Store a string into an array - c++

I have a string variable
string = " this is my string"
I want to store every word separated by a whitespace into an array
array[0]="this";
array[1]="is";
array[2]="my";
array[3]="string";

You could use std::copy with std::istringstream and std::istream_iterator and std::back_inserter:
vector<string> v;
istringstream ss(" this is my string");
copy(istream_iterator<string>(ss), istream_iterator<string>(), back_inserter(v));
LIVE

First you have to include "sstream" class like:
#include <sstream>
Then use below code:
string str= "this is my string"
int len = str.length();
string arr[len];
int i = 0;
stringstream ssin(str);
while (ssin.good() && i < len){
ssin >> arr[i];
++i;
}
for(i = 0; i < len; i++){
cout << arr[i] << endl;
}

For this case, you can use string delimiter concept
strtok(const char * str, int delimiter);
for your case, delimiter is "white space".
program:
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] ="This a sample string";
char * output_after_delimiter_applied;
char *out_arr[4];
int i=0;
printf("The original string is %s\n", str);
output_after_delimiter_applied = strtok (str," ");
while (output_after_delimiter_applied != NULL)
{
out_arr[i] = output_after_delimiter_applied;
output_after_delimiter_applied = strtok (NULL, " ");
// NULL parameter is for continuing the search
// delimiter is "white space".
i++;
}
for(i=0; i < 4 ; i++)
printf("%s\n", out_arr[i]);
return 0;
}
we can add any no. of delimiters with in double quotes. it is like string (delimiter) search.

