# Yacoub Sabatin Muntaser Abulafi Omar Qaraeen. Introduction

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1 Introduction to Computer Engineering Expressions Yacoub Sabatin Muntaser Abulafi Omar Qaraeen 1 Introduction A typical programming scenario: Read input data using scanf function, Perform some calculations involving: Evaluating expressions; Storing results in certain variables by executing assignment statements, and Output result using print function; Read & Output Covered previously; Will focus on the calculation point by combining constants and variables with operators (+,-,/, *, %) to form expressions; 2 1

2 C language has rich and powerful set of operators: Arithmetic Operators: ( ) Brackets; ++ Increment; -- Decrements; * Multiplication; / Division; + Addition (plus); - Subtraction (minus); % Remainder (Mod); = Assignment. 3 Relational Operators: > Greater than; >= Greater than or equal to; < Less than; <= Less than or equal to; == Equal to;!= Not equal to. Logical/Boolean Operators: && AND; OR;! NOT. 4 2

3 The operators precedence (priority): 1. Example: a= * / 2.0; What is the answer?! (Answer=17) Multiplication & Division parts will be evaluated first and then the addition and subtraction parts. Note: To avoid confusion use brackets. The following are two different calculations: a= (2.0 * 5.0) - (6.0 / 2.0) a=( ) * ( ) / 2.0 Level 1: (); Level 2: *, /, %; Level 3: +, -. 5 Evaluate the expression: * / 6? ; Evaluate: 7 / 3 = 2 7 % 3 = / 3 * 20 / 3 = / 3 * 20.0 / 3 = / / 3 = Evaluate: * ( 2 / 3) 3 * /

4 Example: Universal Product Code (UPC) is a bar code for the product, to be able to identify it and know its price. It consists of 12 digits (1, 5, 5, and 1) the 1 st (d) identifies the type of the item, the next 5 identify the manufacturer, the next 5 identify the product, then the final digit is check digit which can tell if the previous 11 digits were read correctly or not, so to re-read them by the scanner once again if there is something wrong. The formula to calculate this digit: first_sum= d + i2 + i4 + j1 + j3 + j5 second_sum= i1 + i3 + i5 + j2 + j4 total = 3 * first_sum+ second_sum check digit=9 -( (total-1)% 10 ) The following example implements this algorithm: 7 /* Computes a Universal Product Bar Code check digit */ #include <stdio.h> main() { int d, i1, i2, i3, i4, i5, j1, j2, j3, j4, j5, first_sum, second_sum, total; printf("enter the first (single) digit: "); scanf("%1d", &d); printf("enter first group of five digits: "); scanf("%1d%1d%1d%1d%1d", &i1, &i2, &i3, &i4, &i5); 8 >> 4

5 printf("enter second group of five digits: "); scanf("%1d%1d%1d%1d%1d", &j1, &j2, &j3, &j4, &j5); first_sum = d + i2 + i4 + j1 + j3 + j5; second_sum = i1 + i3 + i5 + j2 + j4; total = 3 * first_sum + second_sum; } printf("check digit: %d\n", 9-( (total - 1) % 10 )); return 0; 9 Try: 3, 1, 2, 3, 4, 5, 0, 6, 7, 8, 9: Check Digit =?? Assignment Statements Once you've declared a variable you can use it, but not until it has been declared - attempts to use a variable that has not been defined will cause a compiler error! Using a variable means storing something in it You can store a value in a variable by using assignment statements of the form: Variable name (v) = value or an expression(e); Example: a=10; /*stores the value 10 in the int variable a */ 10 5

6 Expression, e, is evaluated and its value is assigned to v, (overwriting the previously stored value; if any) The assignment produces a result which is the value of v after assignment e can be: Constant i = 5; /* i is 5 */ variable j = i; /* j is 5 */ expression k = 10 * i + j; /* k is 55 */ Generally, v is an expression that have an lvalue (a location in the machine s memory). So an assignment i + j = 0; is illegal, because i + j has no lvalue or no place to store the Zero! 11 BE CAREFUL WITH ARITHMATIC!!! What is the answer to this simple calculation? a=10/3 The answer depends upon the data type of the division sides (see the next example). If a declared as type in the answer will be 3; If a is of type floa then the answer will be Two points to note from the above calculation: 1. C ignores fractions when doing integer division! 2. When doing float calculations integers will be converted into float. We will see later how C handles type conversions. 12 6

7 float a=10/3; 2 examples: printf("the answer is: %f",a); Would print: The answer is: float a=10/3.0; printf("the answer is: %f",a); Would print: The answer is: Compound Assignments i = i + 3; An assignment statement that adds 3 to the previous value stored in i Compound assignment operators: (+=, -=, *=, /=, %=) Allows to shorten statements: (i = i + 3;) to (i += 3;) += adds value of right operand to variable to left Other compound assignments: v += e Adds e to v, storing result in v; v -= e Subtracts e from v, storing result in v v *= e Multiplies v by e, storing result in v v /= e Divides v by e, storing result in v 14 v %= e Computes the remainder when v is divided by e, storing result in v; 7

