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We can also use the logical operators to numbers directly and perform simple bit manipulation . The operators are
& Bitwise AND | Bitwise OR ^ Bitwise exclusive or ~ Bitwise one's complement i.e. NOT << Left shift >> Right shift
Example A 8 bit number represents a coded function, bit 3-4 describes the operation to be performed on the number in bits 0- 2 and 5-7 the function i.e. 00 Add 01 Subtract 10 Divide 11 Multiply Write a Program to extract the number and operation
Answer We need to extract the bits 3-4, the answer is bit operators Assume number is in the variable A i.e. 11011011 Mask off the first number i.e. bit 0-2 num1 = a & 0x07; 11011011 and 00000111 00000011 Mask off the second number i.e. bit 5 -7 num2 = a & 0xe0; 11011011 and 11100000 11000000 We need to shift num2 down by 5 places i.e. num2 = num2 >>5; 00000110
This could be done in one instruction i.e. num2 = (a & 0xe0) >> 5; Mask off the operation bits i.e. bit 4-5 operation = (a & 0x18) >> 3; We can then use the switch statement to select each operation switch(operation) { case 0: total = num1+num2; break; case 1: total = num1-num2; break; case 2: total = num1/num2; break; case 3: total = num1*num2; break; } Hence the entire program is #include void main() { char prompt; /*Author : Mr James Mc Carren Company: Staffordshire University Date: 26th August 2012 Version 1.0 Function : To show bit manipulation Modifications: none*/ int num1,num2,operation,total,a; printf("Please enter in the number\n\r"); scanf("%x",&a); num1 = a & 0x07; num2 = (a & 0xe0) >> 5 ; operation = (a & 0x18) >> 3; switch(operation) { case 0: total = num1+num2; break; case 1: total = num1-num2; break; case 2: total = num1/num2; break; case 3: total = num1*num2; break; } printf("The total is %d\n\r",total); printf("Press and key to exit \n\r"); scanf("\n%c",&prompt); }
Which Texts in the text elements of the program helps in changing the displayed names of variables in the parameters statement. Selection text in the elements of the program
Result extends to functions - perceptrons: Thus the dotted lines can be seen as the threshold in perceptrons: whether the weighted sum, S, falls below it, after then the perce
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