Related

Removing all the vowels in a string in c++

I've written a code that removes all vowels from a string in c++ but for some reason it doesn't remove the vowel 'o' for one particular input which is: zjuotps.
Here's the code:
#include<iostream>
#include<string>
using namespace std;
int main(){
string s;
cin >> s;
string a = "aeiouyAEIOUY";
for (int i = 0; i < s.length(); i++){
for(int j = 0; j < a.length(); j++){
if(s[i] == a[j]){
s.erase(s.begin() + i);
}
}
}
cout << s;
return 0;
}
When I input: zjuotps
The Output I get is: zjotps
This is a cleaner approach using the C++ standard library:
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
int main()
{
std::string input = "zjuotps";
std::string vowels = "aeiouyAEIOUY";
auto predicate = [&vowels](char c) { return vowels.find(c) != std::string::npos; };
auto iterator = std::remove_if(input.begin(), input.end(), predicate);
input.erase(iterator, input.end());
cout << input << endl;
}
Edit:
as #RemyLebeau pointed out, std::erase_if can be used which is introduced in c++20 and the answer becomes one line of code:
std::erase_if(input, [&vowels](char c) { return vowels.find(c) != std::string::npos; });
You can develop a solution by adding the matching characters to the new string object. The eliminate() method writes the character to the result object if the characters in the input object doesn't match the characters in the remove object.
#include <iostream>
/**
* #brief This method scans the characters in the "input" object and writes
* the characters not in the "remove" object to the "result" object.
* #param input This object contains the characters to be scanned.
* #param remove This object contains characters that will not match.
* #param result Non-match result data is writed to this object.
*/
void eliminate(std::string input, std::string remove, std::string &result);
int main()
{
std::string input = "zjuotpsUK", remove = "aeiouyAEIOUY", result;
eliminate(input, remove, result);
std::cout << result << std::endl;
return 0;
}
void eliminate(std::string input, std::string remove, std::string &result)
{
for (size_t i = 0, j = 0; i < input.length(); i++)
{
for(j = 0; j < remove.length(); j++)
if(input[i] == remove[j])
break;
if(j == remove.length())
result += input[i];
}
}
In your code here, I replaced s with input_str, and a with vowels, for readability:
for (int i = 0; i < input_str.length(); i++){
for(int j = 0; j < vowels.length(); j++){
if(input_str[i] == vowels[j]){
input_str.erase(input_str.begin() + i);
}
}
}
The problem with your current code above is that each time you erase a char in the input string, you should break out of the vowels j loop and start over again in the input string at the same i location, checking all vowels in the j loop again. This is because erasing a char left-shifts all chars which are located to the right, meaning that the same i location would now contain a new char to check since it just left-shifted into that position from one position to the right. Erroneously allowing i to increment means you skip that new char to check in that same i position, thereby leaving the 2nd vowel in the string if 2 vowels are in a row, for instance. Here is the fix to your immediate code from the question:
int i = 0;
while (i < s.length()){
bool char_is_a_vowel = false;
for(int j = 0; j < a.length(); j++){
if(s[i] == a[j]){
char_is_a_vowel = true;
break; // exit j loop
}
}
if (char_is_a_vowel){
s.erase(s.begin() + i);
continue; // Do NOT increment i below! Skip that.
}
i++;
}
However, there are many other, better ways to do this. I'll present some below. I personally find this most-upvoted code difficult to read, however. It requires extra study and looking up stuff to do something so simple. So, I'll show some alternative approaches to that answer.
Approach 1 of many: copy non-vowel chars to new string:
So, here is an alternative, simple, more-readable approach where you simply scan through all chars in the input string, check to see if the char is in the vowels string, and if it is not, you copy it to an output string since it is not a vowel:
Just the algorithm:
std::string output_str;
for (const char c : input_str) {
if (vowels.find(c) == std::string::npos) {
output_str.push_back(c);
}
}
Full, runnable example:
#include <iostream> // For `std::cin`, `std::cout`, `std::endl`, etc.
#include <string>
int main()
{
std::string input_str = "zjuotps";
std::string vowels = "aeiouyAEIOUY";
std::string output_str;
for (const char c : input_str)
{
if (vowels.find(c) == std::string::npos)
{
// char `c` is NOT in the `vowels` string, so append it to the
// output string
output_str.push_back(c);
}
}
std::cout << "input_str = " << input_str << std::endl;
std::cout << "output_str = " << output_str << std::endl;
}
Output:
input_str = zjuotps
output_str = zjtps
Approach 2 of many: remove vowel chars in input string:
Alternatively, you could remove the vowel chars in-place as you originally tried to do. But, you must NOT increment the index, i, for the input string if the char is erased since erasing the vowel char left-shifs the remaining chars in the string, meaning that we need to check the same index location again the next iteration in order to read the next char. See the note in the comments below.
Just the algorithm:
size_t i = 0;
while (i < input_str.length()) {
char c = input_str[i];
if (vowels.find(c) != std::string::npos) {
input_str.erase(input_str.begin() + i);
continue;
}
i++;
}
Full, runnable example:
#include <iostream> // For `std::cin`, `std::cout`, `std::endl`, etc.
#include <string>
int main()
{
std::string input_str = "zjuotps";
std::string vowels = "aeiouyAEIOUY";
std::cout << "BEFORE: input_str = " << input_str << std::endl;
size_t i = 0;
while (i < input_str.length())
{
char c = input_str[i];
if (vowels.find(c) != std::string::npos)
{
// char `c` IS in the `vowels` string, so remove it from the
// `input_str`
input_str.erase(input_str.begin() + i);
// do NOT increment `i` here since erasing the vowel char above just
// left-shifted the remaining chars in the string, meaning that we
// need to check the *same* index location again the next
// iteration!
continue;
}
i++;
}
std::cout << "AFTER: input_str = " << input_str << std::endl;
}
Output:
BEFORE: input_str = zjuotps
AFTER: input_str = zjtps
Approach 3 of many: high-speed C-style arrays: modify input string in-place
I borrowed this approach from "Approach 1" of my previous answer here: Removing elements from array in C
If you are ever in a situation where you need high-speed, I'd bet this is probably one of the fastest approaches. It uses C-style strings (char arrays). It scans through the input string, detecting any vowels. If it sees a char that is NOT a vowel, it copies it into the far left of the input string, thereby modifying the string in-place, filtering out all vowels. When done, it null-terminates the input string in the new location. In case you need a C++ std::string type in the end, I create one from the C-string when done.
Just the algorithm:
size_t i_write = 0;
for (size_t i_read = 0; i_read < ARRAY_LEN(input_str); i_read++) {
bool char_is_a_vowel = false;
for (size_t j = 0; j < ARRAY_LEN(input_str); j++) {
if (input_str[i_read] == vowels[j]) {
char_is_a_vowel = true;
break;
}
}
if (!char_is_a_vowel) {
input_str[i_write] = input_str[i_read];
i_write++;
}
}
input_str[i_write] = '\n';
Full, runnable example:
#include <iostream> // For `std::cin`, `std::cout`, `std::endl`, etc.
#include <string>
/// Get the number of elements in an array
#define ARRAY_LEN(array) (sizeof(array)/sizeof(array[0]))
int main()
{
char input_str[] = "zjuotps";
char vowels[] = "aeiouyAEIOUY";
std::cout << "BEFORE: input_str = " << input_str << std::endl;
// Iterate over all chars in the input string
size_t i_write = 0;
for (size_t i_read = 0; i_read < ARRAY_LEN(input_str); i_read++)
{
// Iterate over all chars in the vowels string. Only retain in the input
// string (copying chars into the left side of the input string) all
// chars which are NOT vowels!
bool char_is_a_vowel = false;
for (size_t j = 0; j < ARRAY_LEN(input_str); j++)
{
if (input_str[i_read] == vowels[j])
{
char_is_a_vowel = true;
break;
}
}
if (!char_is_a_vowel)
{
input_str[i_write] = input_str[i_read];
i_write++;
}
}
// null-terminate the input string at its new end location; the number of
// chars in it (its new length) is now equal to `i_write`!
input_str[i_write] = '\n';
std::cout << "AFTER: input_str = " << input_str << std::endl;
// Just in case you need it back in this form now:
std::string str(input_str);
std::cout << " C++ str = " << str << std::endl;
}
Output:
BEFORE: input_str = zjuotps
AFTER: input_str = zjtps
C++ str = zjtps
See also:
[a similar answer of mine in C] Removing elements from array in C

