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What is procedure given in class to convert binary to hex?
Split the binary value into groups of 4 bits (half-a-byte). Translate each nybble to its corresponding hex digit. Use the following table to translate each nybble: 0000 = 0x0 0001 = 0x1 0010 = 0x2 0011 = 0x3 0100 = 0x4 0101 = 0x5 0110 = 0x6 0111 = 0x7 1000 = 0x8 1001 = 0x9 1010 = 0xA 1011 = 0xB 1100 = 0xC 1101 = 0xD 1110 = 0xE 1111 = 0xF
How do you convert exe to binary?
An exe is machine code and machine code is written entirely in binary. No conversion is necessary. A hex-editor is the simplest way to view the binary code. The code will be shown in hexadecimal rather than binary, however this actually makes it much easier for humans to interpret the binary code because the conversion from hex to binary is so simple. Each hex digit represents a unique 4-bit binary pattern: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE= 1110 0xF = 1111 Thus the hex value 0x9A translates directly to the 8-bit binary value 10011010. That is, 8 binary digits reduce to just 2 hex digits and therefore makes it much easier for humans to interpret the binary value.
How do you convert the hexadecimal number ff to binary?
Any base that is itself a power of 2 is easily converted to and from binary. With base 4, each digit represents 2 bits. With base 8 (octal), each digit represents 3 bits. And with base 16 (hexadecimal), each digit represents 4 bits. Thus two hexadecimal digits represent an 8-bit binary value. This is convenient because we typically refer to a unit of computer memory as an 8-bit byte, thus every byte value can be represented using just 2 hex digits. If we had a system with a 9-bit byte we'd use 3 octal digits instead. A 24-bit value can either be represented using 6 hex digits or 8 octal digits. To convert a hexadecimal value to binary, we simply consult the following table (note that 0x is the conventional prefix for a hexadecimal value): hex = binary 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE = 1110 0xF = 1111 Here, hexadecimal digit 0xF has the binary value 1111, thus 0xFF would be 11111111. Note that the bit patterns are in the same order as the hexadecimal digits. Thus 0x0F becomes 00001111 and 0xF0 becomes 11110000. Knowing this, we can easily convert binary values into hexadecimal, we simply divide the binary value into groups of 4 bits and convert each group to the corresponding hex digit. Thus 101101001100 becomes B4C (1011=B, 0100=4 and 1100=C). If there aren't enough bits, we simply pad the first group with leading zeroes. We can use a similar technique to convert between octal and binary, we simply divide the bits into groups of 3: octal = binary 00 = 000 01 = 001 02 = 010 03 = 011 04 = 100 05 = 101 06 = 110 07 = 111 Note that a leading 0 is the conventional prefix for octal values. Thus binary value 100010 would be written 042 in octal to avoid confusion with 42 decimal.
How do you convert binary numbers to hexadecimal notation?
Each hexadecimal digit represent four binary bits. Using the table... 0 = 0000 1 = 0001 2 = 0010 3 = 0011 4 = 0100 5 = 0101 6 = 0110 7 = 0111 8 = 1000 9 = 1001 A = 1010 B = 1011 C = 1100 D = 1101 E = 1110 F = 1110 ... replace each hexadecimal digit with its correspnding binary digits. As an example, 37AB16 is 00110111101010112.
What are the binary and decimal values of the ASCII letter K?
