For each of the following expressions add a pair of parenthe

For each of the following expressions add a pair of parenthesis for each operator to indicate the order of precedence. Then evaluate the expression in a series of steps, one line for each operation. The first expression is done for you as an example:

Solution

(a) 5 & 6 && 7 & 8

((5&6) && (7&8))           ------ 5 =101,6= 110 ,110&110 = 100 =4 , 7=111,8=1000 , 111&1000=0-----& bitwise operator

((4) && (0))

(4 && 0)   ----- && relational operator , 4 = true,0=false

0

(b) 7>6 >>2 <<3 <4

((7>6)>>2)<<(3<4))

((1)>>2)<<(1))

((1>>2)<<1)

(0<<1)

1

(c)

1<<2<7>6>>2

((1<<(((2<7)>6))>>2)

((1<<(1>6))>>2)

((1<<0)>>2)

(0>>2)

0

(d) 4>2<<1 == 5-3

(((4>2)<<1) ==( 5-3))

((1<<1)==(2))

(2 == 2)

1

(e) 12&10|5^2+1

((12&10)|((5^2)+1))

((8)|(7+1))    ^--exclusive or 101^010 = 111=7

(8|8)

8

(f) 3+9>>7-2*2

((3+9)>>(7-(2*2)))

((12)>>(7-(4))

(12>>3)

1

(g) 7-9/2*2%3

(7-((9/2)*(2%3)))

(7-(4*(2))

(7-(8))

(7-8)

-1

(h) !0 | 6 & 3^6

(((!0) | 6 )& (3^6))

((1)|6)&(5))

((1|6)&(5))

(7&5)

4

i) 9%7<<5-3|1

((9%7)<<(5-(3|1)))

((2)<<(5-(3))

(2<<2)

8

(j) 3!=2&&4<=5^4

((3!=2)&&(4<=(5^4)))

(1&&(4<=(1)))

(1&&(4<=1))

(1&&(0))

(1&&0)

0

(k) 7-2==5?3:2+2

((((7-2)==5)?3:2)+2)

(((5 ==5)?3:2)+2)

((3)+2)

5