progLT06TypeCasting_Ans

0
int
(double)(j + 1) / i
5.5
double
(double)((j + 1) / i )
5.0
double
2.Write proper C# code for the following formulas. Assuming thatx,y,a,b, andcaredouble,iandjare integers andmandnarefloat, you should first declare the necessary variables before writing the statements.
sum of all digits = digit100 + digit10 + digit1
Lab Exercise
3.Write a program that reads an integer between 0 and 999 and displays the sum of all digits of the input value. For example, if the user enters 932, the sum of all its digits is 14.
or
10 + 9 /(double)i ;
i = 5 +(int)3.4 ;
or
(int) (5 + 3.4);
4.For the given statements below, underline all thevariables/constants/intermediate resultsthat areimplicitlytype casted during execution. You should also indicate to which type it is casted.
int i = 2;
char ch = 'a';
double d = 2.5;
d =i* 0.2f ;//sample answer
(float)
i = ch + 1 ;
(int)
d = d * ( i + 2.5f ) ;
(float)
d = 4 + i ;
d = 0.5 + 34 ;
(double)
a.
double x, a, b, c;
x = Math.Pow(a,2) + 2 * a * b + Math.Pow(c, 2);
b.
double y;
int i, j;
float m, n;
y = (m-n)/(5*(i+j))* Math.Pow(m,2) / Math.Pow(n,3) + i * (m + n);
3.
For each of the statements below, insert explicit type casting where needed to avoid syntax error:
int i = 2;
double d = 2.5;
char ch = 'M';
float f = 2.0f;
Formula to compute sum of all digits:
digit100 = quotient of input / 100
digit10 = quotient of (remainder of input /10) / 10
digit1 = remainder of input / 10
Test your program with the following data:
Input Integer
Sum of All Digits
0
0
999
27
34
7
5
5
Identify the following:
Input:Output:
An integer (input)Sum of all digits
Tutorial Exercise
1.Given the following declarations and initializations, evaluate the expressions below:
int i = 2, j = 10;
double d = 2.5, e = 11.5;
Expression
ch =(char)(ch + 1);//sample answer
f =(float)d * ( i + 0.2f ) ;
or
(float)(d * ( i + 0.2f ));
d = i + f ;// No explicit type casting needed
f =(float)d ;
d = 10 +(double)9 / i ;//expecting d=14.5
Result
ห้องสมุดไป่ตู้Result type
( d + 10 * i ) / i
11.25
double
(int)e * i
22
int
j / i % i
1
int
7 % 4 + (int)(10/d)
7
int
(int)(10.0/d) + 10.0 / (int)d
9.0
double
j % (i * 3) % i