ASCII symbol 'K' has the binary code 01001011. Reading right-to-left, each bit has the following value: bit 0 = 1 = 2^0 * 1 = 1 * 1 = 1 bit 1 = 1 = 2^1 * 1 = 2 * 1 = 2 bit 2 = 0 = 2^2 * 0 = 4 * 0 = 0 bit 3 = 1 = 2^4 * 1 = 8 * 1 = 8 bit 4 = 0 = 2^8 * 0 = 16 * 0 = 0 bit 5 = 0 = 2^16 * 0 = 32 * 0 = 0 bit 6 = 1 = 2^32 * 1 = 64 * 1 = 64 bit 7 = 0 = 2^64 * 0 = 128 * 0 = 0 1 + 2 + 0 + 8 + 0 + 0 + 64 + 0 = 75 Therefore 01001011 is the binary encoding for the decimal value 75. Thus ASCII symbol 'K' has the decimal value 75 Working the other way, we divide the decimal value by 2. The remainder can only be 0 or 1 and we write this remainder down. We continue dividing by 2, writing a 0 or a 1, until the decimal value is 0. We then pad any remaining bits with 0s. Like so: 75 / 2 = 37 r 1 37 / 2 = 18 r 1 18 / 2 = 9 r 0 9 / 2 = 4 r 1 4 / 2 = 2 r 0 2 / 2 = 1 r 0 1 / 2 = 0 r 1 Reading from the bottom up, the remainders are 1001011, which we pad with a leading zero to create the binary value 01001011. Note that we typically use leading zeroes when the number of bits is not an exact multiple of 8, thus creating a binary value of one or more 8-bit bytes. Another way to do the conversion is to use hexadecimal notation. Hexadecimal makes it much easier for humans to interpret binary values because there are 16 digits (0-9 then A-F) and each digit represents a unique 4-bit binary pattern: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE = 1110 0xF = 1111 From this we can see that 01001011 is formed from 4-bit patterns 0100 and 1011, which from the table above equates to 0x4 and 0xB respectively. Thus binary 01001011 is equivalent to 0x4B in hexadecimal. 0xB has the decimal value 11 while 0x4 has the decimal value 4. However, as with decimal, the position of the digit is significant. Instead of increasing powers of 10 we are dealing with increasing powers of 16, thus 0x4B really means 11 x 16^0 + 4 x 16^1, which reduces to 11 x 1 + 4 x 16 or simply 11 + 64 = 75. Working the other way, we divide the decimal value by 16 and use the remainder to determine the hex digit: 75 / 16 = 4 r 11 4 / 16 = 0 r 4 11 in hexadecimal is B, thus 75 in hexadecimal is 0x4B. From the table above, we see that 0x4 is 0100 and 0xB is 1011, thus 75 = 0x4B = 01001011. From this we can see that converting decimal to any base is simply a matter of repeatedly dividing by the base and taking the remainder. If the base is greater than 10, we convert that remainder to its corresponding digit. Converting the other way (to decimal) we need to know the positional value of each significant digit and that is always an increasing power of the base, starting from 0 in the least-significant position.
0x5 0x1 0x2 0x3 0x4 0x6 0x7 0x8?
0x5 = 5 0x1 = 1 0x2 = 2 0x3 = 3 0x4 = 4 0x5 = 5 0x6 = 6 0x7 = 7 0x8 = 8 Is there something specific you would like to know about these values or a question you would like to ask related to them?
What is the degrees difference between 32 Fahrenheit and 8 Celsius?
32 degrees in Fahrenheit =0 Celsius 32-32=0x5/9 -0 C Celsius divide by 5/9 and +32. 8C / 0.55 = 14.5 +32 =46.5 degrees Fahrenheit
What is procedure given in class to convert binary to hex?
Split the binary value into groups of 4 bits (half-a-byte). Translate each nybble to its corresponding hex digit. Use the following table to translate each nybble: 0000 = 0x0 0001 = 0x1 0010 = 0x2 0011 = 0x3 0100 = 0x4 0101 = 0x5 0110 = 0x6 0111 = 0x7 1000 = 0x8 1001 = 0x9 1010 = 0xA 1011 = 0xB 1100 = 0xC 1101 = 0xD 1110 = 0xE 1111 = 0xF
What if a ball is dropped off the roof of a tall building if the ball reaches the ground in 3 seconds how tall is the building?
