C++ Loops - inputting integers until user quits - c++

well I'm trying to get the user to input an integers as much as they want, until they input a negative number. 1st step was to use the ATOF function to convert string to number(which I did), and then allow the user to input integers(I only manage to do once just to see if I can use the atof function correctly.
Any help/tips is appreciated on giving me the right direction.
Here is my code thus far:
#include <iostream>
#include <string>
int main() {
using namespace std;
char buffer[256];
char tempBuff[256] = {'\n'};
double result;
int count = 0;
cout << "Testing " << endl;
cout << "Enter Any integers: ";
cin.getline(buffer,256);
for(int i = 0; i < strlen(buffer); i++)
{
if(isdigit(buffer[i]))
{
tempBuff[count] = buffer[i];
count++;
}
}
if (atof(tempBuff) > 0) {
result = atof(tempBuff) / 2;
}
cout << endl << "The integer you put was: " << tempBuff
<< " And dividing the integers "<< result << endl;
cin.ignore();
return 0;
}

How is atof supposed to know how many valid digits tempBuff contains? The atof function only accepts a C-style string as its input. Otherwise, it has no way to know how many characters are valid.
You can use tempBuff[count] = 0; before the call of atof. A C-style string is terminated by a zero byte.

Related

C++ isdigit() Query for converting a char array to an int array

I am trying to convert an input character array, to an int array in c++.
Inputs would be in a format like: 'M 911843 6', where the first value of the char array is a uppercase letter, which I convert to an ASCII value and -55.
Edit: I should also mention I just want to use the iostream library
The last value of the char array can be a letter or number also.
I want to retain the exact number value of any input in the char array, which is why I convert to an ASCII value and -48, which retains the same number, but stored as an int value:
I use the checkdigit() function to check if the char input is a number or not.
The difficulty I am facing is that the input will always have a blank space at i[1] and i[8] (if we count i[0] as the first value) - so I try to give them an int value of 0 (int of a " " is 0)
Upon several debugging attempts, I found that it is after the blank space is given a 0 value, the output in my for loop keeps outputting the wrong values, I suspect it has something to do with the isdigit() function in my for loop.
If the spaces from M 911843 6 were removed, the int output is usually fine, e.g. a char input of
M9118436 will return an int array of [22][9][1][1][8][4][3][6].
The output with spaces: [22][0][-183][-120][37][-118][-59][72][0][-55]
Ideal output: [22][0][9][1][1][8][4][3][0][6]
The code is listed below, any help or advice would be greatly appreciated, thanks!!
#include <iostream>
using namespace std;
int main() {
char a[10];
int z[10];
int i = 0;
int r; //result of the isdigit check (0 or 1)
cout << "in ";
cin >> a;
for (int i = 0; i < 10; i++) {
r = isdigit(a[i]);
if (r == 0) {
if (i==1 || i==8)
z[i] = 0;
else z[i] = int(a[i]) - 55;
}
else {
z[i] = int(a[i]) - 48;
}
}
cout << z[0] << "\n" << z[1] << "\n"<< z[2]<< "\n" << z[3] << "\n"<< z[4] << "\n"<< z[5] << "\n"<< z[6] << "\n"<< z[7]<< "\n" << z[8] << "\n"<< z[9];
return 0;
}
The problem is that cin >> a; does not read sizeof(a) characters, but up to the first space character and will terminate that with a null.
That means that you array will containt 'M', '\0' and 8 uninitialized characters. You must read the characters one at a time with unformatted reads:
for (auto& c : a) {
cin.get(c);
if (!cin) {
cerr << "Incorrect input\n";
return EXIT_FAILURE;
}
}
Just a follow on from Serge's answer which gave me a good understanding of how strings are read - I solved my problem using cin.getline() function.

