So I want to make a console application that asks you one letter, and a word, and to see if word starts with that letter.
cout<<"Player 1: "<<endl;
cin>> letter;
cin>> word1;
std::string s(word1);
if (s.find(letter) == 0){
std::cout << "String starts with "<< letter<< endl;
}
You need to access first element of the string:
std::string someString = "foo"; // for example purposes
char someChar = 'f';
if (someString[0] == someChar)
std::cout << "First character of this string is " << someChar << "!";
Be careful though, if you fail to read string, you can try to access the first element (with index 0) but it may not exist. Do it like this:
if (someString.length() > 0)
{
// string is not empty. You can use also !someString.empty()
}
Related
I need to write a sentence in Pig Latin form and I am almost done it successfuly except for 1 case and I almost give up
for example :
If my word starts at a\e\o\u\i the word will look like easy -> easyway , apple -> appleway
and if it doesnt start with a letter that I wrote above
it will look like that: box -> oxbay , king -> ingkay
I succeed with the bolded part but in the first part with a\e\o\u\i letter at the beginning , I dont know where to put the w before and need some help with it
This is my code , thanks in advance
#include <iostream>
//Since those are used in ALL function of program, it wont hurt to set it to global
//Else it is considered EVIL to declare global variables
const int maxLine = 100;
char phraseLine[maxLine] = { '\0' };
void pigLatinString();
using namespace std;
void main()
{
// Displayed heading of program
cout << "* You will be prompted to enter a string of *" << endl;
cout << "* words. The string will be converted into *" << endl;
cout << "* Pig Latin and the results displayed. *" << endl;
cout << "* Enter as many strings as you would like. *" << endl;
//prompt the user for a group of words or press enter to quit
cout << "Please enter a word or group of words. (Press enter to quit)\n";
cin.getline(phraseLine, 100, '\n');
cout << endl;
// This is the main loop. Continue executing until the user hits 'enter' to quit.
while (phraseLine[0] != '\0')
{
// Display the word (s) entered by the user
cout << "You entered the following: " << phraseLine << endl;
// Display the word (s) in Pig Latin
cout << "The same phrase in Pig latin is: ";
pigLatinString();
cout << endl;
//prompt the user for a group of words or press enter to quit
cout << "Please enter a word or group of words. (Press enter to quit)\n";
cin.getline(phraseLine, 100, '\n');
}
return;
}
void pigLatinString() //phraseLine is a cstring for the word, maxline is max length of line
{ //variable declarations
char tempConsonant[10];
tempConsonant[0] = '\0';
int numberOfConsonants = 0;
char previousCharacter = ' ';
char currentCharacter = ' ';
bool isInWord = 0;
// for loop checking each index to the end of whatever is typed in
for (int i = 0; i < maxLine; i++)
{
//checking for the end of the phraseline
if (phraseLine[i] == '\0')
{//checking to see if it's in the word
if (isInWord)
{//checking to see that there wasn't a space ahead of the word and then sending the cstring + ay to the console
if (previousCharacter != ' ')
cout << tempConsonant << "ay" << endl;
}
return;
}
// this covers the end of the word condition
if (isInWord)
{// covers the condition of index [i] being the space at the end of the word
if (phraseLine[i] == ' ')
{
// spits out pig latin word, gets you out of the word, flushes the temp consonants array and resets the # of consonants to 0
cout << tempConsonant << "ay";
isInWord = 0;
tempConsonant[0] = '\0';
numberOfConsonants = 0;
}
cout << phraseLine[i] ;
}
else
{//this covers for the first vowel that makes the switch
if (phraseLine[i] != ' ')
{// sets the c string to what is in the phraseline at the time and makes it capitalized
char currentCharacter = phraseLine[i];
currentCharacter = toupper(currentCharacter);
// this takes care of the condition that currentCharacter is not a vowel
if ((currentCharacter != 'A') && (currentCharacter != 'E') &&
(currentCharacter != 'I') && (currentCharacter != 'O') && (currentCharacter != 'U'))
//this sets the array to temporarily hold the consonants for display before the 'ay'
{//this sets the null operator at the end of the c string and looks for the next consonant
tempConsonant[numberOfConsonants] = phraseLine[i];
tempConsonant[numberOfConsonants + 1] = '\0';
numberOfConsonants++;
}
else
{// this sets the boolean isInWord to true and displays the phraseline
isInWord = 1;
cout << phraseLine[i];
}
}
else
{
cout << phraseLine[i] ;
}
}
previousCharacter = phraseLine[i];
}
return;
}
You have two conditions to consider. if your word starts with a vowel, just add "way" to the end of the word, else move the first letter and add "ay" to the end.
This is a task that can be made a lot simpler by using std::string instead of C-strings. This is because you are now no longer concerned with exceeding your length or losing the null character. It also allows easier access to the Standard Library algorithms.