The vertical distance covered by a free falling object is given by the formula: S= ut+0.5at^2, where S is the distance covered (height of the building), u is the initial velocity (for this case it is 0 since the body is released from rest), t is the time taken for the object to hit the ground (it has taken 5 seconds) and a is the acceleration due to gravitational pull (assumed to be 9.8ms^2). Therefore, the height of the building is given by (0x5 +0.5x9.8 x25) which is 122.5m.
How do you convert exe to binary?
An exe is machine code and machine code is written entirely in binary. No conversion is necessary. A hex-editor is the simplest way to view the binary code. The code will be shown in hexadecimal rather than binary, however this actually makes it much easier for humans to interpret the binary code because the conversion from hex to binary is so simple. Each hex digit represents a unique 4-bit binary pattern: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE= 1110 0xF = 1111 Thus the hex value 0x9A translates directly to the 8-bit binary value 10011010. That is, 8 binary digits reduce to just 2 hex digits and therefore makes it much easier for humans to interpret the binary value.
How do you convert hex decimal number in to octal?
First convert the hexadecimal to binary. Every hexadecimal digit corresponds to 4 binary digits: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE = 1110 0xF = 1111 Next, after you have converted each hexadecimal digit into binary digits, convert the binary digits to Octal. Each octal digit corresponds to 3 binary digits: 00 = 000 01 = 001 02 = 010 03 = 011 04 = 100 05 = 101 06 = 110 07 = 111 Make sure that you don't accidentally mix up the digits. If you are computer scientist, then you may also have to worry about endianess (whether the most significant digit comes first or the least significant digit comes first).
Why cant you divide by zero?
Division tells you how many times you can take the divisor away from the dividend to get zero.For example, 18 ÷ 6 = 3 tells you you can take 6 away from 18 three times to get zero:18 - 6 = 1212 - 6 = 66 - 6 = 0When you subtract zero, the dividend doesn't change, so no matter how many times you take zero away, you will never reach zero:18 - 0 = 1818 - 0 = 1818 - 0 = 18...18 - 0 = 18...Thus you cannot divide by zero.
How do you convert the hexadecimal number ff to binary?
Any base that is itself a power of 2 is easily converted to and from binary. With base 4, each digit represents 2 bits. With base 8 (octal), each digit represents 3 bits. And with base 16 (hexadecimal), each digit represents 4 bits. Thus two hexadecimal digits represent an 8-bit binary value. This is convenient because we typically refer to a unit of computer memory as an 8-bit byte, thus every byte value can be represented using just 2 hex digits. If we had a system with a 9-bit byte we'd use 3 octal digits instead. A 24-bit value can either be represented using 6 hex digits or 8 octal digits. To convert a hexadecimal value to binary, we simply consult the following table (note that 0x is the conventional prefix for a hexadecimal value): hex = binary 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE = 1110 0xF = 1111 Here, hexadecimal digit 0xF has the binary value 1111, thus 0xFF would be 11111111. Note that the bit patterns are in the same order as the hexadecimal digits. Thus 0x0F becomes 00001111 and 0xF0 becomes 11110000. Knowing this, we can easily convert binary values into hexadecimal, we simply divide the binary value into groups of 4 bits and convert each group to the corresponding hex digit. Thus 101101001100 becomes B4C (1011=B, 0100=4 and 1100=C). If there aren't enough bits, we simply pad the first group with leading zeroes. We can use a similar technique to convert between octal and binary, we simply divide the bits into groups of 3: octal = binary 00 = 000 01 = 001 02 = 010 03 = 011 04 = 100 05 = 101 06 = 110 07 = 111 Note that a leading 0 is the conventional prefix for octal values. Thus binary value 100010 would be written 042 in octal to avoid confusion with 42 decimal.
How do you convert binary numbers to hexadecimal notation?