C++ convert binary to decimal from string input

I have a problem because i have string input and i want to convert it to decimal.
Here's my code :
#include <iostream>
#include <string>
#include <stdlib.h>
using namespace std;
string inputChecker;
int penghitung =0;
int main(){
string source = "10010101001011110101010001";
cout <<"Program Brute Force \n";
cout << "Masukkan inputan : ";
cin >> inputChecker;
int pos =inputChecker.size();
for (int i=0;i<source.size();i++){
if (source.substr(i,pos)==inputChecker){
penghitung +=1;
}
}
if (source.find(inputChecker) != string::npos)
cout <<"\nData " << inputChecker << " ada pada source\n";
else
cout <<"\nData "<< inputChecker <<" tidak ada pada source\n";
cout <<"\nTotal kombinasi yang ada pada source data adalah " <<penghitung <<"\n";
cout <<"\nDetected karakter adalah " <<inputChecker;
cout <<"\nThe Decimal is :" <<inputChecker;
}
I want to make that last one which is "Decimal" to show converted inputChecker from binary to decimal. Is there any function to easily convert from binary to decimal in c++?
Thanks in advance :))
Use std::strtol with 2 as the base. For example,
auto result = std::strtol(source.c_str(), nullptr, 2);
For brute force:
static const std::string text_value("10010101001011110101010001");
const unsigned int length = text_value.length();
unsigned long numeric_value = 0;
for (unsigned int i = 0; i < length; ++i)
{
value <<= 1;
value |= text_value[i] - '0';
}
The value is shifted or multiplied by 2, then the digit is added to the accumulated sum.
Similar in principle to converting decimal text digits to internal representation.

Super basic code: Why is my loop not breaking?

for(int i=0;i<50;i++,size++)
{
cin >> inputnum[i];
cout << size;
if(inputnum[i] == '.')
{
break;
}
}
The break breaks the input stream but the size keeps outputting.
The output of size is 012345678910111213...474849.
I tried putting size++ inside the loop but it made no difference. And size afterwards will be equal to 50, which means it went through the full loop.
I forgot to explain that I added the cout << size within the loop to debug/check why it outputted to 50 after the loop even if I only inputted 3 numbers.
I suspect that inputnum is an array of int (or some other numeric type). When you try to input '.', nothing actually goes into inputnum[i] - the cin >> inputnum[i] expression actually fails and puts cin into a failed state.
So, inputnum[i] is not changed when inputting a '.' character, and the break never gets executed.
Here's an slightly modified version of your code in a small, complete program that demonstrates using !cin.good() to break out of the input loop:
#include <iostream>
#include <ostream>
using namespace std;
int main()
{
int inputnum[50];
int size = 0;
for(int i=0;i<50;i++,size++)
{
cin >> inputnum[i];
if (!cin.good()) {
break;
}
}
cout << "size is " << size << endl;
cout << "And the results are:" << endl;
for (int i = 0; i < size; ++i) {
cout << "inputnum[" << i << "] == " << inputnum[i] << endl;
}
return 0;
}
This program will collect input into the inputnum[] array until it hits EOF or an invalid input.
What is inputnum ? Make sure t's a char[]!! with clang++ this compiles and works perfectly:
#include <iostream>
int main() {
int size = 0;
char inputnum[60];
for(int i=0;i<50;i++,size++) {
std::cin >> inputnum[i];
std::cout << size;
if(inputnum[i] == '.') {
break;
}
}
return 0;
}
(in my case with the following output:)
a
0a
1s
2d
3f
4g
5.
6Argento:Desktop marinos$
Your code seams OK as long as you're testing char against char in your loop and not something else.. Could it be that inputnum is some integral value ? if so, then your test clause will always evaluate to false unless inputnum matches the numerical value '.' is implicitly casted to..
EDIT
Apparently you are indeed trying to put char in a int[]. Try the following:
#include <iostream>
int main() {
using namespace std;
int size = 0;
int inputnum[50];
char inputchar[50];
for(int i=0;i<50;i++,size++) {
cin >> inputchar[i];
inputnum[i] = static_cast<int>(inputchar[i]); // or inputnum[i] = (int)inputchar[i];
cout << size << endl; // add a new line in the end
if(inputchar[i] == '.') break;
}
return 0;
}
Then again this is probably a lab assignment, in a real program I'd never code like this. Tat would depend on the requirements but I'd rather prefer using STL containers and algorithms or stringstreams. And if forced to work at a lower-level C-style, I'd try to figure out to what number '.' translates to (simply by int a = '.'; cout << a;`) and put that number directly in the test clause. Such code however might be simple but is also BAD in my opinion, it's unsafe, implementation specific and not really C++.

