i wanna put an integer into a string for example:
Code:
string name;
int num=8;
name= "George"
//
I want in the end of George to pas the num variable i ve seen some examples but ive not yet come up with a solution like str.insert();
Code:
Prompt the user to input a string,
and then output the sum of all the digits in the string.
Sample Run 1:
Input -> A111B222C
output -> 9
Sample Run 2:
Input -> ABC123XYZ32100000005555555555zzzzzz
output -> 62
Complete the function myitohex so that its converts 8-byte integer to string based hexadecimals.
Ex: printf("0x%s",myitohex(64,hex)); should printout "0x40" on the screen
char *myitohex(uint8_t i, char *result){
???
return result;
}
I wrote:
#include <stdio.h>
#include <stdint.h>
char *myitohex(uint8_t i, char *result){
*result = i + '0';
[Code] ....
0xp gets printed out which is wrong. I think *result = i + '0'; is wrong. What changes should i do on this row ?
I have an assignment to do..i have done it..but i need to do one more thing. Things sound like this: user inputs 6 integers program needs to check them if there are integer if not it has to output a message for the user if the input is integer it has to go further and work with the input. I have used this structure :
if ( ! ( cin >> temp ) ) {
cout<<"Input is not integer.This program will end ! "<<endl<<endl<<endl;
system("pause");
return 0;
}
Where I declared temp as being int since i started, the problem is after it gets the last input still waits for an input i will attach the source code if needed.
I have a homework that needs to verify if the input of the user is an integer using only loops no if statements Here's the problem:
A program is required that prompts the user for a number. The program will then print a series of asterisks to represent the number. If the user enters a number less than 1, the program stops. For example:
Enter a number: 5
*****
Enter a number: 3
***
Enter a number: 9
*********
Enter a number: 0
All user input must be validated:
- Check for non-numeric input when reading numeric input
- Check that values entered are within the expected range for their purpose, or in range based on the requirements statement
I have been trying to make a very simply programme that checks if the inputted information is an integer or not (i.e: that it contains no other characters).
I have tried using the isdigit function (but this only works for single characters). I have tried cin.clear, cin.ignore (1000) but this doesn't work either..
Any effective way to check if x in the following programme has been entered correctly
#include <iostream>
using namespace std;
int main() {
cout << "Please enter an integer (positive or negative)" << endl;
int x;
cin >> x;
HERE I WOULD LIKE CODE TO CHECK IF THE USERS INPUT IS VALID
}
I have a question about my program.. I would like to make a program that if you put "ABC" on INT, the program will restart again... below are my coding...but it not works..if i put ABC on INT, the "Invalid Mark.. please re-insert mark :" will repeat, repeat and repeat infinitely...
Code:
#include <stdio.h>
void calcul() {int mark;
char * grade;
if (!(scanf("%d",&mark))){printf("
}
[code]....
i am writing a program that requires me to write am input a float and an integer and save it to a txt file. When i try to compile my code i get an error "assignment from incompatible data type". ?
int *intconstant;
float *floatconstant;
int *value;
struct FILE *infileptr, *outfileptr;
infileptr = fopen("/home/brinkmann.brendon/assign14data.txt", "r");
outfileptr = fopen("/home/brinkmann.brendon/assign14report.txt", "w");
[Code]...
I have been tasked with making a diamond out of asterisks based on a given odd integer input. For some reason the bottom half of my diamond will not print. I'm not sure as to why.
Here is my code:
#include "stdafx.h"
#include <iomanip>
#include <iostream>
using namespace std;
int _tmain(int argc, _TCHAR* argv[]){
[Code] ....
Write a full C++ program that inputs three-digit integer, separates the integer into its individual digits and prints the digits separated from one another. For example, if the user types 549, the program should print;
5 4 9
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.
View 2 Replies View RelatedI am trying to make sure no floating point numbers can be input at any time in my roulette program. My code is too long to put in one post so I shall piece it out
#include <iostream>
#include <iomanip>
#include <random>
#include <cstdlib>
#include <ctime>
using namespace std;
// structure of arrays to pass to functions
[Code] .....
int a[20],x;
for(x=0; x<20; x++)
{
[Code]....
This should be the output
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: 1
Input: -1
Input: -1
Input: -1
Input: 1
Input: -1
Input: 1
Input: -1
Stop!
As an assignment for school , I've to write a program to find the sum prime factors of a user input integer.
E.g. 20 = 2 x 2 x 5 , Sum = 2 + 2 + 5 = 9
E.g. 10 = 2 x 5 , Sum = 2 + 5 = 7
My method for finding the result is as follows :
- Divide the number by increasing values if int i , starting from i=2.
- Once I get a value of i that can divide the number without giving me a remainder , I add this value of i to int sum and divide the number by i.
- I will repeat this process until the user input value is equal to 1.
My code is as shown:
#include<stdio.h>
int primecheck(int n); // Function to check if i is prime
int primesum(int n); // Function to sum the values of i that are prime
int main(void) {
int n;
int sum;
printf("Enter a number (> 1): "); //Prompting and scanning user input ,n
scanf("%d",&n);
sum = primesum(n);
[Code] .....
But for some reason I keep getting an incorrect result, it's as if it is missing out the last factor for each case.
