I wrote a function to squire number and try to cover all the input possibilities.
Overall it works fine with numeric input, but it starts a infinite loop of printing statements on screen when I enter alphabetical input.As all we know that inside computer single character like "A or a or b or B" so on is represented by integers and as i learned from my teacher that we can store single characters into a variable with integer data type. i am not talking about strings which means collection of characters . this program create problem with single character !
#include <iostream>
#include <string>
using namespace std;
void squire();
int main() {
squire();
}
void squire() {
double num = 1.0, pow = 1.0, Squire_Number = 1.0;
char ans;
reStart:
cout << "please Enter the Number: \n";
cin >> num;
cout << "please enter the nmber you want to power the number with: \n";
cin >> pow;
if (num > 0 && pow>0) {
for (int i = 1; i <= pow; i++) {
Squire_Number *= num;
}
cout << pow << " power of " << num << " is equil to : " << Squire_Number;
goto option;
}
else
{
cout << "Please enter Positve Integers. \n" ;
option:
cout<< "\nPease type 'Y' to Enter the values again OR type 'c' to Exit ! \n";
cin >> ans;
if (ans == 'y' || ans == 'Y') {
goto reStart;
} else if (ans == 'c' || ans == 'C') {
cout << "thank you for using our function. \n";
}
}
return;
}
Better try to read the input in an std::string, then parse the string to check if you only have numeric characters and then use std::atoi to convert the string in integer. One last recomendation, avoid to use goto instructions, this practice make a code difficult to read.
#include <iostream>
#include <string>
#include <cstdlib>
bool OnlyNumeric(const std::string& numStr)
{
size_t len= numStr.length();
int i;
for (i=0;i<len && numStr[i] <='9' && numStr[i] >='0';i++) ;
return i == len;
}
int main()
{
std::string inputStr;
int num;
do{
std::cout << "Input number:\n";
std::cin >> inputStr;
}
while (!(OnlyNumeric(inputStr) && (num=std::atoi(inputStr.c_str())) ));
std::cout << "Your number is : " << num;
return 0;
}
Related
How can I make this program display an error message if a user enters a negative number, and then prompt for a new positive number? Since this is a decimal to hexadecimal calculator the user should not be able to enter a negative number.
#include <iostream>
#include <math.h>
#include <string>
using namespace std;
int main()
{
bool play_again;
do
{
int dec_num, r;
string hexdec_num = "";
char hex[] = { '0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F' };
cout << "\n\n Convert a decimal number to hexadecimal number:\n";
cout << "---------------------------------------------------\n";
cout << " Input a decimal number: ";
cin >> dec_num;
while (dec_num > 0)
{
r = dec_num % 16;
hexdec_num = hex[r] + hexdec_num;
dec_num = dec_num / 16;
}
cout << " The hexadecimal number is : " << hexdec_num << "\n";
while (true) // loop asking user
{
string user_input;
cout << "again? ('y', 'n'):" << endl;
cin >> user_input;
if (user_input == "y")
{
play_again = true;
break;
}
else if (user_input == "n")
{
play_again = false;
break;
}
else
{
cout << "Invalid Input" << endl;
}
}
} while (play_again);
}
You can solve this problem by adding an if statement after the cin >> dec_num; so you can check its value if it's under the 0 or no
add this line under the line cin >> dec_num;:
if (dec_num < 0)
{
//do whatever you want here
}
Hello this is my first program with a do-while loop and its taken me a little while to get it down. I need to have the user enter 2 numbers, and raise the first number to the second number. I have finally got the coding to ask if "they would like to raise another number by a power?" and when they say yes and enter 2 new numbers the total adds the total from the first 2 numbers entered with the second set of numbers and so on. Can someone help me out with this problem? Here is the coding and a picture to help y'all out!
