What does the variable x contain at the end of each of these computations? If th

Question

What does the variable x contain at the end of each of these computations?  If the code is not legal in Java, say so.

int iCount = 19;

double dCount = 4.3;

int iSize = 3;

double dSize = 2.7;

a.       int x = iCount + iSize;

b.      int x =  (int) dCount + iCount;

c.       double x =  (int) dCount - iCount;

d.      int x = (int) dCount  + - dSize;

e.       int x = iCount % iSize;

f.        double x = iCount / iSize;

g.       int x = iSize * iSize * iSize / iCount % iSize;

h.      double x = dSize + iCount - dCount + -dSize;

i.         double x = iSize - iCount + iSize / dCount;

j.         double x = ((iCount - iSize) + iCount) - (dCount * iSize)/dSize;

PLEASE EXPLAIN!!

Explanation / Answer

explanation

here, int iCount = 19;

double dCount = 4.3;

int iSize = 3;

double dSize = 2.7;

Lets go one by one

a. int x = iCount + iSize;

here, iCount = 19, iSize = 3, so 19 + 3 = 22

b. int x = (int) dCount + iCount;

the value of dCount is 4.3, but when the double value is converted to int, 4.3 is changed to 4.

and value of iCount is 19

so 4 + 19 = 23

c. double x = (int) dCount - iCount;

the value of dCount is 4.3, but when the double value is converted to int, 4.3 is changed to 4.

value of iCount is 19

so 4-19 = -15 (integer)

the value when converted into double becomes -15.0

d. int x = (int) dCount + - dSize;

the value of dCount is 4.3, but when the double value is converted to int, 4.3 is changed to 4.

value of 2.7 so

4+-2.7. the return type is of int type as type of x is integer, but if we subtract this, we get decimal value which is of double type. So it ll throw run time error

e. int x = iCount % iSize;

value of iCount is 19

value of iSize is 3

19%3 is 1

f. double x = iCount / iSize;

value of iCount is 19

value of iSize is 3

19/3 = 6.166. but as both the values are of int type and not double, so it ll be 6.

now 6 is converted to double as x is of double type, so the answer is 6.0

g. int x = iSize * iSize * iSize / iCount % iSize;

all three have same precedece in java and left precedence

equation is 3*3*3/19%3

27/19%3

27/19 (is 1.something but as both are of integer type, decimal ll be removed, and it ll be just 1

1%3 =1

h. double x = dSize + iCount - dCount + -dSize;

dSize = 2.7

iCount = 19

dCount = 4.3

2.7 + 19 - 4.3 +-2.7

= 14.7

i. double x = iSize - iCount + iSize / dCount;

Here / sign has the highest preference

iSize/dCount = 3/4.3 = 0.6976

now the equation is iSize - iCount +0.6976

3-19 +0.6976

which is equal to -15.30232558139535

j. double x = ((iCount - iSize) + iCount) - (dCount * iSize)/dSize;

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