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Lecture III Start programming in Fortran

Lecture III Start programming in Fortran. Yi Lin Jan 11, 2007. A simple example: Roots finding. Finding roots for quadratic aX^2+bX+c=0 Input a, b, c Output root(s): (-b + SQRT(b*b - 4*a*c))/(2*a) (-b - SQRT(b*b - 4*a*c))/(2*a). Roots finding (cont.). start.

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Lecture III Start programming in Fortran

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  1. Lecture III Start programming in Fortran Yi Lin Jan 11, 2007

  2. A simple example: Roots finding • Finding roots for quadratic aX^2+bX+c=0 • Input a, b, c • Output root(s): (-b + SQRT(b*b - 4*a*c))/(2*a) (-b - SQRT(b*b - 4*a*c))/(2*a)

  3. Roots finding (cont.) start Initialize a=1, b=-5, c=6 w = (b*b – 4*a*c) R1 = (-b + sqrt(w))/(2*a) R2 = (-b - sqrt(w))/(2*a) End

  4. Declare variables, Must at the head of block Statements: initialize variables with values statements: calculation statements:Output results Roots finding (cont.) PROGRAM ROOTSFINDING INTEGER a, b, c REAL x, y, z, w, r1, r2 a=1 b=-5 c=6 ! To calculate b*b – 4*a*c x = b*b y = a*c z = 4*y w = x-z r1 = (-b +SQRT(w))/(2*a) r2 = (-b – SQRT(w))/(2*a) WRITE(*, *) r1, r2 END PROGRAM

  5. Keywords in Fortran90 variables constants operators Roots finding (cont.) PROGRAM ROOTSFINDING INTEGER a, b, c REAL x, y, z, w, r1, r2 a=1 b=-5 c=6 ! To calculate b*b – 4*a*c x = b*b y = a*c z = 4*y w = x-z r1 = (-b +SQRT(w))/(2*a) r2 = (-b –SQRT(w))/(2*a) WRITE(*, *) r1, r2 END PROGRAM

  6. Variables and constants • Constants (e.g., 1, -5, 6, 4, 2) • A variable (e.g., a, b, c, x, y, z, w) is a unique name which a FORTRAN program applies to a word of memory and uses to refer to it. • Naming convention • alphanumeric characters (letters, numerals and the underscore character) • The first character must not be a letter. • No case sensitivity • Don’t use keywords in Fortran as variables’ names

  7. Variable data types • INTEGER • E.g., a, b, c • REAL • E.g., x, y, z, w • LOGICAL • COMPLEX • CHARACTER • Examples: • Data type is important • INTEGER::a=1,b=-5,c=6 • REAL::r1, r2 • IMPLICIT NONE needed to prevent errors. • Comment out “IMPLICIT NONE” • write(*, *) r1, r3 ! You want to print r1 and r2, but by mistakes you type r3

  8. Declarations PROGRAM RootFinding IMPLICIT NONE INTEGER :: a, b, c REAL :: x,y,z,w REAL :: r1, r2 . . . END PROGRAM RootFinding Declarations tell the compiler • To allocate space in memory for a variable • What “shape” the memory cell should be (i.e. what type of value is to be placed there) • What name we will use to refer to that cell

  9. INTEGER a, b, c REAL x, y, z, w A=1 B=-5 C=6 INTEGER:: a=1, b=-5, c=6 REAL:: x, y, z, w Declare variables same

  10. Variable precision 1 byte = 8 bits • INTEGER(KIND=2)::a OR INTEGER(2)::a 11111111

  11. program test implicit none integer(1)::a=127 integer(2)::b=128 write(*,*) "a=", a, “b=“,b End PROGRAM Output a=127 b=-128 Example

  12. Variable precision (cont.) • REAL(KIND=4) • CHARACTER • KIND=1 only

  13. Arithmetic Expressions An arithmetic expression is formed using the operations: + (addition), - (subtraction) * (multiplication), / (division) ** (exponentiation) If the operands are integers, the result will be an integer value If the operands are real, the result will be a real value If the operands are of different types, the expression is called mixed mode.

  14. Arithmetic expressions • Variable operator variable operator …. • E.g., 5 + 2 *3 • E.g., b**2 – 4*a*c • It has a value by itself • E.g. The value of expression 5 + 2*3 is 11

  15. Simple Expressions • 1 + 3 --> 4 • 1.23 - 0.45 --> 0.78 • 3 * 8 --> 24 • 6.5/1.25 --> 5.2 • 8.4/4.2 --> 2.0 rather than 2, since the result must be of REAL type. • -5**2 --> -25 • 12/4 --> 3 • 13/4 --> 3 rather than 3.25. Since the operands are of INTEGER type, so is the result. The computer will truncate the mathematical result to make it an integer. • 3/5 --> 0 rather than 0.6.

