 1 1 0 1 1

  15. Bit Manipulation unsigned char n = ‘\xa6’; n 10100110 ~n 01011001 (1s complement: flip bits) n | ‘\x65’ 10100110 turn on bit in result if | 01100101 on in either operand 11100111

  16. Bit Manipulations n & ‘\x65’ 10100110 turn on bit in result if & 01100101 on in both operands 00100100 n ^ ‘\x65’ 10100110 turn on bit in result if ^ 01100101 on in exactly 1 operand 11000011

  17. Bit Manipulations (Left Shift) char n = ‘\x18’; 00011000 n << 1 00110000 shift 1 to left (like * 2) n << 2 01100000 shift 2 to left (like * 4) (MSBs disappear off left end) n can be either signed or unsigned – no difference Significance can be lost if data shifted too far left Left Shift 0

  18. Bit Manipulations (Right Shift) unsigned char n = ‘\x18’; 00011000 n >> 1 00001100 shift 1 to right (like / 2) n >> 2 00000110 shift 2 to right (like / 4) (LSBs disappear off right end) Integer truncation of divide by 2n just like with / Unsigned Right Shift 0

  19. Bit Manipulations (Right Shift) signed char n = ‘\xa5’; 10100101 n >>1 11010010 (bring in a 1 from the left) n >>2 11101001 (bring in two 1’s from left) The rule is to bring in a copy of the sign bit! • Bringing in extra copies of the sign bit from the left end is another form of "sign extension" Signed Right Shift

  20. Bit Manipulations (Right Shift) • For a signed variable, negative value shifted right by 2 is still a negative value • ‘\xa5’ == 10100101 • ‘\xa5’ >> 2 == 11101001 == ‘\xe9’ • Same result as divide by 4 (22 = 4)

  21. Bit Manipulations (Right Shift) • When right shifting a signed value: Different rounding results for positive vs negative Result stays negative – does not become positive 1>>1 = 0 Rounds down -(1>>1) = 0 Rounds down then applies minus (-1)>>1 = -1 Applies minus then rounds down -1>>1 = -1 Applies minus then rounds down • Rounding behavior not exactly same as /2

  22. Converting Decimal to Binary • From 11710 get binary by halving the number successively (rounding down) and writing down the digit 1 when have an odd result, 0 when result is even (consider number itself the first result): 117 1 LSB 58 0 29 1 14 0 7 1 3 1 1 1 0 0 MSB • Read UP and add any leading 0’s: 01110101 = ‘\x75’

  23. Unsigned Binary to Decimal • Treat each bit position n that contains a one as adding 2n to the value. Ignore zeros because they can’t add anything to the value. Bit 0 LSB 1 1 (= 20) Bit 1 0 0 Bit 2 1 4 (= 22) Bit 3 1 8 (= 23) Bit 4 0 0 Bit 5 1 32 (= 25) Bit 6 0 0 Bit 7 MSB 1 128 (= 27) Total 173

  24. Signed Binary to Decimal • Treat each bit position n that contains a one as adding 2n to the value except treat the MSB as subtracting 2n. (Ignore zeros as before) Bit 0 LSB 1 1 (= 20) Bit 1 0 0 Bit 2 1 4 (= 22) Bit 3 1 8 (= 23) Bit 4 0 0 Bit 5 1 32 (= 25) Bit 6 0 0 Bit 7 MSB 1 -128 (= -27) Total -83

  25. Converting Hex to Decimal • Treat each digit n as adding 16n to the value. Ignore zeros because they can’t add anything to the value. Digit 0 LSBs 0 0 Digit 1 2 32 (= 2 * 161) Digit 2 b 2816 (= 11 * 162) Digit 3 0 0 Digit 4 0 0 Digit 5 1 1048576 (= 1 * 165) Digit 6 0 0 Digit 7 MSBs 0 0 Total 1051424

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