Functions a group of declarations and statements that is assigned a name effectively, a named statement block usually has a value a sub-program when we write our program we always define a function named main inside main we can call other functions which can themselves use other functions, and so on…

Example - Square #include double square(double a) { return a*a; } int main(void) { double num; printf("enter a number\n"); scanf("%lf",&num); printf("square of %g is %g\n",num,square(num)); return 0; } This is a function defined outside main Here is where we call the function square

Why use functions? they can break your problem down into smaller sub-tasks easier to solve complex problems they make a program much easier to read and maintain abstraction – we don’t have to know how a function is implemented to use it generalize a repeated set of instructions we don’t have to keep writing the same thing over and over

Characteristics of Functions return-type name(arg_type1 arg_name1, arg_type2 arg_name2, …) { function body; return value; } double square(double a) { return a*a; } int main(void) { … }

Return Statement Return causes the execution of the function to terminate and usually returns a value to the calling function The type of the value returned must be the same as the return-type defined for the function (or a ‘lower’ type) If no value is to be returned, the return-type of the function should be set to ‘void’

Factorials galore #include int factorial(int n) { int i, fact = 1; for (i=2; i<=n; i++) fact *= i; return fact; } int main(void) { int num; printf("enter a number\n"); scanf("%d",&num); printf("%d!=%d\n",num,factorial(num)); return 0; }

A Detailed Example Write a program that receives a nominator and a denominator from the user, and displays the reduced form of the number. For example, if the input is 6 and 9, the program should display 2/3.

Example – solution (step I) #include int main(void) { int n, d; printf("Please enter nominator and denominator: "); scanf("%d%d", &n, &d); Calculate n’s and d’s Greatest Common Divisor printf("The reduced form of %d/%d is %d/%d", n, d, n/gcd, d/gcd); return 0; }

Example – solution (step II) #include int main(void) { int n, d, g; printf("Please enter nominator and denominator: "); scanf("%d%d", &n, &d); g = gcd(n, d); printf("The reduced form of %d/%d is %d/%d", n, d, n/g, d/g); return 0; }

Example – solution (step III) /* Returns the greatest common divisor of its two parameters. Assumes both are positive. The function uses the fact that the if r = mod(y, x) then the gcd of y and x equals the gcd of x and r. */ int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; }

GCD – step by step int main(void) { int n, d, g; printf("Please enter … : "); scanf("%d%d", &n, &d); g = gcd(n, d); printf("The reduced form…", n, d, n/g, d/g); return 0; } ndg 69---

GCD – step by step int main(void) { int n, d, g; printf("Please enter … : "); scanf("%d%d", &n, &d); g = gcd(n, d); printf("The reduced form…", n, d, n/g, d/g); return 0; } ndg 69---

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 69---

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 69---

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 696

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 396

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 366

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 366

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 363

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 063

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 033

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 033

GCD – step by step int gcd(int x, int y) { int tmp; while(x > 0) { tmp = x; x = y % x; y = tmp; } return y; } xytmp 033

GCD – step by step int main(void) { int n, d, g; printf("Please enter … : "); scanf("%d%d", &n, &d); g = gcd(n, d); printf("The reduced form…", n, d, n/g, d/g); return 0; } ndg 693

Example – Complete Solution gcd.c

Exercise Input – An integer n Output – The n’th fibonacci number Note – Use an appropriately defined function

Solution fibonacci_func.c

Exercise Write a program that gets a positive integer from the user and prints all the prime numbers from 2 up to that integer. (Use a function that returns 1 if its parameter is prime, 0 otherwise)

Solution is_prime_func.c

Exercise Newton was the first to notice that for any positive n, and when x 0 =1, the following series converges to sqrt(n) – Use this fact to write a program that accepts a positive number and outputs its square root Hint – the thousandth element of Newton’s series is a good-enough approximation

Solution sqrt.c

The Great Void Sometimes there’s no reason for a function to return a value In these cases, the function return type should be ‘void’ If the ‘return’ keyword is used within such a function it exits the function immediately. No value needs be specified Calling ‘return’ in a function returning void is not obligatory If the function receives no parameters, the parameter list should be replaced by ‘void’

Example void ShowHelp(void) { printf("This function explains what this program does…\n"); printf("Yadayadayada"); /* …. */ } int main(void) { char choice; printf("Please enter your selection: "); scanf("%c", &choice); if (choice==‘h’) ShowHelp(); else if /* Program continues … */ }

Pass-by-value Function arguments are passed to the function by copying their values rather than giving the function direct access to the actual variables A change to the value of an argument in a function body will not change the value of variables in the calling function Example – add_one.c

add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } ab 341 Main() memory state

add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } ab 341 Main() memory state

add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } ab 341 Main() memory state b 1 add_one memory state

ab 341 Main() memory state add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } b 2 add_one memory state

ab 341 Main() memory state add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } b 2 add_one memory state

add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } ab 21 Main() memory state

add_one – step by step int add_one(int b) { b=b+1; return b; } int main(void) { int a=34,b=1; a=add_one(b); printf("a = %d, b = %d\n", a, b); return 0; } ab 21 Main() memory state

Riddle me this #include int factorial(int n) { int fact = 1; while (n>1) { fact *= n; n--; } return fact; } int main(void) { int n; printf("enter a number\n"); scanf("%d",&n); printf("%d!=%d\n", n, factorial(n)); /* What will this print? */ printf( "n = %d\n", n); return 0; }

Scope of variables A variable declared within a function is unrelated to variables declared elsewhere, even if they have the same name A function cannot access variables that are declared in other functions Example – scope.c

Wrong way to do it int add_one(int b) { a=b+1; } int main(void) { int a=34,b=1; add_one(b); printf("a = %d, b = %d\n", a, b); return 0; }

Function Declaration Most software projects in C are composed of more than one file We want to be able to define the function in one file, and to use it in all files For this reason, the function must be declared in every file in which it’s called, before it’s called for the first time the declaration contains: the function name the data types of the arguments (their names are optional) the data type of the return value

Function Declaration #include int factorial(int a); /* Function Declaration! */ int main(void){ int num; printf("enter a number\n"); scanf("%d",&num); printf("%d!=%d\n",num,factorial(num)); return 0; } int factorial(int a){ int i,b=1; for(i=1; i<=a; i++) b=b*i; return b; }

Function Declaration stdio.h actually contains a large set of function declarations The #include directive tells the compiler to insert these declarations into the file, so that these functions could be called

The math library A collection of mathematical functions Need to include the header file math.h (#include ) Use functions of the library, e.g. double s,p; s=sqrt(p); Declared in math.h : double sqrt ( double x );

The math library sin(x), cos(x), tan(x) x is given in radians asin(x), acos(x), atan(x) log(x) sqrt(x) pow(x,y) – raise x to the yth power. ceil(x), floor(x) …and more

Exercise Write a function that uses the formula  2 /6=1/1+1/4+1/9+1/16+…+1/n 2 (where n goes to infinity) in order to approximate . The function should accept an argument n which determines the number of terms in the formula. It should return the approximation of . Write a program that gets an integer n from the user, and approximate  using n terms of the above formula.

Solution pi.c

Exercise Modify the previous function that approximates . The function should accept an argument specifying the desired accuracy, and keep adding terms until the contribution of the next term drops below this level. Write a program that gets a (small) double, epsilon, from the user, and approximates  within this function.

Solution pi_eps.c