#include <iostream>
using namespace std;
int main()
{
int num;
int pow;
int p;
int power = 1;
char yesno = 'y' || 'Y';
do
{
cout << "Enter a number: ";
cin >> num; "\n";
cout << "Enter the power to raise: ";
cin >> pow; "\n";
for (p = 1; p <= pow; p++)
{
power = power * num;
}
cout << "The total is: " << power << endl;
cout << "\n\n";
cout << "Would you like to raise another number by a power? [Y/N]";
cin >> yesno;
} while (yesno != true);
}
The problem of the ever-increasing answer is that power is not being reset inside the do-while loop, so the last value is being carried forward into the next loop. You need reset it at the top of the loop.
Another problem with the code is that the exit condition would never occur.
Try this instead:
int main()
{
int num;
int pow;
int p;
int power;
char yesno;
do
{
power = 1; // <<<<<< reset power here
cout << "Enter a number: ";
cin >> num; "\n";
cout << "Enter the power to raise: ";
cin >> pow; "\n";
for (p = 1; p <= pow; p++)
{
power = power * num;
}
cout << "The total is: " << power << endl;
cout << "\n\n";
cout << "Would you like to raise another number by a power? [Y/N]";
cin >> yesno;
} while (yesno == 'y' || yesno == 'Y'); // <<<<< test for 'yes' response
}
When you reach line } while (yesno != true); and loop back to do {, the variable power still holds the previous num^pow. You will need to assign power = 1 after do {.
#include <iostream>
// you also need
#include <cmath> // for pow()
using namespace std;
int main()
{
// int num; Declare variables where they're used. As locally as possible.
// int pow;
// int p;
// int power = 1;
// char yesno = 'y' || 'Y'; I don't know what you were trying to do here
// the expression 'y' || 'Y' will always be true
// and evaluate to some value different from null
// wich will be assigne to yesno. But with no con-
char yesno; // sequences since it later gets overwritten by
do // cin >> yesno; There is no need to initialize
{ // this variable.
cout << "Enter a number: ";
int num;
cin >> num; "\n"; // the statement "\n"; has no effect.
cout << "Enter the power to raise: ";
int pow;
cin >> pow; "\n"; // again. no effect.
// for (p = 1; p <= pow; p++) as user4581301 has pointed out in the
// comments it is more ... natural in C
// to loop from 0 to < max:
int power = 1; // now its time to declare and define power ;)
for(int p = 0; p < pow; ++p) // notice that you can declare variables
{ // in the init-statement of a for-loop
// power = power * num; shorter:
power *= num;
}
cout << "The total is: " << power << /* endl; + 2 x '\n' gives: */ << "\n\n\n";
// cout << "\n\n";
cout << "Would you like to raise another number by a power? [Y/N]";
cin >> yesno;
// } while (yesno != true); that condition will most likely always be true
// since the user would have a hard time to input
// a '\0' character, which would evaluate to false
// better:
} while(yesno == 'y' || yesno == 'Y' );
}
done.
Without clutter:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
char yesno;
do {
cout << "Enter a number: ";
int num;
cin >> num;
cout << "Enter the power to raise: ";
int pow;
cin >> pow;
int power = 1;
for(int p = 0; p < pow; ++p)
power *= num;
cout << "The total is: " << power << "\n\n\n";
cout << "Would you like to raise another number by a power? [Y/N]";
cin >> yesno;
} while(yesno == 'y' || yesno == 'Y' );
}
I would like to read numbers into a static array of fixed size 10, but the user can break the loop by entering character E.
Here's my code:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int myArray[10];
int count = 0;
cout << "Enter upto 10 integers. Enter E to end" << endl;
for (int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
cin >> myArray[i];
if (myArray[i] != 'E')
{
cout << myArray[i] << endl;
count++;
}
else
{
break;
}
}
cout << count << endl;
system("PAUSE");
return 0;
}
However, I get the following results while entering E:
Enter upto 10 integers. Enter E to end
Enter num 1:5
5
Enter num 2:45
45
Enter num 3:25
25
Enter num 4:2
2
Enter num 5:E
-858993460
Enter num 6:-858993460
Enter num 7:-858993460
Enter num 8:-858993460
Enter num 9:-858993460
Enter num 10:-858993460
10
Press any key to continue . . .
