Example: What are the final values of the variables listed below after all the s

Question

Example: What are the final values of the variables listed below after all the statements are executed.

#define SQRT(x) x*x

#define P(x, y) (x > y) ? x + 2 : y * 3

#define Q(x, y) x / y

#define NN 100

int a, b, d, f, g, h, n, aa, bb = 90;

double c, e, m;

int main(void)

{

int i = 5, j = -6, k = 9, z = 10, x = 1, y = 20;

a = NN / SQRT(5);

b = i < 2* j – i || i >=j;

c = (NN - 5) / 2;

d = P(i, -2*i);

e = k = --i;

f = j ? 20 : 5 + z;

g = (k == i – j + 5) * i;

h = Q(j + 20, SQRT(i + 1));

m = NN && z + 2 ;

n = z++ - j; x = !NN || k * 3;

y %= (int)(50.0/i);

return(0); }

a = ------- b = -------- c = ------- d = ------- e = -------- f = ------- g = ------- h = -------- m = ------- n = ------- i = -------- j = ------- n = ------- x = -------- y = ------- z = -------- aa = -------- bb = -------

please show step by step

Explanation / Answer

1)

a=NN/SQRT(5)=100/5*5=20*5=100

2)

b=i<2*j-i||i>=j (since i>=j is true as 5>=-6 and returns 1 so b=1)

3)

c=(NN-5)/2=(100-5)/2=95/2=47

4)

d=P(i,-2*i)=P(5,-10)=(5>-10) and it returns true so x+2 is evaluated and 5+2=7 is stored in d

5)

e=k=--i=--5=4 so 4 is assigned to k and e and i value becomes 4

6)

f=j?20:5+z=-6 is not equal to 0 and returns true so 20 is stored in f

7)

g=(k==i-j+5)*i=(4==4-(-6)+5)*4=(4==15)*4=0*4=0

8)

h=Q(j+20,SQRT(i+1))=Q(-6+20,SQRT(4+1))=-6+20/4+1*4+1=-6+5+4+1=4

9)

m=NN&&z+2=100&&10+2=100&&12=1

10)

n=z++-j=10-(-6)=10+6=16 and z value becomes 11

11)

x=!NN||k*3=!100||k*3=0||4*3=0||12=1

12)

y%=(int)(50.0/i)=(int)(50.0/4)=12

y=y%12=20%12=8

13)

z=11

aa=0 and bb=90

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