Each hexadecimal digit represent four binary bits. Using the table... 0 = 0000 1 = 0001 2 = 0010 3 = 0011 4 = 0100 5 = 0101 6 = 0110 7 = 0111 8 = 1000 9 = 1001 A = 1010 B = 1011 C = 1100 D = 1101 E = 1110 F = 1110 ... replace each hexadecimal digit with its correspnding binary digits. As an example, 37AB16 is 00110111101010112.
What are the binary and decimal values of the ASCII letter K?
ASCII symbol 'K' has the binary code 01001011. Reading right-to-left, each bit has the following value: bit 0 = 1 = 2^0 * 1 = 1 * 1 = 1 bit 1 = 1 = 2^1 * 1 = 2 * 1 = 2 bit 2 = 0 = 2^2 * 0 = 4 * 0 = 0 bit 3 = 1 = 2^4 * 1 = 8 * 1 = 8 bit 4 = 0 = 2^8 * 0 = 16 * 0 = 0 bit 5 = 0 = 2^16 * 0 = 32 * 0 = 0 bit 6 = 1 = 2^32 * 1 = 64 * 1 = 64 bit 7 = 0 = 2^64 * 0 = 128 * 0 = 0 1 + 2 + 0 + 8 + 0 + 0 + 64 + 0 = 75 Therefore 01001011 is the binary encoding for the decimal value 75. Thus ASCII symbol 'K' has the decimal value 75 Working the other way, we divide the decimal value by 2. The remainder can only be 0 or 1 and we write this remainder down. We continue dividing by 2, writing a 0 or a 1, until the decimal value is 0. We then pad any remaining bits with 0s. Like so: 75 / 2 = 37 r 1 37 / 2 = 18 r 1 18 / 2 = 9 r 0 9 / 2 = 4 r 1 4 / 2 = 2 r 0 2 / 2 = 1 r 0 1 / 2 = 0 r 1 Reading from the bottom up, the remainders are 1001011, which we pad with a leading zero to create the binary value 01001011. Note that we typically use leading zeroes when the number of bits is not an exact multiple of 8, thus creating a binary value of one or more 8-bit bytes. Another way to do the conversion is to use hexadecimal notation. Hexadecimal makes it much easier for humans to interpret binary values because there are 16 digits (0-9 then A-F) and each digit represents a unique 4-bit binary pattern: 0x0 = 0000 0x1 = 0001 0x2 = 0010 0x3 = 0011 0x4 = 0100 0x5 = 0101 0x6 = 0110 0x7 = 0111 0x8 = 1000 0x9 = 1001 0xA = 1010 0xB = 1011 0xC = 1100 0xD = 1101 0xE = 1110 0xF = 1111 From this we can see that 01001011 is formed from 4-bit patterns 0100 and 1011, which from the table above equates to 0x4 and 0xB respectively. Thus binary 01001011 is equivalent to 0x4B in hexadecimal. 0xB has the decimal value 11 while 0x4 has the decimal value 4. However, as with decimal, the position of the digit is significant. Instead of increasing powers of 10 we are dealing with increasing powers of 16, thus 0x4B really means 11 x 16^0 + 4 x 16^1, which reduces to 11 x 1 + 4 x 16 or simply 11 + 64 = 75. Working the other way, we divide the decimal value by 16 and use the remainder to determine the hex digit: 75 / 16 = 4 r 11 4 / 16 = 0 r 4 11 in hexadecimal is B, thus 75 in hexadecimal is 0x4B. From the table above, we see that 0x4 is 0100 and 0xB is 1011, thus 75 = 0x4B = 01001011. From this we can see that converting decimal to any base is simply a matter of repeatedly dividing by the base and taking the remainder. If the base is greater than 10, we convert that remainder to its corresponding digit. Converting the other way (to decimal) we need to know the positional value of each significant digit and that is always an increasing power of the base, starting from 0 in the least-significant position.
How do you convert a decimal into a BCD in C plus plus?