Creating an array

write a program that let's the user enter 10 numbers into an array. The program should then display the largest number as and the smallest number stored in the array.
I am very confused on this question that was on a previous exam and will be on the final. Any help would be appreciated! This is what I had on the test and got 3/15 points, and the code was almost completely wrong but I can post what I had if necessary, thanks! For creating the array, i can at least get that started, so like this?
#include <iostream>
using namespace std;
int main()
{
int array(10); // the array with 10 numbers, which the user will enter
cout << "Please enter 10 numbers which will be stored in this array" << endl;
cin >> array;
int smallest=0; //accounting for int data type and the actual smallest number
int largest=0; //accounting for int data type and the actual largest number
//-both of these starting at 0 to show accurate results-
And then on my test, i started using for loops and it got messy from there on out, so my big problem here i think is how to actually compare/find the smallest and largest numbers, in the best way possible. I'm also just in computer science 1 at university so we keep it pretty simple, or i like to. We also know binary search and one other search method, if either of those would be a good way to use here to write code for doing this. Thanks!
Start by declaring an array correctly. int array(10) initializes a single integer variable named array to have the value 10. (Same as saying int array = 10)
You declare an array of 10 integers as follows:
int array[10];
Anyway, two simple loops and you are done.
int array[10];
cout << "Enter 10 numbers" << endl;
for (int x = 0; x < 10; x++)
{
cin >> array[x];
}
int smallest=array[0];
int largest=array[0];
for (int x = 1; x < 10; x++)
{
if (array[x] < smallest)
{
smallest = array[x];
}
else if (array[x] > largest)
{
largest = array[x];
}
}
cout << "Largest: " << largest << endl;
cout << "Smallest: " << smallest << endl;
You can actually combine the two for loops above into a single loop. That's an exercise in an optimization that I'll leave up to you.
In this case, you don't actually have to do a binary search, or search the array. Since you will be receiving the input directly from the user, you can keep track of minimum and maximum as you encounter them, as show below. You know the first number you receive will be both the min and max. Then you compare the next number you get with those ones. If it's bigger or smaller, you store it as the max or min respectively. And then so on. I included code to store the number in an array, to check errors and to output the array back to the user, but that's probably not necessary on an exam due to the limited time. I included it as a little bit of extra info for you.
#include <cctype> // required for isdigit, error checking
#include <cstdlib> // required for atoi, convert text to an int
#include <iostream> // required for cout, cin, user input and output
#include <string> // required for string type, easier manipulation of text
int main()
{
// The number of numbers we need from the user.
int maxNumbers = 10;
// A variable to store the user's input before we can check for errors
std::string userInput;
// An array to store the user's input
int userNumbers[maxNumbers];
// store the largest and smallest number
int max, min;
// Counter variables, i is used for the two main loops in the program,
// while j is used in a loop for error checking
int i;
unsigned int j;
// Prompt the user for input.
std::cout << "Please enter " << maxNumbers << " numbers: " << std::endl;
// i is used to keep track of the number of valid numbers inputted
i = 0;
// Keep waiting for user input until the user enters the maxNumber valid
// numbers
while (i < maxNumbers)
{
// Get the user's next number, store it as string so we can check
// for errors
std::cout << "Number " << (i+1) << ": ";
std::cin >> userInput;
// This variable is used to keep track of whether or not there is
// an error in the user's input.
bool validInput = true;
// Loop through the entire inputted string and check they are all
// valid digits
for (j = 0; j < userInput.length(); j++)
{
// Check if the character at pos j in the input is a digit.
if (!isdigit(userInput.at(j)))
{
// This is not a digit, we found an error so we can stop looping
validInput = false;
break;
}
}
// If it is a valid number, store it in the array of
// numbers inputted by the user.
if (validInput)
{
// We store this number in the array, and increment the number
// of valid numbers we got.
userNumbers[i] = atoi(userInput.c_str());
// If this is the first valid input we got, then we have nothing
// to compare to yet, so store the input as the max and min
if (i == 0)
{
min = userNumbers[i];
max = userNumbers[i];
}
else {
// Is this the smallest int we have seen?
if (min < userNumbers[i])
{
min = userNumbers[i];
}
// Is this the largest int we have seen?
if (max < userNumbers[i])
{
max = userNumbers[i];
}
}
i++;
}
else
{
// This is not a valid number, inform the user of their error.
std::cout << "Invalid number, please enter a valid number." << std::endl;
}
}
// Output the user's numbers to them.
std::cout << "Your numbers are: " << userNumbers[0];
for (i = 1; i < maxNumbers; i++)
{
std::cout << "," << userNumbers[i];
}
std::cout << "." << std::endl;
// Output the min and max
std::cout << "Smallest int: " << min << std::endl;
std::cout << "Largest int: " << max << std::endl;
return 0;
}

