1. Arithmetic Calculations
Chapter 7
Arithmetic Calculations
Basic arithmetic operators:
+ addition
- subtraction
* multiplication
/ division
% remainder (or modulus). Same precedence and associativity as * and /
Examples:
17 % 3 = 2 (operands must be integers)
100 / 6 = 16 Division of two integers! (The answer is not or 17!)
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2. More examples with % and /
Chapter 7
More examples with % and /
Given num is an integer. Decide if it is even or odd?
if (num % 2)
printf(“Odd”);
else
printf(“Even”);
Or, we could do the following:
if ((num % 2) == 0)
Given that a and b are integers. Let a=10, b=3.
What is a / b? Answer: 3 (Result of integer division)
What is a % b? Answer: 1 (Remainder of division)
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3. Example 1 Given the following declarations and assignments:
Chapter 7
Example 1
Given the following declarations and assignments:
int a, b, c, d, e;
Let a=10, b=20, c=15, d=8, and e=40
What is result of the following expression?
(a + b / (c – 5)) / ((d + 7) / (e - 37) % 3)
Answer:
(a + b / 10) / ((d + 7) / (e - 37) % 3)
(a + b / 10) / (15 / (e - 37) % 3)
(a + b / 10) / (15 / 3 % 3)
(a + 2) / (15 / 3 % 3)
12 / (15 / 3 % 3)
12 / (5 % 3)
12 / 2 = 6
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4. Example 2 Given the following declarations and assignments:
Chapter 7
Example 2
Given the following declarations and assignments:
double a, b, c, d, e;
Let a=10.0, b=20.0, c=15.0, d=8.0, and e=40.0
What is result of the following expression?
(a + b / (c – 5.0)) / ((d + 7.0) / (e – 37.0) / 3.0)
Answer:
(a + b / 10.0) / ((d + 7.0) / (e – 37.0) / 3.0)
(a + b / 10.0) / (15.0 / (e – 37.0) / 3.0)
(a + b / 10.0) / (15.0 / 3.0 / 3.0)
(a + 2.0) / (15.0 / 3.0 / 3.0)
12.0 / (15.0 / 3.0 / 3.0)
12.0 / (5.0 / 3.0)
12.0 / = 7.2
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5. Example 3 Given the following algebraic expression:
Chapter 7
Example 3
Given the following algebraic expression:
Write the corresponding C expression:
(a*a*a + b*b*b) / (c*c – d*d)
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6. Increment and Decrement Operators
Chapter 7
Increment and Decrement Operators
Increment and decrement are unary operators
In C, for integer variables, such as count, we can write:
(Post-increment): count++ instead of count=count+1
(Pre-increment): ++count instead of count=count+1
(Post-decrement): count-- instead of count=count-1
(Pre-decrement): --count instead of count=count-1
Example:
int count=3
printf(“%d”, count++); /* prints 3! */
printf(“%d”, count); /* prints 4 */
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7. Comparison of Prefix and Postfix Increments
Chapter 7
Comparison of Prefix and Postfix Increments
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8. Compound Assignment Operators
Chapter 7
Compound Assignment Operators
Compound assignment operators:
+= assignment sum
-= assignment difference
*= assignment product
/= assignment division
%= assignment remainder
Examples:
a += b; /* a = a + b; */
a *= b; /* a = a * b; */
Suppose a=10 and b=12. Then,
a *= ( b %= 7); /* a = 50, b = 5*/
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9. Chapter 7
Some Rules
Rules for assigning a type to arithmetic expressions that involve integers and doubles:
If one or more operators in an arithmetic expression are of type double, the result of the expression is also of type double
If all operands in an arithmetic expression are of type integer, the result of the expression is also of type integer
The type of the entire statement and the type of the value stored in the variable to the left of the assignment operator are the same as the type of the variable on the left
Examples:
double first=4.7; int second=27;
second = first + second; /* first + second is 31.7! */
However, second=31 since 31.7 is converted (and truncated) to integer!!
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10. Examples of Mixed Expressions
Chapter 7
Examples of Mixed Expressions
double x;
double b=12.5;
int a=7;
x = a / 3 + b;
What is the value of x?
Evaluate a / 3 = 7 / 3 = 2
Evaluate 2 + b = 14.5
So, x = 14.5
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11. Explicit Type Conversions
Chapter 7
Explicit Type Conversions
Explicit type conversions are done by casting:
The form of a cast operation is (Type) Expression
double first = 4.7;
int second = 27;
(int)(first + second) is 31
first = (int)first + second; /* first is 31.0 */
first = (int)first % second; /* first is 4.0 */
first = second % (int)second; /* first is 3.0 */
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12. Mathematical Library Functions
Chapter 7
Mathematical Library Functions
Declarations are found in <math.h> and <stdlib.h>
ceil(x) ceil(4.2)=5, ceil(-5.7)=-5
floor(x) floor(4.2)=4, floor(-5.7)=-6
abs(x) abs(-12)=12, abs(-12.7)=12
fabs(x) fabs(-12)=12.0, fabs(-12.8)=12.8
sqrt(x) sqrt(2)=
pow(x, y) pow(2, 3)=8
cos(x) cos(60* /180)=0.5
sin(x) sin(30* /180)=0.5
tan(x) tan(45* /180)=1.0
exp(x) exp(2.1)=
log(x) log(2)=
log10(x) log10(1000)=3.0
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13. Example 3 (modified) Given the following algebraic expression:
Chapter 7
Example 3 (modified)
Given the following algebraic expression:
Write the corresponding C expression using pow().
(pow(a,3)+pow(b,3)) / (pow(c,2)–pow(d,2))
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14. Arithmetic Errors and Inaccuracies
Chapter 7
Arithmetic Errors and Inaccuracies
Division by zero (generally results in a run-time error)
Arithmetic overflow
When two numeric values are added or multiplied, result of operation may be in excess of max value that can be represented by the number of bits allocated for the type of target variable
Arithmetic underflow
Arithmetic operation results in a value that is less than the smallest value that can be stored for that data type. The computer stores a value of zero instead.
Representational inaccuracies
Precision limitations of floating-point data types
4.0/3.0= (round-off errors)
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