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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); }
How many address bits are required to represent a 32 K memory ? Ans. 32K = 25 x 210 = 215, Hence 15 address bits are needed; Only 16 bits can address this.
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