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Programming C for Engineers. An exercise is posted on the web site! Due in one week Single submission. Selection Statements. Selects statements to execute based on the value of an expression The expression is sometimes called the controlling expression Selection statements:
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Programming C for Engineers • An exercise is posted on the web site! • Due in one week • Single submission
Selection Statements • Selects statements to execute based on the value of an expression • The expression is sometimes called the controlling expression • Selection statements: • if statement • switch statement
Selection statements: if • used to execute conditionally a statement or block of code. if(expression) statement • If expression is true, statement is executed (what is true?). • statement can be replaced by a block of statements, enclosed in curly braces.
An example /* This program displays the absolute value of a number given by the user */ #include <stdio.h> int main() { double num; printf("Please enter a real number: "); scanf("%lf", &num); if (num<0) num = -num; printf("The absolute value is %g\n", num); return 0; }
if-else statement if (expression) statement1 else statement2 • if expression is true, statement1 is executed. • if expression is false, statement2 is executed • both statements can be (and very often are) replaced by blocks of statements (“compound statements”)
An example (fragment) int first, second, min; /* … */ if (first < second) { min = first; printf ("The first number is smaller than the second.\n"); } else { min = second; printf ("The second number is smaller than the first\n"); } printf("The smaller number is equal to %d\n", min);
True or false • In C, every expression has a numeric value • An expression is ‘true’ when its value is non-zero • If it is zero, it is false • Therefore, in the following – if (expression) statement statement is executed if expression is non zero.
More about operators • In C, every expression has a numeric value • When using arithmetical operators (+, -, *, /) this is straightforward • The value of A+B is the sum of A and B • And so on…
More about operators • Expressions with relational operators (<, <=, >, >=, etc.) have values as well (intuitively, we are used to thinking about them as ‘true’ or ‘false’) • A < B evaluates to zero if A is larger than or equal to B, and some non-zero value if A is smaller than B • The exact non-zero value varies (and is not important for that matter)
Relational operators • They are: • A == B (Note the difference from A = B!!!!!) • A != B • A < B • A > B • A <= B • A >= B • The value of the expression is non-zero if it’s true, zero if it’s false
An example int a, b; printf("Enter two numbers\n"); scanf("%d%d", &a, &b); if (a == b) { printf("The numbers equal %d\n", a); printf("The expression a == b is %d\n", a == b); } else { printf("The numbers are not equal\n"); printf("The expression a == b is %d\n", a == b); }
The assignment operator = • The assignment operator is also an operator. Hence, expressions involving it have a numeric value. • This value equals to whatever appears on the right of the assignment operator • For example: • (x = 4) evaluates to 4 • (y = 0) evaluates to 0
A very common mistake • Very often a programmer might confuse between the equality operator and the assignment operator: • if (x==4) … • if (x=4) … • The second is usually a mistake, but legal in C so the compiler doesn’t warn us about it!
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); }
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); }
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); } i= 2
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); }
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); } (i==4) = 0
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); }
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); } i= 2
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); }
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); } (i=4) = 4
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); }
Example /* This program exemplifies the difference between the equality relational operator and the assignment operator. */ #include <stdio.h> int main() { int i = 2; printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i); } i= 4
Logical operators • Allows to evaluate two or more expressions - • !A – ‘not’ - True when A is not, and vice versa. • A && B – ‘and’ - True when both A and B are true • A || B – ‘or’ (inclusive or) - True when either A or B (or both) are true
A silly example #include <stdio.h> int main(void) { int grade; printf("Please enter your grade: "); scanf("%d", &grade); if (grade < 0 || grade > 100) printf("This is not a valid grade!\n"); else printf("This is indeed a grade.\n"); return 0; }
else if • if statements distinguish between exactly 2 cases and execute different code in each case • The else-if construction allows for a multi-way decision
else if if (expression) statement else if (expression) statement elseif (expression) statement else statement
An example if (grade >= 90) printf ("A\n"); else if (grade >= 80) printf ("B\n"); else if (grade >= 70) printf ("C\n"); else if (grade >= 60) printf ("D\n"); else printf ("F\n");
Validating input • When getting input from the user, it is highly recommended to check whether it is valid. • If it’s not, you should display an appropriate message and return a non-zero value. • For example – if (grade < 0 || grade > 100) { printf(“Invalid input!\n”); return 1; }
The return keyword • For now, used to terminate the program and return a value to the operating system • If the program is successful the return value should be zero; non-zero otherwise • The exact nature of this keyword will become clear in the future
Exercise • Input – • An English letter • Output – • If input is a lowercase letter – the corresponding uppercase letter • If input is an uppercase letter - corresponding lowercase letter • Notes – • You do not need the specific ASCII values!! • Remember to check for input validity!