How can I fix this code in the simplest way?
cin fails for parsing character 'E' to int. The solution would be to read string from user check if it is not "E" (it is a string not a single char so you need to use double quotes) and then try to convert string to int. However, this conversion can throw exception (see below).
Easiest solution:
#include <iostream>
#include <cmath>
#include <string> //for std::stoi function
using namespace std;
int main()
{
int myArray[10];
int count = 0;
cout << "Enter upto 10 integers. Enter E to end" << endl;
for (int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
std::string input;
cin >> input;
if (input != "E")
{
try
{
// convert string to int this can throw see link below
myArray[i] = std::stoi(input);
}
catch (const std::exception& e)
{
std::cout << "This is not int" << std::endl;
}
cout << myArray[i] << endl;
count++;
}
else
{
break;
}
}
cout << count << endl;
system("PAUSE");
return 0;
}
See documentation for std::stoi. It can throw exception so your program will end suddenly (by termination) that is why there is try and catch blocks around it. You will need to handle the case when user puts some garbage values in your string.
Just use:
char myArray[10];
because at the time of taking input console when get character then try to convert char to int which is not possible and store default value in std::cin i.e. 'E' to 0 (default value of int).
Use below code:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
char myArray[10];
int count = 0;
cout << "Enter upto 10 integers. Enter E to end" << endl;
for (int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
cin >> myArray[i];
if (myArray[i] == 'E')
{
break;
}
else
{
cout << myArray[i] << endl;
count++;
}
}
exitloop:
cout << count << endl;
system("PAUSE");
return 0;
}
Output:
Enter upto 10 integers. Enter E to end
Enter num 1:1
1
Enter num 2:E
1
sh: 1: PAUSE: not found
If you debug this, you will find all your myArray[i] are -858993460 (=0x CCCC CCCC), which is a value for the uninitialized variables in the stack.
When you put a E to an int variable myArray[i]. std::cin will set the state flag badbit to 1.
Then when you run cin >> myArray[i], it will skip it. In other words, do nothing.
Finally, you will get the result as above.
The problem is that attempting to read E as an int fails, and puts the stream in an error state where it stops reading (which you don't notice because it just doesn't do anything after that) and leaves your array elements uninitialized.
The simplest possible way is to break on any failure to read an integer:
for(int i = 0; i < 10; i++)
{
cout << "Enter num " << i + 1 << ":";
if (cin >> myArray[i])
{
cout << myArray[i] << endl;
count++;
}
else
{
break;
}
}
If you want to check for E specifically, you need to read a string first, and then convert that to an int if it's not E.
As a bonus, you need to handle everything that's neither int nor E, which complicates the code a bit.
Something like this:
int count = 0;
string input;
while (cin >> input && count < 10)
{
if (input == "E")
{
break;
}
istringstream is(input);
if (is >> myArray[count])
{
cout << myArray[count] << endl;
count++;
}
else
{
cout << "Please input an integer, or E to exit." << endl;
}
}
I am writing a game, where it should do below
Randomly generate a number. Check whether it is a Prime or not. Store "y" if its a Prime else "n"
I will ask the user whether it is prime or not. If he answers correctly I will proceed. Otherwise I will say you lost.
