Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 3112-2•Know how if and switch C statements control the sequence of execution of statements.•Be able to use relational and logical operators in the conditional part of an if or a switch statement.•Related Chapters: ABC Chapter 4.1-4.7 & 4.1612-31. Problem DefinitionWrite a program that reads a number and computes the square root if the number is non-negative. 2. Refine, Generalize, Decompose the problem definition(i.e., identify subproblems, I/O, etc.)Input = real numberOutput=real number3. Develop Algorithm(processing steps to solve problem)12-4FlowchartPrint “enter value”Read value value >= 0.0Print sqrt(value)TrueFalse12-5/* C Program to compute the square root of a positive number */#include <stdio.h>#include <math.h>void main(void){ double value; /* Declare variables. */ /* request user input */ printf("Please enter a non-negative number :"); /* read value */ scanf("%lf", &value); /* Output the square root. */ if (value >= 0.0)printf("square_root(%lf) = %lf \n", value , sqrt(value));}12-6if(expression)statement;• if expression evaluates to true, the statement is executed; otherwise execution passes to the next statement in the program.• if Selection Structureif(value >= 0.0);printf("square_root(%lf) = %lf \n", value,sqrt(value));/* Error! Don’t put semicolon here *//* This is an example of a logical error */12-71. Problem DefinitionModify the previous program to notify the user when the input is invalid.12-8FlowchartPrint “enter value”Read value value >= 0.0Print sqrt(value);TrueFalsePrint “invalid input”12-9/* C Program to compute the square root of a positive number */#include <stdio.h>#include <math.h>void main(void){ double value; /* Declare variables. */ /*request user input*/ printf(”Please enter a non-negative number :”); scanf("%lf", &value); /* read value */ /* Output the square root. */ if (value >= 0.0) printf("square_root(%lf) = %lf \n", value,sqrt(value)); else printf("invalid user input, please enter non-negative value\n");}12-10- in header file math.hArguments (parameters) for each of the following functions are assumed to be of type double. If not, a type double copy is made for use in the function. To compile a program that contains math functions you need to use the -lm (Lm not 1m )option for gcc. > gcc file.c -lmSee next pagefabs (x) - |x| (not the same as the abs(x) function) sqrt (x) - square root of x pow (x, a) - xa exp (x) - ex (e = 2.718281828 …) log (x) - ln x = loge x log10 (x) - log10 x sin (x) - sine function (x in radians) cos (x) - cosine function (x in radians) tan (x) - tangent function (x in radians) ceil (x) - smallest integer >= x floor (x) - largest integer <= x12-12if (expression)statement1;elsestatement2;-if expression evaluates to true, statement1 is executed and execution skips statement2 -If expression evaluates to false, execution skips statement1 , statement2 is executed12-13We can also execute multiple statements when a given expression is true:if (expression) {statement1;statement2;statementn;}Example -if(b < a){temp = a;a = b;b = temp;}or...if (expression) {statement1;statementn;}else { statement1; statementm;}(what does this code do?)......12-141. Problem DefinitionModify the previous program to compute the following:You must check that the value is legal, i.e. value >= 1.0 or value <= -1.0 0.12value12-15FlowchartPrint “enter value”Read value value >= 1.0 or value <= -1.0Print sqrt(value*value -1.0);TrueFalsePrint “invalid input”12-16/* Compute the square root of value*value-1.0 */#include <stdio.h>#include <math.h>void main(void){ double value; /* Declare variables. */ /* request user input*/ printf("Please enter value >= 1.0 or <= -1.0 :"); scanf("%lf", &value); /* read value */ /* Output the square root. */ if ((value >= 1.0) || (value <= -1.0)) printf("square_root(%f) = %f \n", value,sqrt(value*value - 1.0)); else { printf("invalid user input\n"); printf("input should be a value >= 1.0 or <= -1.0 \n"); }}12-17In logical expressions (which evaluate to true or false), we can use the following Relational operators:RelationalOperatorType of Test == equal to (don’t use =) != not equal to > greater than >= greater than or equal to < less than <= less than or equal to12-18A B A && BA || BTRUE TRUE TRUE TRUETRUE FALSE FALSE TRUEFALSE TRUE FALSE TRUEFALSE FALSE FALSE FALSEA !ATRUE FALSEFALSE TRUE12-19if ( 5 ) printf("True"); /* prints True */In C the value for False is represented by the value zero and True is represented by any nonzero value. The value False can be any zero value such as the number 0 or 0.0 or null character ‘ \0 ’ or the NULL pointer.Example 2:int x = 0; /* x declared as an integer variable */ /* and initialized to the value 0 */ if (x = 0) /* note the error, == should be used */ printf(" x is zero\n"); /*message not printed, why?*/Example 1:12-20Avoid using == to test real numbers for equality!Exampledouble x = .3333333333; /* ten digits of 3s */if (x == 1.0/3.0) printf("equal!\n");else printf(" not equal!\n"); /* prints not equal! */if ( fabs(x - 1.0/3.0) < 1.0e-10 ) printf("equal!\n"); /* prints equal! */else printf(" not equal!\n");12-211. Problem DefinitionWrite a program that returns a letter grade based on a quiz score. The input will be the integer score from a 10 point quiz. The letter grades are assigned by:9 - 10 “A”7 - 8 “B”5 - 6 “C”3 - 4 “D”< 3 “F” 2. Refine, Generalize, Decompose the problem definition(i.e., identify subproblems, I/O, etc.)Input = integer scoreOutput=character “grade”3. Develop Algorithm(processing steps to solve problem)12-22FlowchartPrint “enter score”Read score score == 10 || score == 9Print “A”TrueFalse(continued on next slide)(skip else part of statement)12-23False score == 8 || score == 7Print “B”;True(skip else part of statement)(continued on next slide)False12-24False score == 6 || score == 5Print “C”;True(skip else part of statement)(continued on next slide)False12-25False score == 4 || score == 3Print “D”TrueFalsePrint “F”12-26/* C Program to compute