The following program shows one way to convert to BCD using a variation of packed BCD 8421 encoding. Each digit is represented by 4 bits in the range 0x0 to 0x9. 0xa denotes a minus symbol (for negative values) while 0xb denotes a decimal point for floating point values. The unused value 0xf is used to pad a BCD to a full byte when the number of symbols is odd. Values 0xc, 0xd and 0xe are ignored, but could be used for other purposes such as when representing exponents or fractions/ratios. Values may be any length, the BCD being represented by a vector of type unsigned char. Since the value of a BCD could easily exceed the range of a long double or a 64-bit integer, conversion to these types is disabled. BCD values are constructed from an input string and converted back to a string for output. Arithmetic operations can be overloaded to handle a BCD (using decimal notation arithmetic), however this has been left as an exercise for the reader. #include<iostream> #include<sstream> #include<vector> #include<exception> using uchar = unsigned char; class BCD_8421 { private: std::vector<uchar> v; static const uchar lo_mask {0x0f}; static const uchar hi_mask {0xf0}; public: BCD_8421 (void); BCD_8421 (const BCD_8421&); BCD_8421 (BCD_8421&&); BCD_8421 (const std::string&); BCD_8421& operator= (const BCD_8421&); BCD_8421& operator= (BCD_8421&&); BCD_8421& operator= (const std::string&); static bool is_valid (const std::string& s); operator std::string (void) const; friend std::ostream& operator<< (std::ostream&, const BCD_8421&); }; BCD_8421::BCD_8421 (void): v(1, 0) {} BCD_8421::BCD_8421 (const BCD_8421& bcd): v (bcd.v) {} BCD_8421::BCD_8421 (BCD_8421&& bcd): v (std::move(bcd.v)) {} BCD_8421::BCD_8421 (const std::string& s): v{} { *this = s; } BCD_8421& BCD_8421::operator= (const BCD_8421& bcd) { v=bcd.v; return *this; } BCD_8421& BCD_8421::operator= (BCD_8421&& bcd) { v=std::move(bcd.v); return *this; } BCD_8421& BCD_8421::operator= (const std::string& s) { v.clear(); uchar c {}; bool msb {false}; bool minus {false}; bool point {false}; bool error {false}; bool digit {false}; for (auto i=s.begin(); i!=s.end(); ++i) { if (!msb) c = hi_mask; uchar x; switch (*i) { case ('0'): if (!digit) continue; case ('1'): case ('2'): case ('3'): case ('4'): case ('5'): case ('6'): case ('7'): case ('8'): case ('9'): digit = true; x = *i - '0'; break; case ('.'): if (!point) { point = !point; x = 10; break; } else error = !error; case ('-'): if (!minus && i==s.begin()) { minus = !minus; x = 11; break; } default: error = !error; } if (error) { v.clear(); v.push_back (hi_mask); std::stringstream ss; ss << "Error: invalid argument in BCD_8421.operator= ("" << s << "")"; throw std::invalid_argument (ss.str()); } if (msb) v.push_back ((c & lo_mask) | (x << 4)); else c |= x; msb = !msb; } if (msb && c) v.push_back (c); else if (!digit) { v.clear(); v.push_back (hi_mask); } return *this; } bool BCD_8421::is_valid (const std::string& s) { const std::string valid_chars {"0123456789-."}; if (s.find_first_not_of (valid_chars) != valid_chars.npos) return false; auto f = s.find ('-'); if (f != s.rfind('-') (f && f!=s.npos)) return false; if (s.find ('.') != s.rfind('.')) return false; return true; } BCD_8421::operator std::string (void) const { std::stringstream ss; bool digit {false}; bool point {false}; for (auto i : v) { bool msb {false}; do { uchar c {}; uchar x = msb ? i >> 4 : i & lo_mask; switch (x) { case (0x0): case (0x1): case (0x2): case (0x3): case (0x4): case (0x5): case (0x6): case (0x7): case (0x8): case (0x9): c = '0' + x; digit = true; break; case (0xa): c = '.'; point = true; break; case (0xb): c = '-'; break; default: break; } if (point && !digit) ss << '0'; if (c) ss << c; msb = !msb; } while (msb); } if (ss.str().back()=='.') ss << '0'; return ss.str(); } std::ostream& operator<< (std::ostream& os, const BCD_8421& bcd) { return os << static_cast<std::string> (bcd); } int main() { for (unsigned loop=0; loop<10; ++loop) { std::string input; while (true) { std::cout << "Enter a numeric value: "; std::cin >> input; if (BCD_8421::is_valid (input)) break; std::cerr << "Invalid input.\n"; } try { BCD_8421 a = input; std::cout << "You entered the value: " << a << std::endl; } catch (std::range_error& e) { std::cerr << e.what() << std::endl; } } }
Where can you find an opcode sheet for 8085 microprocessor?
