1. Lecture 8

2. Outline
Sizeof
Type casting
Constants
Examples

3. sizeof
The sizeof keyword returns the number of bytes of the given expression or type
returns an unsigned integer result
sizeof variable_Identifier;
sizeof (variable_Identifier);
sizeof (Data_Taype);
Example:
int x;
printf("size of x = %d", sizeof x);//4
printf("size of long long = %d", sizeof(long long));//8
printf("size of x = %d", sizeof (x));//4

4. Type Casting What is the casting?
When the type of variable and value are not the same
Example: Assigning double value to integer variable

5. Implicit casting Implicit We don’t say it But we do it
carried out by compiler automatically
char f2= 50e6; /*cast from double to char */
int i= 98.01;/*cast from double to int */
float f = 65.6;
int i = f; //f = (int)f;
char c = i; // c = (char)i;

6. Explicit casting Explicit And we do it
We say it
And we do it
carried out by programmer using casting
int i=(int)98.1;/*cast from double to int */
char c = (char) 90;/* cast from int to char */
int k, i = 7;
float f = 10.14;
char c = 'B';
k = (i + f) % 3; // error
k = (int)(i + f) % 3;

7. Type Casting char c = 'A'; short s = 1; int i; float f; double d;
i = (c s);
(int) (char) (short)
(short)
(int)
d = (c / i) (f * d) (f i);
(char) (int) (float) (double) (float) (int)
int double float
double

8. Casting effects Casting from small types to large types
There is not any problem
No loss of data
int i;
short s;
float f;
double d;
s = 'A'; // s = 65
i = 'B'; // i = 66
f = 4566; // f =
d = 5666; // d =

9. Casting effects (cont’d)
Casting from large types to small types
Data loss is possible
Depends on the values
float double short
f = 65536;
d = 65536;
s = 720; //
// 720
char short
c = (char)
65536;
// c = 0
// s = 0
s = (short) 65536; int i =
int i =
1.22;
1e23;
// i = 1
// i = ???

10. Casting effects (cont’d)
Casting to Boolean
If value is zero  false
If values is not zero  true
bool bool
b2 = 'a', b3 = -9,
b5 =
b4 = 4.5;
//true
//false
0, b6 = false;
b7 = '\0';

11. Constant Variables Constants
Do not want to change the value
Example: pi = 3.14
We can only initialize a constant variable
We MUST initialize the constant variables (why?!)

12. Constant const [Data_Type] Identifier = constant_value;
const p = 3; // const int p = 3;
const p;
p = 3.14; // compile error
const p = 3.14; // p = 3 because default is int
const float p = 3.14;

13. Constant
const int STUDENTS = 38; const long int MAX_GRADE =20; int i; i = MAX_GRADE; STUDENTS = 39;//ERROR

14. Definitions Another tool to define constants
Definition is not variable
We define definition, don’t declare them
Pre-processor replaces them by their values before compiling
#define STUDENTS 38 int main(void){
int i;
i = STUDENTS
STUDENTS = 90; //ERROR! What compiler sees: 38 = 90

15. Constant
#define Identifier constant_value #define PI 3.14 #define ERROR "Disk error " #define ONE 1 #define TWO ONE + ONE

16. #include <stdio.h> // (preprocessor )
#define PI // PI constant (preprocessor )
// calculating area of circle
int main() {
/* variable definition */
float Radius;
float Area = 0;
// get radius of circle form user
printf("Enter Radius :\n");
scanf("%f", &float );
// calculating area of circle
Area = PI * Radius * Radius;
printf(“Area = %f", Area );
system("Pause");
return 0; }

17. Examples Precedence Rules

18. int num = 0, i = 0; for (i = 0; i < 5; i++) { printf("%d", ++num);}
printf("%d", num++);
12345
01234 _ 18

19. b=(a++>0) ? printf("%d\n",a):printf("%d\n",a);
int b,a = 0;
b=(a++>0) ? printf("%d\n",a):printf("%d\n",a);
b = (a>0)?printf("%d\n",++a):printf("%d\n",a);
printf("%d\n" , a) ;
b=++a+(a>0 )?printf("%d\n",a++):printf("%d\n",a);
printf("%d\n" ,a) ; 1 _ 1 2 19

20. int a = 0; int b = 5; b = ( a++ > 0 ) ? ++a : ++b;
printf("%d %d\n" , a, b);
int a = 5;
int b = 13;
a = a * b % b / a;
b = (++a > 0 ) ? a : b++ ;
1 6
1 1 _ 20

21. int j,k,t=1,i=2; for(k=0;k<5;k++){ j = i++ + t; printf("%d",j); }
34567 _ 21

22. int a = 40, b = 30, c; c = !a <= !(c = b); printf("%d", c);
int i =1;
int j = 3*(4+i++);
printf("%d %d\n", i,j); 1
2 15 _ 22

23. int n = 5;int i = 0; while(i++ <= n) printf("%d " , i);
1 2 3 _
1 2 3 23

24. printf("%d %d %d %d", i, j, k, m); printf("%d",1 + 5&4 == 3);
int i=-3, j=2, m, k=0;
m=++i && ++j || ++k;
printf("%d %d %d %d", i, j, k, m);
printf("%d",1 + 5&4 == 3);
m=++i && ++j && k++; _ 24

25. printf("%d %d %d", i, ++i, i++); int i = 0, j = 0; j = i++ + ++i;
printf("%d %d",i,j);
3 3 1
2 2 _ 25

26. int i=1; printf("%d", i++ + ++i); int i=3; printf("%d",++i + ++i);
printf("%d\n", i); 4 10 1
C has the concept of undefined behavior, i.e. some language constructs are syntactically valid but you can't predict the behavior when the code is run. _ 26