If Value Is Not Within Range Prompt Again C++

Functions in C++

A role is a lawmaking module that performs a single task. Some examples such as sorting, search for a given particular, and invert a square matrix. Once a office is created it is tested extensively. Later this, information technology becomes a part of the library of functions. A user can use such a library role equally many times as needed. This idea improves software robustness and also reduced code development time. Functions are classified into 2 categories: system defined, and user defined. Examples for a organisation defined math library functions are: sqrt, exp, and log. In this course, we only focus our attention to user defined functions

User defined functions

As the name suggests these functions are defined and created past users. Functions in C++ agree with the notion of functions in mathematics. For example, they let functional composition such as f(f(p),q,f(f)). Yet, C++ functions either render ane value or no value. Consider the following instance. This is a user defined function that computes the value of factorial of a small integer.

            // This the factorial function int fact(int n){  //This function maps input integer north to another  int ans =1;	// ans is the local variable of the function for(int i= 2; i<= north; i++)	// i also a local scratch pad variable ans *=i; render ans;	   // function returns the answer to the caller }        

The integer variable n is the input to the function and information technology is also chosen the parameter of the function.

If a part is defined after the main() function then its paradigm must be specified int fact (int);

at the elevation in club to avoid forward reference error beingness flagged past the compiler. A image definition specifies the name of the part, number and type of each parameter and the information blazon of the output produced.

All pieces of the code such as a role epitome, a calling main program and the complete function definition are shown in the following

#include <iostream> using namespace std; int fact(int northward); // function image  int main(){ int nv,ans; 	cout<<" Input n:"; 	cin>> nv; 	ans = fact(nv);// Function is called 	cout<<"Answer is:"<<ans<<endl; } 	 int fact(int northward){  // This office maps input int to another int 	int ans =i; // ans is the local variable of the function 	for(int i= 2; i<= n; i++)	// i also a local sractch pad variable 	     ans *=i; 	return ans; // function returns the answer to the caller }        

Typical outputs that event from the execution of this lawmaking are shown beneath:

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Function examples

Example 1

Write a function in order to make up one's mind the maximum of two integers x and y Answer:

//this function finds the bigger number of the two input variables int max2(int x, int y){ int result; 		if ( x > y )//if "10" is bigger than "y" 			event =ten;//output x 		else 			result = y;//if non bigger than "y" output "y" 		return result; }        

This function accepts two integers x and y as its input. It chooses x as the result if ten is larger than y; otherwise y is chosen for the result.

#include <iostream> using namespace std; int max2(int north,int m); //role prototype int main() {     int x,y,ans; 	cout<<" Input x, y:"; 	cin>> x >> y; 	ans = max2(10, y); // Office is called 	cout<<"Answer is:"<<ans<<endl;     return 0; } int max2(int x, int y){     int effect;     if ( x > y ) 		result =ten; 	else 		result = y; 	return result; }        

Instance two:

Write a function to observe the maximum of the three integers x, y and z. Answer:

int max3(int ten, int y, int z){ int result=10; 	if(result < y) //if 10 is smaller than y 		outcome = y; //output y 	if (effect < z) //if x is smaller than z 		result= z; //output z     else 	render result; }        

To start with x is causeless to be the result. Then the result is compared against y and z. if y or z is larger than the outcome y or z is chosen for the result respectively.

#include <iostream> using namespace std;  int max3(int due north,int k,int o); //function epitome  int principal() {     int x,y,z,ans; 	cout<<" Input ten, y, z:"; 	cin>> x >> y >> z; 	ans = max3(x, y, z); // Role is called 	cout<<"Answer is: "<<ans<<endl;      return 0; }  int max3(int x, int y, int z){ int outcome=ten; 	if(upshot < y) //if x is smaller then y 		result = y; //output y 	if (issue < z) //if 10 is smaller and so z 		event= z; //output z     else  	return result; }        

The aforementioned function can composed around max2 function and this is shown below:

int max3(int x, int y, int z){ int result; result=max2(x,max2(y,z)); //this shows the max2 role  						 //inside the max 3 function 	render event; }        

This calculation shows the utilise of functional composition that is max(x, y, z) = max(10, max(y,z))

#include  <iostream> using namespace std;  int max2(int ten,int y); int max3(int north,int m,int o); //function prototype  int main() {     int x,y,z,ans; 	cout < <" Input x, y, z:"; 	cin>> 10 >> y >> z; 	ans = max3(x, y, z); // Office is chosen 	cout < <"Respond is: " < <ans < <endl;      render 0; } int max2(int x, int y){ int result; 		if ( x > y ) //if "x" is bigger than "y" 			result =x; //output ten 		else 			result = y; //if not bigger than "y" output "y" 		return upshot; }  int max3(int ten, int y, int z){ int result; 	result=max2(ten,max2(y,z)); 	return result; }        

Instance 3

Write a function that will return how many iii digit natural numbers are divisible by either iv or five merely not both.

int div45(){ int count, i;//initialize the count and the pointer     for(i=100,count =0; i < 1000; i++) //this goes through all the #'s   //that are 3 digits 		if ((i % 4 == 0 && i % v !=0 ) || 		   (i % 4 != 0 && i % five == 0)) //to prevent double counting 							   //i%v == 0 and i%4 == 0 finds 							   //the numbers divisible by four 							   //or 5 		   count++; 	return count; }        

Answer: this function commencement goes through all the possibilities for three digit numbers first, so it checks if that current "i" value is divisible by 4 OR 5 while at the same time checking that information technology is not double counting the other number.

Instance four

Write a function that will return truthful if the given number due north is a prime number, otherwise it must return false.

bool isPrime(int north){ int i; 	for(i=2; i < n/2; i++) 		if ( due north% i == 0) 			return false; 	render true; }        

Answer: This function checks for all divisors from 2 upward to and including north/ii -1. Information technology will get out beforehand if it finds a divisor earlier completion of the loop. If the code completes the loop means the number does not have a divisor and therefore information technology is a prime number.

Instance 5

Write a function that will return true if the given number is a palindrome otherwise it must return false. A number is said to exist a palindrome if it reads the same whether it is scanned from left to right or vice-versa. For instance, 506605 is a palindrome while 5123 is not.

bool isPalind(int northward){ int nr=0; int org=northward; 	while ( n != 0){ 		nr = x * nr + n % 10; 		north /= 10; 	} 	render ( org == nr); }        

Respond: The logic behind a palindrome is to reverse the given number n. Information technology is done past the while loop. The given number n is a palindrome if and only if the original northward and its reversal are identical.

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Source: https://www.cpp.edu/~elab/ECE114/Functions.html

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