ACI n |CE|*****| 7|Add with Carry Immediate |A=A+n+CY | |ADC r |8F|*****| 4|Add with Carry |A=A+r+CY(21X)| |ADC M |8E|*****| 7|Add with Carry to Memory |A=A+[HL]+CY | |ADD r |87|*****| 4|Add |A=A+r (20X)| |ADD M |86|*****| 7|Add to Memory |A=A+[HL] | |ADI n |C6|*****| 7|Add Immediate |A=A+n | |ANA r |A7|****0| 4|AND Accumulator |A=A&r (24X)| |ANA M |A6|****0| 7|AND Accumulator and Memory|A=A&[HL] | |ANI n |E6|**0*0| 7|AND Immediate |A=A&n | |CALL a |CD|-----|18|Call unconditional |-[SP]=PC,PC=a| |CC a |DC|-----| 9|Call on Carry |If CY=1(18~s)| |CM a |FC|-----| 9|Call on Minus |If S=1 (18~s)| |CMA |2F|-----| 4|Complement Accumulator |A=~A | |CMC |3F|----*| 4|Complement Carry |CY=~CY | |CMP r |BF|*****| 4|Compare |A-r (27X)| |CMP M |BF|*****| 7|Compare with Memory |A-[HL] | |CNC a |D4|-----| 9|Call on No Carry |If CY=0(18~s)| |CNZ a |C4|-----| 9|Call on No Zero |If Z=0 (18~s)| |CP a |F4|-----| 9|Call on Plus |If S=0 (18~s)| |CPE a |EC|-----| 9|Call on Parity Even |If P=1 (18~s)| |CPI n |FE|*****| 7|Compare Immediate |A-n | |CPO a |E4|-----| 9|Call on Parity Odd |If P=0 (18~s)| |CZ a |CC|-----| 9|Call on Zero |If Z=1 (18~s)| |DAA |27|*****| 4|Decimal Adjust Accumulator|A=BCD format | |DAD B |09|----*|10|Double Add BC to HL |HL=HL+BC | |DAD D |19|----*|10|Double Add DE to HL |HL=HL+DE | |DAD H |29|----*|10|Double Add HL to HL |HL=HL+HL | |DAD SP |39|----*|10|Double Add SP to HL |HL=HL+SP | |DCR r |3D|****-| 4|Decrement |r=r-1 (0X5)| |DCR M |35|****-|10|Decrement Memory |[HL]=[HL]-1 | |DCX B |0B|-----| 6|Decrement BC |BC=BC-1 | |DCX D |1B|-----| 6|Decrement DE |DE=DE-1 | |DCX H |2B|-----| 6|Decrement HL |HL=HL-1 | |DCX SP |3B|-----| 6|Decrement Stack Pointer |SP=SP-1 | |DI |F3|-----| 4|Disable Interrupts | | |EI |FB|-----| 4|Enable Interrupts | | |HLT |76|-----| 5|Halt | | |IN p |DB|-----|10|Input |A=[p] | |INR r |3C|****-| 4|Increment |r=r+1 (0X4)| |INR M |3C|****-|10|Increment Memory |[HL]=[HL]+1 | |INX B |03|-----| 6|Increment BC |BC=BC+1 | |INX D |13|-----| 6|Increment DE |DE=DE+1 | |INX H |23|-----| 6|Increment HL |HL=HL+1 | |INX SP |33|-----| 6|Increment Stack Pointer |SP=SP+1 | |JMP a |C3|-----| 7|Jump unconditional |PC=a | |JC a |DA|-----| 7|Jump on Carry |If CY=1(10~s)| |JM a |FA|-----| 7|Jump on Minus |If S=1 (10~s)| |JNC a |D2|-----| 7|Jump on No Carry |If CY=0(10~s)| |JNZ a |C2|-----| 7|Jump on No Zero |If Z=0 (10~s)| |JP a |F2|-----| 7|Jump on Plus |If S=0 (10~s)| |JPE a |EA|-----| 7|Jump on Parity Even |If P=1 (10~s)| |JPO a |E2|-----| 7|Jump on Parity Odd |If P=0 (10~s)| |JZ a |CA|-----| 7|Jump on Zero |If Z=1 (10~s)| |LDA a |3A|-----|13|Load Accumulator direct |A=[a] | |LDAX B |0A|-----| 7|Load Accumulator indirect |A=[BC] | |LDAX D |1A|-----| 7|Load Accumulator indirect |A=[DE] | |LHLD a |2A|-----|16|Load HL Direct |HL=[a] | |LXI B,nn |01|-----|10|Load Immediate BC |BC=nn | |LXI D,nn |11|-----|10|Load Immediate DE |DE=nn | |LXI H,nn |21|-----|10|Load Immediate HL |HL=nn | |LXI SP,nn|31|-----|10|Load Immediate Stack Ptr |SP=nn | |MOV r1,r2|7F|-----| 4|Move register to register |r1=r2 (1XX)| |MOV M,r |77|-----| 7|Move register to Memory |[HL]=r (16X)| |MOV r,M |7E|-----| 7|Move Memory to register |r=[HL] (1X6)| |MVI r,n |3E|-----| 7|Move Immediate |r=n (0X6)| |MVI M,n |36|-----|10|Move Immediate to Memory |[HL]=n | |NOP |00|-----| 4|No Operation | | |ORA r |B7|**0*0| 4|Inclusive OR Accumulator |A=Avr (26X)| |ORA M |B6|**0*0| 7|Inclusive OR Accumulator |A=Av[HL] | |ORI n |F6|**0*0| 7|Inclusive OR Immediate |A=Avn | |OUT p |D3|-----|10|Output |[p]=A | |PCHL |E9|-----| 6|Jump HL indirect |PC=[HL] | |POP B |C1|-----|10|Pop BC |BC=[SP]+ | |POP D |D1|-----|10|Pop DE |DE=[SP]+ | |POP H |E1|-----|10|Pop HL |HL=[SP]+ | |SUB r |97|*****| 4|Subtract |A=A-r (22X)| |SUB M |96|*****| 7|Subtract Memory |A=A-[HL] | |SUI n |D6|*****| 7|Subtract Immediate |A=A-n | |XRA r |AF|**0*0| 4|Exclusive OR Accumulator |A=Axr (25X)| |XRA M |AE|**0*0| 7|Exclusive OR Accumulator |A=Ax[HL] | |XRI n |EE|**0*0| 7|Exclusive OR Immediate |A=Axn | |XTHL |E3|-----|16|Exchange stack Top with HL|[SP]<->HL |SBB r |9F|*****| 4|Subtract with Borrow |A=A-r-CY | |SBB M |9E|*****| 7|Subtract with Borrow |A=A-[HL]-CY | |SBI n |DE|*****| 7|Subtract with Borrow Immed|A=A-n-CY