Solution #include <stdio.h> int main( ) { char c; printf("Please enter an english letter: "); scanf("%c", &c); if (c <= 'z' && c >= 'a') printf("%c in uppercase is %c\n", c, c-'a'+'A'); else if (c <= 'Z' && c >= 'A') printf("%c in lowercase is %c\n", c, c-'A'+'a'); else { printf("%c is not an english letter!\n", c); return 1; } return 0; }
The ?: operator • expr1 ? expr2 : expr3 • Nicer way to write: • (expr1)? expr2 : expr3 • If expr1 is true (non-zero), expr2 is evaluated. Otherwise, expr3 is evaluated
The ?: operator #include <stdio.h> int main() { int i, j, min; printf("Please enter two numbers: "); scanf("%d%d", &i, &j); min = (i < j)? i : j; printf("The minimum between %d and %d is %d\n", i, j, min); return 0; }
The switch statement • a multiway conditional statement • similar to if-else if-else • allows the selection of an arbitrary number of choices based on an integer value • switch (expression) {case const-expr: statementscase const-expr: statements…default:statements}
The switch statement • expression must have an integer value • when the switch statement is executed: • the expression is evaluated • if a case matches the value of the expression, the program jumps to the first statement after that case label • otherwise, the default case is selected • the default is optional
That grade example again switch (grade/10) { case 10: case 9: printf ("A\n"); break; case 8: printf ("B\n"); break; case 7: printf ("C\n"); break; case 6: printf ("D\n"); break; default: printf ("F\n"); }
Give me a break • when the switch transfers to the chosen case, it starts executing statements at that point • it will “fall through” to the next case unless you “break out” • break causes the program to immediately jump to the next statement after the switch statement
A more interesting example #include <stdio.h> int main( ){ double n1, n2, res; char op; printf("Please enter two numbers: "); scanf("%lf%lf", &n1, &n2); printf("Please enter an arithmetical operator (+, -, * or /): "); scanf(" %c", &op); /* The space before the %c will be explained... */
A more interesting example switch(op) { case '+': res = n1+n2; break; case '-': res = n1-n2; break; case '*': res = n1*n2; break; case '/': /* We're not checking for division by zero for clarity... */ res = n1/n2; break; default: printf("%c is an invalid arithmetical operator!\n", op); return 1; } /* Display the expression and its result */ printf("%g %c %g = %g\n", n1, op, n2, res); return 0; }
Exercise • Write a program that accepts a number between 1 and 100 from the user. If there is a coin of that value in cents, it should display its name. Otherwise, it should report that there is no such coin • 1 = cent, 5 = nickel, 10 = dime, 25 = quarter, 100 = dollar • Remember to check for the validity of the input!
Solution /#include <stdio.h> int main() { int num; printf("Please enter a number from 1 to 100: "); scanf("%d", &num); /* Make sure the input is valid */ if (num<1 || num>100){ printf("Invalid input!\n"); return 1; } /* Display the correct coin name, or a default message if there's no such coin */ switch (num) { case 1: printf("It's a cent!\n"); break; case 5: printf("It's a nickel!\n"); break; case 10: printf("It's a dime!\n"); break; case 25: printf("It's a quarter!\n"); break; case 100: printf("It's a whole dollar!\n"); break; default: printf("It's not a coin!\n"); } return 0; }
Debugging • It is virtually impossible to program without errors • Syntax errors are detected by the compiler • However, often a program has no syntax errors and compiles, but still doesn’t perform as desired
Debugging • Debuggers are software tools designed to help find software bugs • Both Visual C and the lcc compiler include a debugger
Debugging • The debugger allows us to – • Execute the program one line at a time • At each step see the values of all variables and expressions • Run the program up to a pre-specified point • And more…