8 Example int a, b, c, d, e; a = 14; b = a + 1; //Now, b=15 b *= b; //b=225 c = b / 10-2; //c=20 d = a c; //Now, d=9 c += d; //c=29 e = c + d; //e=38 e /= 2; planets. 15 //e=19 printf( %d%d%d%d%1.0f DNA is a Main-belt Asteroid discovered by C. Juels & P. Holvorcem at Fountain Hills.\n,a-9,b/45,c-24,d-4,e/4.0); Asteroid: The belt is the region of the Solar System located roughly between the orbits of the planets Mars and Jupiter. It is occupied by numerous irregularly shaped bodies called asteroids or minor What s the output?? What are the final values of X, Y, & Z: X = Y = Z = 10;/* The = operator is right associative, so this assignment is equivalent to: X=(Y=(Z=10)); */ X += 15; Y -= 15; Z %= 3; X =?, Y =?, Z =? Watch out an unexpected results in chained assignments and type conversion: int i; float f; f=i=33.3; i is assigned the value 33 (because: int), then f is assigned 33.0 (not 33.3, as you might think)! 16 8

9 17 Increment & Decrement Another way to increment variables is using: ++ (increment) (add 1 to operand) and -- (decrement) (subtract 1 from operand) ++ and -- can be used as: prefix operators (++i and --I) Variable is changed before the expression is evaluated. postfix operators (i++ and i--) Variable is not changed until after it is used in the expression. Simple Assignment Compound Increment i = i + 1 i += 1 ++i j = j - 1 j - = 1 --j That is: A prefix ++i, the variable is incremented (or decremented) and then used: int i = 10; i is incremented to 11, & int j = ++i; then j is set to 11 (the value of i) A postfix i++, the variable is used and then incremented. (increments i, but returns the prior, non-incremented value) int i = 10; j is set to 10 (the value of i) and int j = i++; then i is incremented to

10 The expressions: k = 2 * ++i Add one to i, Store the result back in i, Multiply i by 2, and Store that result in k. k = 2 * i++ Take i's old value and multiply it by 2, Increment i, and Store the result of the multiplication in k. The -- operator works in the same way. 19 Example 1: i = 1; printf ( i is %d \n, ++i); /* prints i is 2 */ printf ( i is %d \n, i); /* prints i is 2 */ Example 2: i = 1; printf ( i is %d \n, i++); /* prints i is 1 */ printf ( i is %d \n, i); /* prints i is 2 */ Example 3: a= 3; printf("a=%d, a+1=%d \n", a, ++a); /* incremented before evaluation, then passed */ /* will print out a=4, a+1=4 */ 20 10

11 21 If the increment operator comes after the named variable, then the value of the statement is calculated after the increment occurs: So for the statement: a= 3; printf("a=%d, a+1=%d \n", a, a++);»would display a=3 a+1=3 Now, a will be set to 4, Try printf("a=%d", a);»the result will be: a=

12 If Expression contains more than One operator and NO parentheses are used to clarify expression Then can USE operator Precedence & Associatively to clarify any ambiguity! Example: Operator Precedence & Associativity Does x+y*k equals x+(y*k) or (x+y)*k? 23 Recalling Precedence: Highest - + (unary) * / % Lowest + - (binary) Unary operation: An operation with only one operand: Operand One of the inputs (such as 3, 6) of an operator (such as +); Ex.: = 9 i.e. an operation with a single input

13 In C language, the following operators are unary: Increment / Decrement : ++x, x++, --x, x-- (Doesn t work on expressions; needs l-value) Address: &x Indirection: *x Positive / Negative: +x, -x One's complement: ~x Logical negation:!x (ex. printf( %d,!!3); would print 1) Sizeof operator: sizeof x, sizeof(type-name) Cast: (type-name) cast-expression 25 Ex. printf( %f,(float)(10/3)); Ex. printf( %f,(float)10/3); If expression contains operators of the same precedence, then we have to use Associativity Left Associativity: Operator groups from L to R Ex: 26 Associativity Binary Arithmetic Operators (+ - * / %): i j + k (i j) + k I * j / k (I * j) / k Right Associativity: Operator groups from R to L Ex: Unary Arithmatic Operators (- +) -+i -(+i) 13

14 27 Notes In compound assignment operators: i+=j; equivalent to i=i+j; i=+j; compiles BUT equivalent to i=(+j) which merely copies the value of j into i! Compound assignment operators have the same properties as the = operator (associativity) i+=j+=k; means i+=(j+=k); Equality & Assignment: x = 5; /* sets value of x to 5 */ if (x == 5) /* returns true/false */ x += 3; /* x is now is 8 */ if (x!= 5) { Printf ( x is not 5 \n ); } x *= 2; /* x is now 16 */ 28 14

15 1. What is the value of x in each of the following: 1.x = % 6 / 2 + 5? 2.x = y = 4; x *=3 + y; 2. What is the output in each case: 1.printf( %d, 3 * /18); 2.printf( %f,193/19/pow(3.0, 2.0)); 3.printf( %d, *( 2 / 3)); 4.printf( %f, 3 * /18.0); 5.printf( %f,193/19.0/pow(3.0, 2.0)); 29 Examples 6.printf( %d, 5 % / 3-8); 15

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