CMPE 150 Introduction to Computing If and Switch

CMPE 150: Introduction to Computing If and Switch statements

If statement • The "if statement" is used to break the sequential flow of execution. • Enforces branching in execution according to the result of an expression. – There are two possible paths to take. – The expression determines which path should be taken. Read age T F age<=25 ? Message: "You are young" Message: "You are mature" Rest of the program CMPE 150 – Introduction to Computing 1

If statement • Syntax: if (int_expr) stat_block 1 else stat_block 2 Text in green is optional where stat_block is one of the following: • a single statement stat; • the null statement ; • a group of statements enclosed in braces { stat 1; . . . statn; } CMPE 150 – Introduction to Computing 2

If statement stat 0; if (expr) { stat 1; stat 2; } else stat 3; stat 4; stat 0 T expr ? stat 1 F stat 3 stat 2 stat 4 Notice the indentation. CMPE 150 – Introduction to Computing 3

If statement stat 0; if (expr) { stat 1; stat 2; } stat 4; stat 0 T expr ? F stat 1 stat 2 stat 4 CMPE 150 – Introduction to Computing 4

If statement • Read a number and state whether it is odd or even. int num; scanf("%d", &num); printf("%d is an ", num); if (num%2!=0) printf("odd "); else printf("even "); printf("number. n"); CMPE 150 – Introduction to Computing 5

If statement • Read one character as input; check if it is a digit; if so, convert it to an integer and display the number that is two times the input number; o/w display an error message. char ch; int num; scanf("%c", &ch); if ((‘ 0’<=ch) && (ch<=‘ 9’)) { num=ch-’ 0’; printf("Two times the input is %d n", 2*num); } else printf("Input is not a digit! n"); CMPE 150 – Introduction to Computing 6

Nested-if statements • Remember the syntax: if (int_expr) stat_block 1 else stat_block 2 • Statement block contains statements. – "if" is also a statement. – So, you can "nest" one if statement in another. • This structure is called nested-if statements. • You can nest as much as you want. CMPE 150 – Introduction to Computing 7

Nested-if statements stat 0; if (expr 1) if (expr 2) { stat 1; stat 2; } else if (expr 3) { stat 4; stat 5; } else stat 6; else stat 7; stat 8; stat 0 T T expr 1 ? F expr 2 ? T stat 1 stat 2 F expr 3 ? stat 4 F stat 7 stat 6 stat 5 stat 8 CMPE 150 – Introduction to Computing 8

Nested-if statements stat 0; if (expr 1) if (expr 2) { stat 1; stat 2; } else if (expr 3) { stat 4; stat 5; } else stat 6; else stat 7; stat 8; stat 0 T expr 1 ? F IF INDENTATION IS NOT CORRECT, IT IS MISLEADING T F expr 2 ? T stat 1 stat 2 expr 3 ? stat 4 F stat 7 stat 5 stat 8 CMPE 150 – Introduction to Computing 9

Nested-if statements • Remember that the "else" part is optional. – Thus, some of the if statements may not have an "else" part. – Then, it is a little bit tricky to find to which "if" the "else" belongs. • Note that indentation is completely ignored by the compiler, so it does not help. • The trick is the following: – "else" belongs to the nearest incomplete "if" CMPE 150 – Introduction to Computing 10

Else-if statements if (age<=1) printf("infant"); if (age<=3) printf("toddler"); if (age<=10) printf("child"); if (age<=18) printf("adolescent"); if (age<=25) printf("young"); if (age<=39) printf("adult"); if (age<=65) printf("middle-aged"); else printf("elderly"); CMPE 150 – Introduction to Computing 11

Else-if statements • Else-if is a variation of nested-if. • An inner if statement is executed iff all previous if statements have failed. – Thus, executing an inner if statement implies that all previous expressions were false. • Syntax: if (int_expr 1) stat_block 1 else if (int_expr 2) stat_block 2. . . else stat_blockn CMPE 150 – Introduction to Computing 12

Else-if statements if (age<=1) printf("infant"); else if (age<=3) printf("toddler"); else if (age<=10) printf("child"); else if (age<=18) printf("adolescent"); else if (age<=25) printf("young"); else if (age<=39) printf("adult"); else if (age<=65) printf("middle-aged"); else printf("elderly"); CMPE 150 – Introduction to Computing 13

Else-if statements • Alternative would be: if (age<=1) printf("infant"); if ((1<age) && (age<=3)) printf("toddler"); if ((3<age) && (age<=10)) printf("child"); if ((10<age) && (age<=18)) printf("adolescent"); if ((18<age) && (age<=25)) printf("young"); if ((25<age) && (age<=39)) printf("adult"); if ((39<age) && (age<=65)) printf("middle-aged"); if (65<age) printf("elderly"); CMPE 150 – Introduction to Computing 14

Ternary operator • Ternary operator is similar to the "if" statement. But it is an operator, not a statement. • Syntax: int_expr ? value 1 : value 2 • Eg: a = (b>c) ? b : c; k = (n!=0) ? m/n : 0; CMPE 150 – Introduction to Computing 15

Ternary operator • Note that it is not possible to know at compile time whether value 1 or value 2 will be used. – Therefore, the type of the expression is the type of the larger value. • Eg: In the expression below, if the value of b is 9 a = b / (b%2)? 2: 3. 0; the value of a is 4. 5 (not 4), because we perform a float division (not integer division) CMPE 150 – Introduction to Computing 16