#include <algorithm>
#include <iostream>
#include <string>
std::string make_pig_latin(const std::string& word) {
std::string vowels("aeiou");
std::string newWord(word);
if (newWord.find_first_not_of(vowels) == 0) {
// Word starts with a consanant
std::rotate(newWord.begin(), newWord.begin() + 1, newWord.end());
newWord += "ay";
} else {
newWord += "way";
}
return newWord;
}
int main() {
std::cout << make_pig_latin("apple") << '\n'
<< make_pig_latin("box") << '\n'
<< make_pig_latin("king") << '\n'
<< make_pig_latin("easy") << '\n';
}
The function above highlights how you can go about structuring your conversion. You just need to know if your word starts with a vowel or not, and take the appropriate action.
Output:
appleway
oxbay
ingkay
easyway
I did not get the impression that you have to care about words like 'phone'.
Looking through your code, you should try to do a better job at separating your concerns. Pig Latin is easier done one word at a time, but you have string splitting code and a lot of "not Pig Latin" code in your Pig Latin function. Your main can handle getting input. You should probably have a separate function to break the line up into individual words, using std::vector to hold the words would be best since it can grow on demand and doesn't have to know a specific capacity up front. You then iterate through your array of words and translate them individually. Depending on what your actual requirements are, it's possible that you don't even have to store the translated words, just print them directly to the screen.
Here's the same program, but now it can separate words. Note how the pig latin function doesn't have to change (much, I added upper-case vowels just because I didn't want to bothered converting words) in order for the added functionality to be added.
#include <algorithm>
#include <iostream>
#include <iterator>
#include <sstream>
#include <string>
#include <vector>
std::string make_pig_latin(const std::string& word) {
std::string vowels("aeiouAEIOU");
std::string newWord(word);
if (newWord.find_first_not_of(vowels) == 0) {
// Word starts with a consanant
std::rotate(newWord.begin(), newWord.begin() + 1, newWord.end());
newWord += "ay";
} else {
newWord += "way";
}
return newWord;
}
int main() {
std::string phrase(
"A sentence where I say words like apple box easy king and ignore "
"punctuation");
std::istringstream sin(phrase);
std::vector<std::string> words(std::istream_iterator<std::string>(sin), {});
for (auto i : words) {
std::cout << make_pig_latin(i) << ' ';
}
}
Output:
Away entencesay hereway Iway aysay ordsway ikelay appleway oxbay easyway ingkay andway ignoreway unctuationpay
First of all i am new in programming.
I was asked to create a program that prompts the user to insert a word and then I translate it into some sort of fake language.
So i make the following:
firstLetter of the new word is the last char of the original word
secondLetter is ncy
thirdLetter is the entered word without the first and last char
fourthLetter is nan
fifthLetter is the first character of the word
e.g. user enters= dog
new word is: gncyonand
My code is this but its failing and I assume is because the string does not exist yet(the user still has to insert it). Please help:
**
#include <iostream> //for cin and cout
#include <string> //for string data
using namespace std;
int main()
{
//I add a welcome message:
std::cout << "*************************************************\n"
<< " Welcome to Nacy-latin converter program\n"
<< "*************************************************\n\n";
// I declare first string:
std:: string userWord; //the word the user imputs
std::string firstLetter= userWord.substr(-1,0); //last char of the entered word
std::string secondLetter = "ncy";
std::string thirdLetter= userWord.substr(1, userWord.length() - 1); //last char of the entered word
std::string fourthLetter = "nan"; //just nan
std::string fifthLetter= userWord.substr(0,1); ; //the first char of the userWord
//I ask the user to imput data:
cout << "Hey there!";
cout << endl<<endl;
cout << "Please enter a word with at least two letters and I will converted into Nacy-latin for you:\n";
//return data to the user:
cout<<"The word in Nancy-Latin is:" <<firstLetter << secondLetter << thirdLetter <<fourthLetter <<fifthLetter<<'\n';
// Farewell message
cout << "\nThank you for the 'Nancy-latin' converter tool!\n";
// system(“pause”);
return (0) ;
}
**
Did you use Python before? std::string does not allow negative indexes. You can use a mix of front(), back(), and substr() string methods to get individual pieces and then use the C++ class std::stringstream to build your new string.
std::stringstream ss;
ss << userWord.back() << "ncy";
ss << userWord.substr(1, userWord.size() - 2);
ss << "nan" << userWord.front();
std::cout << ss.str();
Do not forget to check user input for at least two characters.
Alternative to make new word in place.
std::swap(userWord.front(), userWord.back());
userWord.insert(1, "ncy");
userWord.insert(userWord.size() - 2, "nan");
std::cout << userWord;
I've looked everywhere, but I cannot find a solution to exactly why this happens in my situation.
I'm making a simple string function that asks for a string, and prints out the length.
However, I get an "Invalid Null Pointer" assertion error when I run the compiled version. I have had no errors when compiling, but the error comes up when I run it.
This is the function causing the problem:
string getString()
{
string wordInput;
cout << "Enter a word that has AT LEAST four (4) letters! ";
getline(cin, wordInput);
while (wordInput.length() <= 3)
{
cout << "Enter a word that has AT LEAST four (4) letters! ";
getline(cin, wordInput);
}
return 0;
}
The while loop isn't a problem. I commented it out and I still got the same error. How is initializing word input, cout, and getline giving me the error?