Code sample I tried is as follows:
cout<<"\n \n a no. of prime no.s will be displaed to you ,you will have to decide if it is prime or not within the given time\n \n ";
char ch1,ch2;
for(int i=1;i<=20;i++)
{
int a= rand( ) % 20;
cout<< "The number is:";
cout<<a;
for(int l=2;l<=a-1;l++)
{
if(a%l==0)
ch2='n';
else
ch2='y';
}
cout<<ch2;
cout<<"\n\n Enter ""y"" If the number is Prime Else Please enter ""n""\n";
cin>>ch1;
if(ch1 == ' ')
{
cout<<"\n \n u lost";
break;
}
else if(ch1==ch2)
continue;
else if(ch1!=ch2)
{
cout<<"\n \n u lost";
break;
}
}
My code is not determining whether it is Yes or No. It is everytime storing "Y" only
You made a couple of mistakes
#include <math.h>
#include <iostream>
using namespace std;
int main() {
char ch1, ch2;
for (int i=i; i<=20; i++) {
int a = rand () % 20;
cout << "The number is: " << a << endl;
ch2 = 'y';
//You only need to loop until its square root; after that is repetition
for (int l=2; l<=sqrt(a); l++) {
//You need to break if a divisor is found, else it is always going to say it's prime
if (a%l==0){
ch2 = 'n';
break;
}
}
cout << ch2 << endl;
cout << "Enter 'y' if number is prime else 'n'" << endl;
cin >> ch1;
if (ch1 != ch2) {
cout << "You lost" << endl;
break;
}
}
}
I also did some cleanup of your code
May be this one helps
#include <iostream>
#include <ctype.h>
#include <string.h>
using namespace std;
int main()
{
int n, i;
char res='y';
char ans;
n =rand( ) % 20;
cout << "The Number is:" << n << endl;
for(i = 2; i <= n / 2; ++i)
{
if(n % i == 0)
{
res='n';
break;
}
}
cout << "Enter 'y' if it is a prime number and 'n' if it is a non-prime number"<< endl;
cin >> ans;
ans = tolower(ans);
if (ans==res)
{cout << "You win";}
else
{cout << "You Lost";}
return 0;
}
is it possible, say your trying to do calculations so the primary variable type may be int... but as a part of the program you decide to do a while loop and throw an if statement for existing purposes.
you have one cin >> and that is to take in a number to run calculations, but you also need an input incase they want to exit:
Here's some code to work with
#include <iostream>
using namespace std;
int func1(int x)
{
int sum = 0;
sum = x * x * x;
return sum;
}
int main()
{
bool repeat = true;
cout << "Enter a value to cube: " << endl;
cout << "Type leave to quit" << endl;
while (repeat)
{
int input = 0;
cin >> input;
cout << input << " cubed is: " << func1(input) << endl;
if (input = "leave" || input = "Leave")
{
repeat = false;
}
}
}
I'm aware they wont take leave cause input is set to int, but is it possible to use a conversion or something...
another thing is there a better way to break the loop or is that the most common way?
One way to do this is read a string from cin. Check its value. If it satisfies the exit condition, exit. If not, extract the integer from the string and proceed to procss the integer.
while (repeat)
{
string input;
cin >> input;
if (input == "leave" || input == "Leave")
{
repeat = false;
}
else
{
int intInput = atoi(input.c_str());
cout << input << " cubed is: " << func1(intInput) << endl;
}
}
You can read the input as a string from the input stream. Check if it is 'leave' and quit.. and If it is not try to convert it to a number and call func1.. look at atoi or boost::lexical_cast<>
also it is input == "leave" == is the equal operator. = is an assignment operator.
int main() {
cout << "Enter a value to cube: " << endl;
cout << "Type leave to quit" << endl;
while (true)
{
string input;
cin >> input;
if (input == "leave" || input == "Leave")
{
break;
}
cout << input << " cubed is: " << func1(atoi(input.c_str())) << endl;
}
}
you can use like
int input;
string s;
cint>>s; //read string from user
stringstream ss(s);
ss>>input; //try to convert to an int
if(ss==0) //not an integer
{
if(s == "leave") {//user don't want to enter further input
//exit
}
else
{
//invalid data some string other than leave and not an integer
}
}
else
{
cout<<"Input:"<<input<<endl;
//input holds an int data
}