Remove chars and append them in the end of the string ( C++ )

How can I erase the first N-th characters in a given string and append them in the end. For example if we have
abracadabra
and we shift the first 4 characters to the end then we should get
cadabraabra
Instead of earsing them from the front which is expensive there is another way. We can rotate them in place which is a single O(N) operation. In this case you want to rotate to the left so we would use
std::string text = "abracadabra";
std::rotate(text.begin(), text.begin() + N, text.end());
In the above example if N is 4 then you get
cadabraabra
Live Example
You can try an old-fashioned double loop, one char at a time.
#include "string.h"
#include "stdio.h"
int main() {
char ex_string[] = "abracadabra";
int pos = 4;
char a;
int i, j;
size_t length = strlen(ex_string);
for (j = 0; j < pos; j++) {
a = ex_string[0];
for (i = 0; i < length - 1; i++) {
ex_string[i] = ex_string[i + 1];
}
ex_string[length-1]=a;
}
printf("%s", ex_string);
}
string str = "abracadabra" //lets say
int n;
cin>>n;
string temp;
str.cpy(temp,0,n-1);
str.earse(str.begin(),str.begin()+n-1);
str+=temp;
Reference
string::erase
string::copy
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
string str;
cin >> str;
int number;
cin >> number;
string erasedString = str;
string remainingString = str.substr(number + 1, strlen(str));
erasedString.erase(0, number);
return 0;
}

C++ Program to print the longest word of the string?

