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If both input bits are 1, only then AND operation will generate 1 as an output bit. However, the content of the source operand remains unchanged. It ANDs each bit of source operand with the destination operand and stores the result back into the destination operand. Both source and destination operands cannot be a memory location. The source can be an immediate, register, or a memory location and the destination can be either a register or a memory location. The AND instruction perform logical AND operation between two operands. 8086 Arithemtic Right Shift Instruction.Right Shift Instruction Assembly Example 1.It might take some time, but if you want to practice, I highly recommend you to give it a try. The microcontroller or microprocessor can understand only the binary language like 0’s or 1’s therefore the assembler convert the assembly language to binary language and store it the memory to perform the tasks. The assembly programming language is a low-level language which is developed by using mnemonics. You just need to increase the number of elif statements to cover all the possibilities. Assembly Level Programming 8086 Assembly Level Programming 8086. The logic is "simple": Scissors cuts paper, paper covers rock, rock crushes lizard, lizard poisons Spock, Spock smashes scissors, scissors decapitates lizard, lizard eats paper, paper disproves Spock, Spock vaporizes rock, and as it always has, rock crushes scissors. You could add other variations of the game like Rock, Paper, Scissors, Lizard, Spock. Try to change the messages in the print() functions, change the number of possible turns. If you want to truly understand what is going on in this code, the best thing you can is to modify it and see what happens. > Computer wins! > You scored: 1 point(s) <<< > You win! > You scored: 2 point(s) > rock, paper, scissors: rock Print(f'> You scored: point(s) > rock, paper, scissors: rock The last if/else block make a simple test to check which one scored the most points and prints the winner. The number of either player_wins or computer_wins is increased by 1 each turn, except when there is a tie.Īt the end of the loop, another if statement will check at each turn if either player reached the necessary points to win and breaks the loop if this is true.
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The series of if and elif statements will test all the possible combinations of the player’s option against the computer’s option.

Inside there is another while loop to get the player input (rock, paper, or scissors), and this loop breaks only if the player typed a valid option contained in the list options.Ĭomputer = random.choice(options) uses the random module and the choice() function to randomly choose an option from the list options. The while loop is infinite and breaks only when either player gets the minimum amount of wins needed (2 or 3 depending on your choice of the number of turns). Player_wins and computer_wins are initialized with 0 and are the counters of how many victories the player and the computer have respectively. The same logic applies when you choose 3 turns. Notice that 5/2 is 2.5 and we can’t win half a game, so we use int() to round down to 2 and add 1, giving us 3 as the number of wins needed out of 5 turns. The expression necessary_wins = int(turns/2) + 1 gives the minimum amount of victories to win the game. If the input is valid and either 3 or 5, the program moves on. If the input is valid, but not 3 or 5, the loop will restart using continue so the player can the number of turns again. The input() function is inside a while loop that only breaks if you pick a valid integer, otherwise, it will catch the error with the try/except block and print "Invalid choice", restarting the while loop so the player can choose the number of turns again. We initialize the game with a list of the options: rock, paper, and scissors.



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Here are the things you need to know to understand the code of this game: As a working environment is chosen shortly presented.
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Last week we learned how to make a Guessing Game.
