C++ :: Return The Digit At User Specified Index Of Integer
Oct 10, 2014
Return the digit at the user specified index of an integer. If the integer has n digits and the index is NOT in the range 0 <=index <n return -1 Start the digit numbering at 0. Example, if user input is 4 (index) and the integer equals 123456790 the return value for the function is 5 (start index at 0) ; if user input is 40 (index) and the integer equals 123456790 the return value for the function is -1
#include <iostream>
#include <istream>
#include <cstdlib>
#include <cassert>
#include <string>
using namespace std;
int getIndex(int, int);
#include <cstdlib> #include <iostream> #define TRUE 1 #define FALSE 0 using namespace std; typedef int Bool;
[Code] ....
Gives repeated digits in an integer but only in one condition : Only if the repeated digit is the result of n%10 where n is the integer the user writes. If the repeated digit is not the result of n%10 , then the compiler gives a wrong result.
so the question is : how to make this code gives the repeated digit in an integer (regardless the fact that the repeated digit is the result of n%10 or not and especially with making the minimum of changes on the code)????????? ?????
Assignment: Take an integer keyed in from the terminal and extract and display each digit of the integer in English. Ex. 932 --> nine three two
Code:
/*This program takes an integer keyed in from the terminal and extracts and displays each digit of the integer in English.*/ #include<stdio.h> int main(void) { //DECLARE VARIABLES int num; }
[code]....
I don't know how the program works if the integer is more than one digit.
I need understanding the logic behind this function. The function is supposed to "Return a pointer to the character at the index given" . For example, what exactly are we declaring at "(char * const c, int index)"? where does "index" come from or equal to in the body of the function?
Code: 1 2 3 4 5 6 7 8
char * GetValueAtIndex(char * const c, int index) { int i = 0; char *p = c; while (i++ != index) p++; return p; }
I want this code to return the value of the bit at position bitIndex. It can be either false or true. The problem is, that it always returns false, even thought I enter 16 as my number, so the 5th bit should be true.
I need to create a program that lists off each digit of an integer and then display the sum off all the digits in that integer. I know that sepereatly the sum function i wrote works. But the first part which i try to list off the digits work but in reverse order which i dont know how to correct. and for some reason that i cant figure out this is affecting the sum output.
#include <iostream> using namespace std; int digcount (int x) {;
I have been working on some C++ code that doesn't seem to be going right. I'm wanting it to read a (three-digit) integer representing the value to be encrypted, a (one-digit) integer representing the encryption key, encrypt the value and print the encrypted value. The encrypting method used is that each digit in the given number is replaced by ((the sum of that digit plus key) modulo 10) then the first and last “encrypted” digits are swapped.
For example, if the number entered was 216 and the key given was 7, after applying the encryption procedure described the first digit (2) would become 9, the middle digit (1) would become 8 and the last digit (6) would become 3. The first and last encrypted digits are then swapped. The program displays the encrypted number: that is 389 in this case.
#include <iostream> using namespace std; int main() { int isolateDigits(); int replaceDigits(); int swapDigit1withDigit3();
The problem states that i need to accept any pin number between "00000" and "99999". I need to read the PIN the user enters as an integer. The problem is that if the PIN is read as an integer, the PIN "01111" will be "1111" which is invalid and the pin "00001" would be read as "1" which is also invalid. How would I go about fixing this problem for PIN numbers that start with a "0"? Again I cannot read the PIN as a char array or string.
convert a positive integer code into its english name equivalent for digit. A valid code is of size between four (4) to six (6) digits inclusive. A zero is not allowed in the code.
example : if the input is 234056 the output is : INVALID CODE (PRESENCE OF ZERO) if the input is 23456 the output is : TWO THREE FOUR FIVE SIX if the input is 9349 the output is : NINE THREE FOUR NINE if the input is 245 the output is : INVALID CODE (3 DIGITS) if the input is 2344567 the output is : INVALID CODE (7 DIGITS)
step 1 : input code step 2 : count the number of digits in the code step 3 : if there is a zero in the code, "INVALID CODE (PRESENCE OF ZERO)" go to step 4 step 4 : if number of digits is mode or equal than 4 and less or equal than 6, go to step 5 else display the following message "INVALID CODE (<number of digits> DIGITS) step 5 : call a function called digit_name to convert each digit into its equivalent english name. display the result step 6 : print the digits in reverse order eg; if input is 13453, reverse order is 35431
Here is my code to find the index of a string array whose string is equal to the query string. I have checked the program can return the correct index, but the cout result is totally wrong.