  16. Evaluating Complex Expressions Evaluate operators in order of precedence. • First evaluate operators of higher precedence 3 * 4 – 5  7 3 + 4 * 5  23 • For operators of the same precedence, use associativity. Exponentiation is right associative, all others are left associative • Expressions within parentheses are evaluated first

  17. Arithmetic operators: precedence 2+3*4 = ? = 5 * 4 = 20 Or = 2+12 = 14 (2+3)*4 = 20

  18. Arithmetic operators: associativity associativity resolves the order of operations when two operators of the same precedence compete for three operands: 2**3**4 = 2**(3**4) = 2**81, 72/12/ 3 = (72/12)/3 = 6/3 = 2 30/5*3 = (30/5)*3 = 18 if *,/ associativity is from right to left 30/5*3 = 30/(5*3) = 2

  19. Examples 2 * 4 * 5 / 3 ** 2 --> 2 * 4 * 5 / [3 ** 2] --> 2 * 4 * 5 / 9 --> [2 * 4] * 5 / 9 --> 8 * 5 / 9 --> [8 * 5] / 9 --> 40 / 9 --> 4 The result is 4 rather than 4.444444 since the operands are all integers.

  20. Examples 100 + (1 + 250 / 100) ** 3 --> 100 + (1 + [250 / 100]) ** 3 --> 100 + (1 + 2) ** 3 --> 100 + ([1 + 2]) ** 3 --> 100 + 3 ** 3 --> 100 + [3 ** 3] --> 100 + 27 --> 127 Parentheses are evaluated first

  21. Examples 1.0 + 2.0 * 3.0 / ( 6.0*6.0 + 5.0*44.0) ** 0.25 --> 1.0 + [2.0 * 3.0] / (6.0*6.0 + 5.0*44.0) ** 0.25 --> 1.0 + 6.0 / (6.0*6.0 + 5.0*55.0) ** 0.25 --> 1.0 + 6.0 / ([6.0*6.0] + 5.0*44.0) ** 0.25 --> 1.0 + 6.0 / (36.0 + 5.0*44.0) ** 0.25 --> 1.0 + 6.0 / (36.0 + [5.0*44.0]) ** 0.25 --> 1.0 + 6.0 / (36.0 + 220.0) ** 0.25 --> 1.0 + 6.0 / ([36.0 + 220.0]) ** 0.25 --> 1.0 + 6.0 / 256.0 ** 0.25 --> 1.0 + 6.0 / [256.0 ** 0.25] --> 1.0 + 6.0 / 4.0 --> 1.0 + [6.0 / 4.0] --> 1.0 + 1.5 --> 2.5

  22. Mixed Mode Expressions If one operand of an arithmetic operator is INTEGER and the other is REAL • the INTEGER value is converted to REAL • the operation is performed • the result is REAL 1 + 2.5  3.5 1/2.0  0.5 2.0/8  0.25 -3**2.0  -9.0 4.0**(1/2)  1.0 (since 1/2  0)

  23. Example of Mixed Mode 25.0 ** 1 / 2 * 3.5 ** (1 / 3)  25.0 ** 1 / 2 * 3.5 ** (1 / 3)  25.0 ** 1 / 2 * 3.5 ** (1 / 3)  25.0 ** 1 / 2 * 3.5 ** ([1 / 3])  [25.0 ** 1] / 2 * 3.5 ** 0  25.0 / 2 * [3.5 ** 0]  25.0 / 2 * 1.0  [25.0 / 2] * 1.0  12.5 * 1.0  12.5

  24. Statements • Assignment statement: Variable = expression e.g., b = -5 (never the other way around, show an example) • General statement: INTEGER a, b, c write (*,*) ‘Hello world!’

  25. Assignment Statement The assignment statement has syntax: variable = expression Semantics • Evaluate the expression • If the type of result is the same as the type of the variable store the result in the variable • Otherwise convert the value to the type of the variable and then store it • If the value is REAL and the variable is INTEGER remove the decimal part (truncate) • If the value is INTEGER and the variable is REAL convert the value to a decimal • The original value of the variable is destroyed

  26. Assignment statements examples REAL::X=3.5 X=2 * 2.4 ! X=4.8 INTEGER X=3 X=2 * 2.4 ! X=4 X=2 * 2.6 ! X=5

  27. Declare variables, Must at the head of block statements: calculation statements:Output results Roots finding revisited PROGRAM ROOTSFINDING INTEGER::a=1, b=-5, c=6 REAL x, r1, r2 ! To calculate b*b – 4*a*c x = b**2- 4*a*c r1 = (-b +SQRT(x))/(2*a) r2 = (-b – SQRT(x))/(2*a) WRITE(*, *) r1, r2 END PROGRAM

  28. Example in a previous final PROGRAM Q1 IMPLICIT NONE INTEGER :: I, J, K REAL :: A, B, C A = 3.2 + 4 B = 22/7 C = 22.0/7.0 I = 3.7 + MOD(78,5) J = 45/3*3+1-3*4 K = C WRITE(*,*) I,J,K,A,B,C END PROGRAM Q1 a) 6 34 3 7.200 3.000 3.143 b) 7.200 3.000 3.143 6 34 3 c) 6.700 34.000 3.143 7.200 3.000 3.143 d) 6 34 3 7.200 3.143 3.143

  29. What does the following program output? PROGRAM midterm REAL A,B,C INTEGER I,J,K A = 3.5 I = A J = 5.25 K = I*2 B = A*I C = J/3 WRITE (*,*) A,B,C,I,J,K END PROGRAM midterm A. 3.5 10.5 1. 3 5 7 B. 3.5 12.25 1. 3 5 6 C. 3.5 10.5 1. 3 5 6 D. 3.5 10.5 1.75 3 5 6 E. None of the above Example in midterm05

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