Example 1 • Write a code segment that detects whether a number is divisible by 6. if ((num%2==0) && (num%3==0)) printf("%d is divisible by 6 n", num); CMPE 150 – Introduction to Computing 17

Example 2 • Write a code segment that detects whether a number is divisible by 3 or 6. if (num%3==0) if (num%2==0) printf("%d is divisible by 6 n", num); else printf("%d is divisible by 3 n", num); CMPE 150 – Introduction to Computing 18

Example 3 • Write a program that reads two real numbers and checks if they are equal. (Assume first number is smaller. ) #include <stdio. h> #define EPSILON 0. 00001 float r 1, r 2; int main() { scanf("%f %f", &r 1, &r 2); if ((r 2 -r 1)<=EPSILON) printf("The numbers are (almost) equivalent"); return 0; } CMPE 150 – Introduction to Computing 19

Example 4 • Write a program that reads a 3 -digit number from the input char-by-char, and displays its square. #include <stdio. h> char c 1, c 2, c 3; int num; int main() { scanf("%c%c%c", &c 1, &c 2, &c 3); num = (c 1 -'0')*100; num += (c 2 -'0')*10; num += c 3 -'0'; num *= num; printf("%d", num); return 0; } CMPE 150 – Introduction to Computing 20

Switch statement • If you have multiple cases depending on different values of the same integer expression, switch is easier to use. • Syntax: switch (int_expr) { case constant_int_value 1: stat(s); case constant_int_value 2: stat(s); . . . default: stat(s); } • You may have zero or more statements in each case. CMPE 150 – Introduction to Computing 21

Break statement • Switch statement actually gathers many statements of several cases. – The case labels denote the specific statement from which the execution of this group of statements begins. – All statements till the end of the group are executed sequentially. • To separate the cases, break statement is used. – breaks the sequential execution of the statements and immediately jumps to the end of the switch statement. CMPE 150 – Introduction to Computing 22

Break statement stat 0; switch (expr) { case value 1: stat 1; stat 2; case value 2: stat 3; stat 4; stat 5; case value 3: stat 6; break; case value 4: stat 7; stat 8; } stat 9; stat 0 stat 1 stat 2 stat 3 stat 4 stat 5 stat 6 stat 7 stat 8 stat 9 If expr happens to be value 2 CMPE 150 – Introduction to Computing 23

Example 6 • Write a program that reads a character ('a', 'p', or 'v') and radius R. It displays the area or perimeter of a circle with radius R, or the volume of a sphere. #include <stdio. h> #define PI 3. 141592654 char ch; float R; int main() { scanf("%c%f", &ch, &R); switch(ch) { case 'a': printf("Area of circle = %fn", PI*R*R); break; case 'p': printf("Perimeter of circle = %fn", 2*PI*R); break; case 'v': printf("Volume of sphere = %fn", (4/3)*(PI*R*R*R)); (4. 0/3) break; default: printf("Invalid inputn"); break; } return 0; } CMPE 150 – Introduction to Computing 24

Switch statement • Define the days of the week as an enumerated type and display them as strings. enum day_type {MON=1, TUE, WED, THU, FRI, SAT, SUN} day; scanf("%d", &day); switch (day) { case SUN: printf("Sundayn"); break; case WED: printf("Wednesdayn"); break; case TUE: printf("Tuesdayn"); break; case THU: printf("Thursdayn"); break; case FRI: printf("Fridayn"); break; case SAT: printf("Saturdayn"); break; case MON: printf("Mondayn"); break; default: printf("Incorrect dayn"); break; } CMPE 150 – Introduction to Computing 25

Switch statement • Note that without the "break" statement, execution traverses all cases until the end of the switch statement. • This allows implementation of OR (if you use it properly. • Eg: switch (number) { case 1: case 3: case 5: printf("Odd number n"); break; case 0: case 2: case 4: printf("Even number n"); break; } CMPE 150 – Introduction to Computing 26

Switch statement • As long as the cases are separated with "break"s, their order is not relevant. • "default" is optional. If the default case is not specified and none of the cases holds, no statement is executed; this is not an error. • It is a good practice to put a break even after the last case. CMPE 150 – Introduction to Computing 27

Example 5. 1 • Write a program that defines a type for letter grades (AA, BB, CC, DD, F as 4, 3, 2, 1, 0), reads a grade and displays corresponding letter grade. #include <stdio. h> enum grade_type {F, DD, CC, BB, AA}; int main() { enum grade_type g; scanf("%d", &g); printf("%d", g); return 0; } CMPE 150 – Introduction to Computing 28

Example 5. 2 • But, we don't want to display a number. We want to see the letter grade. #include <stdio. h> enum grade_type {F, DD, CC, BB, AA}; int main() { enum grade_type g; scanf("%d", switch (g) { case AA: case BB: case CC: case DD: case F: default: } return 0; &g); printf("AA n"); break; printf("BB n"); break; printf("CC n"); break; printf("DD n"); break; printf("F n"); break; printf("Invalid n"); break; } CMPE 150 – Introduction to Computing 29

Example 5. 3 • Could we do it without defining an enumerated type? #include <stdio. h> int main() { int g; scanf("%d", switch (g) { case 4: case 3: case 2: case 1: case 0: default: } return 0; &g); printf("AA n"); break; printf("BB n"); break; printf("CC n"); break; printf("DD n"); break; printf("F n"); break; printf("Invalid n"); break; } CMPE 150 – Introduction to Computing 30