Computer Science, asked by janhavidubey1, 2 months ago

1) int x=5;
int y=x++;

2) int p=0;
p--;
q=(p++)+p;

3) int a=-10, b=++a;
b=++a;
a=++b + --b;

4) int x=1,y=2,z=++x + ++y;
x + =y;
z += y--;

5) double x1 = 1.1;

double x2 = --x1;

x2 = x1-- + x1-- + x1;


6) int m = 10, n = 20, p = 21;

n--;

m += n;

boolean res = (m >= p) || (n <= p) && !(m==n);



Plz find the output of thr follwing sums with a proper method​

Answers

Answered by BrainlyProgrammer
10

Question 1:

int x=5;

int y=x++;

Answer 1:

Given, x=5

therefore, y= x++

Note : x++ is postfix increment (first calculate then increases the value)

so, y= 5++

y=5. (Now x will become 6)

________

Question 2:

int p=0;

p--;

q=(p++)+p;

Answer 2:

  • Initially, p=0
  • p--; here p remain 0, it will not decrease
  • q=(p++)+p (Now p changes to -1)
  • q= (-1++)+p. (now p changes to 0 (p++))
  • q=-1+0
  • q=-1

________

Question 3:

int a=-10, b=++a;

b=++a;

a=++b + --b;

Answer 3:

  • Initially, a=-10
  • b=++a. (a will immediately change to -9)
  • b=-9
  • b=++a (again a will change to -8)
  • b=-8
  • a= ++b + --b
  • Here b will first change to -7
  • a= -7 + --b
  • Now b will change to -8
  • a= -7-8 = -15
  • a=-15

_____

Question 4:

int x=1,y=2,z=++x + ++y ;

x + =y;

z += y--;

Answer 4:

  • Initially, x=1,y=2

z=++x + ++y

z= 2 + 3

z= 5. (Now x=2,y=3)

x+=y

x=x+y

x= 2+3= 5

z += y--;

z= z+ y--

z= 5+ 3 Now y will become 2

z= 8

_______

Question 5:

double x1 = 1.1;

double x2 = --x1;

x2 = x1-- + x1-- + x1;

Answer 5:

  • x1= 1.1
  • x2= --x1 = --1.1 = 1.0

  • x2 = x1-- + x1-- + x1
  • x2= 1.0 +0.9 + 0.8
  • x2= 2.7

_________

Question 6:

int m = 10, n = 20, p = 21;

n--;

m += n;

boolean res = (m >= p) || (n <= p) && !(m==n);

Answer 6:

  • m=10
  • n=20
  • p=21
  • n--;

n will become 19

  • m+=n
  • m=m+n
  • m= 10 +19
  • m= 29

boolean res = (m >= p) || (n <= p) && !(m==n) ;

Note: Boolean variable returns the answer in True/False

boolean res = (m >= p) || (n <= p) && !(m==n)

Order of Logical operator is NAO 1.not 2.and 3.or

Therefore,

(m >= p) || (n <= p) && !(m==n)

(29>=21 ) || (19<=21) && !(m==n)

(29>=21 ) || (19<=21) && !(F)

(29>=21 ) || T && T

T || T && T

T|| T

T

Therefore, res will return True.

Note: F= False, T=True

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