Recognize "010" and similar numbers as a palindrome using modulus in C++

First post! This is my second semester with "Advanced C & C++" so any help is GREATLY appreciated. I've already scoured as much of stackoverflow and a few other resources to try and help me understand what I'm doing (or not doing) with this slew of logically inept code.
The goal of this program is to recognize whether or not a 'number' given by the user is a palindrome. Sounds simple enough right?! Ugh...well this is what I have been stuck on:
#include <iostream>
using std::cout;
using std::cin;
#include <string>
using std::string;
#include <cstdlib>
int main()
{
//variable declarations
string buffer;
//user input
cout << "Enter a number to see if it is a palindrome[Q to quit]: ";
getline(cin, buffer);
//looooop
while(buffer != "Q" && buffer !="q")
{
int userNum, length, sum = 0, temp;
userNum = atoi(buffer.c_str());
for(temp = userNum; userNum !=0; userNum=userNum/10)
{
length = userNum % 10;
sum = sum*10+length;
}
if(temp==sum)
cout << temp << " is a palindrome!!\n\n";
else
cout << buffer << " is NOT a palindrome!\n\n";
cout << "Enter a number to see if it is a palindrome[Q to quit]: ";
getline(cin, buffer);
}
}
The problem arises when input of "010", or "400" is given. "400" is essentially "00400" in this case and both should be seen as a palindrome.
A better approach would be to get trailing zeros for the given number as below:
int noOfTrailingZeros = str.length;
while(str[--noOfTrailingZeros]=='0');
noOfTrailingZeros = str.length - noOfTrailingZeros;
Or the integer way as:
int noOfTrailingZeros = str.length;
while(num%10==0)
{
noOfTrailingZeros++;
num/=10;
}
Now, check for the input string whether it has the same number of zeros befire the number or not as:
int counterZeros = 0;
while(str[counterZeros++]=='0');
check these 2 numbers and if trailing zeros are more than the zeros at beginning, add that many at the beginning and pass that string to palindrome function.
First of all, to recognize a palindrome, you don't have to do atoi. Just pass from the start to the middle checking if
buffer[i] == buffer[length - i]
Second, use the atoi to make sure it is a number and you're done.
Other way is to compare the string with itself reversed:
string input;
cout << "Please enter a string: ";
cin >> input;
if (input == string(input.rbegin(), input.rend())) {
cout << input << " is a palindrome";
}