#include <iostream> #include <string> using namespace std;
I'm new to C programming and in my first computer science class. Our assignment was to write a program that displays each digit of an integer in English. I wrote the following but cannot figure out why it won't display zeros. When I execute and type in the number 1,000, I get "zero."
Code: #include <stdio.h> int main (void) { int x, y = 0; printf ("This program displays each digit of an integer in English.
I have an assignment which requires me to do the following:
Required to write a function that finds an integer in an array and returns its corresponding index. The function must be called findNumber. It must have FOUR parameters:
- The first parameter is the array to be searched - The second parameter is the integer to be found within the array - The third parameter is the size of the array - The fourth parameter is an integer that indicates whether the array is sorted. A value of 1 means the array is sorted; a value of zero means the array is not sorted.
Since a function can only return one value(To return the position of a required integer in an array in this instance) I have tried to make use of pointers to try and return a value stating whether the array is sorted or not.This is my code : (It compiles perfectly but it does not produce any outputs)
Code:
#include <stdio.h> #define SIZE 10 size_t findNumber(int *sort, const int array[],int key,size_t size); int main(void){ int a[SIZE]; size_t x;
I am trying to write up something to have a user to enter a four digit number. Only four digits, Ex: 0001, 0116, or 9999. There is no getting around the selection 0001 or 0002. I understand if not done correctly, the first three 0's will be ignored.
I've been just playing around with what I have below but I just don't remember how to do it nor can I find a good example online to figure it out. I know it is not correct just typing to try to remember. I am aware that it is gibberish right now, this is just me brainstorming.
int number; cout<<("Please enter the four digit number(Ex: 0001): "); cin>> setw(2) >> number; cout<<)"Please enter four digit date. Two digits for month and two digits for year: "); cin>> date; if (number< || number > 30) cout << "Invalid choice. Try again." << endl; cin.clear();
Im trying to create a program that has the user input a 5 digit number. If it's between 10000 & 99999, it will do one thing..(just saying 'yes' for now. Outside those numbers will prompt the user to input again. However, if the user inputs the exact digits 76087, it should display 'term'.
This current code is displaying 'term' whenever the user inputs the 5 digits.
Code:
#include <iostream> using namespace std; int main() { int pin; cout << "Welcome to Movie Food Enter your 5-digit pin code: " ;
So I have a simple calculator that does a few operations (+ - * / %) Pretty basic stuff
I declared int x, y for the numbers, char operation, and float result.
the code is based on switch(operation)
The program is running alright, but when I divide 8/7 it returns 1 as the result, I tried changing the x and y to float but that won't work because of the case '%'
I also tried making local float variables in the case'/' but it won't compile "E2126 Case bypasses initialization of a local variable"
How can I make the division work and return a float value?
Write a function which will take 3 arguments. The function needs to return a new number which is formed by replacing the digit on a given position in the number with a digit which is carried as an argument (the position in the number is counted from right to left, starting with one). Write a main program which will print the newly formed number.
Examples: A function call of 2376, 3 and 5 should return the number 2576 A function call of 123456, 4 and 9 should return the number 129456
What I succeeded to do so far: Figure out the logic for swapping the digit and write working code for it (in the main function).
What I failed to do so far: Write a function which will return the desired result.
What is my problem: I tried writing a function to do this, but as you see from my calculations, my result is divided in 3 parts. I don't know how to return more variables from a function.
Code:
#include <stdio.h> int main() { int inputNumber, swapPosition, swapDigit; scanf("%d%d%d", &inputNumber, &swapPosition, &swapDigit); int i, numberPart1 = inputNumber; for (i = 1; i <= swapPosition; i++)
I know that if I just use "return" by itself the warning goes away but fails to exit when the error occurs. I also believe this may not be the correct use of stderr. But I need the program to exit when an error has occurred.
Firstly, I'd like to say that I'm relatively new to functions and my course standards state that I'm now supposed to relay information from the user to the program via functions, this means that I'm no longer able to use 'printf/scanf' combos in the main function.
Is it possible to do it via functions? I've tried the following method for example, but to no avail.
Code: #include <stdio.h> int getlowR(); int main(void) { getlowR();
[Code] ....
I think the following works but then lowRange is native to the function, isn't it? How can I use it outside the function? For example, how would I go about actually printing the value acquired by the function in main?
I'm having a problem with the two while statements in my UDF.
- 1. It will run both while loops twice...?
- 2. It now goes into a continuous loop.
- 3. When it did work, it would only return 1 value to the main()...?
#include <iostream> // for use of "cin" & "cout", endl... #include <iomanip> // for formatting setw function #include <cmath> // for the general math computations #include <string> // for creating descriptive strings #include <sstream> // used to convert a string to an integer //user defined function int userValue (int);