Eg. 20 = 2 x 2 x 5 , the result I get is 4 , which is 2+2
Eg. 40 = 2 x 2 x 2 x 5 , the result I get is 6 , which is 2+2+2;
I've looked through my code numerous times.
i was trying to solve a problem in SPOJ and what i wanted to do is to accept an input number from the user and to convert it into a array of integer.
Code:
#include<stdio.h>
int * toarray(int *num);
int main(void)
{
int testCases;
}
[code]....
But whenever i try to print the array in the main function i get only two value and the rest address
Code:
1//number of testCases
23456 //input number
6
2293452
4
2293700
1974439125
Process returned 0 (0x0) execution time : 4.152 s
Press any key to continue. However, if i tried to print the array from within the function, it prints the numbers just fine.
print the array elements from the main program, so that i would be able to go on with the rest of it
I'm writing the program below to take an integer input for temperature and display a cheesy message for different ranges for each day of the week.
I tried using two if statements, the first to determine if the input is valid, and the else display that it was invalid and to try again.
The second set of if (if else) statements would evaluate a number input and display the cheesy message based on the value, or if it was out of range, to make you re input that days temperature.
I understand that the else statement is linked to the closest if statement, unless brackets are added to prevent this. I thought my brackets were correct, but I keep getting [Error]'else' without previous if'. The red lines are the if and else statement I'd like to be linked, and tried to do so by enclosing the 2nd if statement in brackets.
The assignment doesn't require the error checking, but if I don't use some form of it I believe it will just loop. Again I have to use the last 4 lines due to instructors rules, and have to use scanf and not strings or arrays.
This program will ask the user what the temperature will be for each day of the week. The integer input from the keyboard will be tested in a series of IF tests and a message will be printed depending on the temperature.
Code] ....
#include <stdio.h>
#define MAX 100
#define MIN 0
int main() {
int day = 1;
int temp;
[Code] ....
Write a C++ application program to accept a signed decimal integer as input and output the equivalent 2s complement version in 16-bit binary. Include a space between every four bits in the output string. The input will only be processed by the application if it falls in the valid range that can be represented in 2s complement format with 16 bits. The range of a decimal number from - to + is -32768 to 32767.
View 3 Replies View RelatedI have to write a program where the user will input integer numbers. How many numbers they enter is unknown, therefor you should use a repetition structure for the input. When the user is done, they will enter -1 to exit.
Create a dynamic array if the size=2( the initial size must be 2) Repeat until user enters -1.
I have to do this without using vectors.
This is what i have, I cannot figure out what to put in main. I was thinking of a do-while?
Code:
#include <iostream>
using namespace std;
void resize(int *[], int);
int main() {
int *listDyn;
int size=2;
[code].....
Very new to programming, and I know that there must be another way on inputting a string into each array cells not by just inputting it one by one, but as a whole. My code at the meantime is: [URL]
View 1 Replies View RelatedI'm trying to convert a string into a integer and when I call stoi(line) it causes the program to crash.
int main() {
vector<int> numbers;
int i;
string line;
ifstream myfile ("example.dat");
[Code] ....
The file being read from looks like:
3
12
23
34
12 34
12 12
34 23
23 23
How do I read integers to a string and store then into a vector array?
I need to add two input digits i.e. "1234567878887777" and "123344543543543"
I need to convert below value char *time="00.00.05".
I need to convert this to hh/mm/ss ....
struct time
{ int t;
int h,m,s;
};
int main() {
time t;
int totalsecond;
char A[10],B[10];
[Code] ....
It gives error at line no 12.
I'm trying to create a program that takes an integer input and displays each digit on a new line from right to left.
An example of this would be:
Enter an integer: 657
Digit (1): 7
Digit (2): 5
Digit (3): 6
So far I have this:
#include <stdio.h>
#include <math.h>
int main() {
int i, input, output;
printf("Enter an integer: ");
scanf("%d", &input);
if(input == 0)
printf("Digit (1): 0");
[Code] ....
It works perfectly fine, but I just learned that I am not allowed to use the #include <math.h> so I have to take out floor, log, and pow.
I'm pretty sure I have to do some sort of while or for loop to get the results, but how to do it.
I have problem with string compare. I want to compare the string user input with a string in binary. And I don't know how to do it. Problem in function login();Here is the code: And you also can download file in attachment too..
Code:
#include<conio.h>#include<dos.h>
#include<stdio.h>
#include<process.h>
#include<string.h>
char nsb=1;
char ch, password[20], passlogin[20], inputpass[20], checked[20];
FILE *fp;
}
[code]....
I wrote a program that reads a list from a file and stores it in a string type vector. Now, I want the user to input a word so that the program can search the vector to see if that word already exists. I have used every possible way of reading input from the console and storing it in order to compare with the vector but it never results in a match. When I print the input string and the vector string they are exactly the same thing (or at least print to the console as if they were). I've tried using getline; using cin direct to a string var; using cin to a char array and then casting to string using string str(arr); I even added a newline at the end just in case and STILL I cannot get a match.
vector <string> currentSet; //read a list in from a file and has 9 items in it
cin.ignore();
string line;
getline(cin, line);
if(line == vector[0]){//if printed to console line is HEAT and vector[0] is HEAT
cout<<"match"<<endl;
}