Here is my whole code so far (not finished). I tried running the string by itself too, the getKeyLetter function isn't a problem.
#include <iostream>
#include <string>
#include <cassert>
using namespace std;
char getKeyLetter()
{
char keyLetter;
string convertFromString;
cout << "Enter a SINGLE character! ";
getline(cin, convertFromString);
while (convertFromString.length() > 1)
{
cout << "Enter a SINGLE character! ";
getline(cin, convertFromString);
}
assert(convertFromString.size() == 1);
keyLetter = convertFromString[0];
return 0;
}
string getString()
{
string wordInput;
cout << "Enter a word that has AT LEAST four (4) letters! ";
getline(cin, wordInput);
while (wordInput.length() <= 3)
{
cout << "Enter a word that has AT LEAST four (4) letters! ";
getline(cin, wordInput);
}
return 0;
}
int main()
{
getKeyLetter();
getString();
return 0;
}
First, in your GetKeyChar() function, writing:
char ch;
cout << "Enter a single character: ";
cin >> ch;
will give you the first character the person types into the command prompt. So, typing "check" will have ch = c.
Second, as eran said, at the end of your functions, you have
return 0;
Unless you want both functions to return a char and string respectively, make them void GetKeyLetter() and void GetString(). Or, if you do want to return something, have them return ch (from my example) and return wordInput.
Only int main(), per standard, needs return 0, to show you that it exited correctly. the variable type you put in front of your functions is what variable you plan on returning. 0 is an int, so that's what it returns based on convention. As was pointed out, a return is not necessary in main. If you want your functions to return values, do this in your main.
string str;
char ch;
ch = GetKeyLetter();
str = GetString();
return 0;
And have your functions return the char and string value you want them to.
I need help changing my current program so that it displays the letters already entered by the user, and displays the letters again immediately before prompting the user to enter another letter. what I have so far is below.
int main()
{
char another = 'Y';
string message = "";
while (toupper(another) == 'Y')
{
cout << "Enter a message: ";
getline(cin, message);
for (int x = 0; x < message.length(); x += 1)
cout << message.substr(x) << endl;
cout << endl << "Another message (Y/N)? ";
cin >> another;
cin.ignore(100, '\n');
}
system("pause");
return 0;
}
If you want all the characters (you entered at any time) printed, you can do the following:
Start off with two empty strings. One is a buffer, that stores the string that's currently added and the other holds all of the strings already added, like so:
string buf = "";
string messages = "";
Read your characters into your buffer, via:
getline(cin, buf);
Append the string to the other messages, already entered:
messages.append(buf);
Append a string delimiter to your messages, so you know which sequence of characters (including whitespaces) belong to the message you entered:
messages.append(";");
(BTW: Using "-quotes here is really important, to let the compiler know you are comparing strings, not characters, as there is no string::append(char s)-method defined, only string::append(string s).)
Iterate through your messages-string, and check if the character at position x is equal to ';' (Using '-quotes here is also important, because string::operator[] return a character not a string!!!). For instance you might code:
for (int x = 0; x < messages.length(); x++) {
//Test if string delimiter is reached, if so, jump to next line.
if(messages[x] == ';') {
std::cout << "\n";
}
//Else just print the string:
else {
std::cout << messages[x];
}
}
Test if another message should be entered.
OR you create a linked list of strings. Adding new strings to the list, every time you enter a new string. This might be the more elegant way to do this, however it is slightly more involved. (I'm assuming you're fairly new to programming, if not I apologize!). Check Wikipedia or cplusplus.com for more info on linked lists!
Hope I could answer your question,
lindebear
I'm doing a method in which I load a notepad file with random data in it, and I have to read character by character. Then I check whether this character is a Digit or a Letter and so on and so forth.
I invented a method called isDigit (ch) in which it accepts the character that is being loaded and checks whether it is between 0 and 9. If it is a Digit it is added to a newly created char tempString.
The Problem is that whenever I add to this char tempString the first number comes out OK but the rest (in my dummy file I have from 1 - 5) it comes out as garbage.
What could be the reason please? Down below you have the code and the output
Coding
Token Lexer::getNextToken()
{
char ch;
char tempString;
std::ifstream input("dummyData.txt");
while (input)
{
input.get(ch);
if(isDigit(ch))
{
tempString += ch;
cout << tempString << endl;
Col++;
}
else if (ch == '\n')
{
cout << "\nNew Line\n" << endl;
Row++;
Col = 1;
}
else
{
cout << "Error" << endl;
}
Offset ++;
cout << " Row " << Row << " Column " << Col << " Offset: " << Offset << endl;
}
return Token::tkDigit;
}
You declare tempString as a single character, rather than a string. Adding a character to it doesn't turn it into a string and append the character; it adds the ASCII value of ch to the value already in tempString, producing some other character value.
Try std::string tempString;
You use char tempString as a buffer, but instead to store data in it you increment its value which results as a rubbish. Notice that char which presents a digit is not a digit value you expect. This way you go outside printable string boundary.
tempString is defined as a char. I believe you want it to be a std::string.
In addition you should use standard functions instead of rewriting from scratch. See std::isdigit.