#include <iostream>
#include <string>
using namespace std;
int main()
{
string s;
getline(cin , s) ; #input of string from user
int counter = 0;
int max_word = -1;
int len = s.length(); #length of string
string max = " ";
string counter_word = " ";
for (int i = 0; i < len; i++)
{
if(s[i] != ' ')
{
counter++;
}
if(s[i] == ' ' || i == len - 1)
{
if(counter > max_word)
{
max_word = counter;
//handling end of string.
if(i == len - 1)
max = s.substr(i + 1 - max_word, max_word); #sub string command that prints the longest word
else
max = s.substr(i - max_word, max_word);
}
counter = 0;
}
}
cout << max_word << " " << max << endl; #output
return 0;
}
The current output is '4 This' on entering the string "This is cool".
How do I get it to print '4 This; Cool' ?
On running it in Linux through the terminal, it gives me the error
" terminate called after throwing an instance of 'std::out_of_range' what(): basic_string::substr Aborted (core dumped) "
If I have understood you correctly then you mean the following
#include <iostream>
#include <sstream>
#include <string>
int main()
{
std::string s;
std::getline( std::cin, s );
std::string::size_type max_size;
std::string max_word;
std::string word;
std::istringstream is( s );
max_size = 0;
while ( is >> word )
{
if ( max_size < word.size() )
{
max_size = word.size();
max_word = word;
}
else if ( max_size == word.size() )
{
max_word += "; ";
max_word += word;
}
}
std::cout << max_size << ' ' << max_word << std::endl;
}
If to enter string
This is cool
then the output will be
4 This; cool
The basic idea here is to add each character to a temporary (initially empty) string until a space is encountered. At each instance of a space, the length of the temporary string is compared to the length of the "maxword" string, which is updated if it is found to be longer. The temporary string is emptied (reset to null using '\0') before proceeding to the next character in the input string.
#include <iostream>
#include <string>
using namespace std;
string LongestWord(string str) {
string tempstring;
string maxword;
int len = str.length();
for (int i = 0; i<=len; i++) {
if (tempstring.length()>maxword.length())
{
maxword=tempstring;
}
if (str[i]!=' ')
{
tempstring=tempstring+str[i];
}
else
{
tempstring='\0';
}
}
return maxword;
}
int main() {
cout << LongestWord(gets(stdin));
return 0;
}
#include <iostream>
using namespace std;
string longestWordInSentence(string str) {
// algorithm to count the number of words in the above string literal/ sentence
int words = 0;
for (int i = 0; i < str.length(); i++) {
if (str[i] == ' ') {
words++;
}
}
// incrementing the words variable by one as the above algorithm does not take into account the last word, so we are incrementing
// it here manually just for the sake of cracking this problem
words += 1; // words = 5
// words would be the size of the array during initialization since this array appends only the words of the above string
// and not the spaces. So the size of the array would be equal to the number of words in the above sentence
string strWords[words];
// this algorithm appends individual words in the array strWords
short counter = 0;
for (short i = 0; i < str.length(); i++) {
strWords[counter] += str[i];
// incrementing the counter variable as the iterating variable i loops over a space character just so it does not count
// the space as well and appends it in the array
if (str[i] == ' ') {
counter++;
}
}
// algorithm to find the longest word in the strWords array
int sizeArray = sizeof(strWords) / sizeof(strWords[0]); // length of the strWords array
int longest = strWords[0].length(); // intializing a variable and setting it to the length of the first word in the strWords array
string longestWord = ""; // this will store the longest word in the above string
for (int i = 0; i < sizeArray; i++) { // looping over the strWords array
if (strWords[i].length() > longest) {
longest = strWords[i].length();
longestWord = strWords[i]; // updating the value of the longestWord variable with every loop iteration if the length of the proceeding word is greater than the length of the preceeding word
}
}
return longestWord; // return the longest word
}
int main() {
string x = "I love solving algorithms";
cout << longestWordInSentence(x);
return 0;
}
I have explained every line of the code in great detail. Please refer to the comments in front of every line of code. Here is a generalized approach:
Count the number of words in the given sentence
Initialize a string array and set the size of the array equal to the number of words in the sentence
Append the words of the given sentence to the array
Loop through the array and apply the algorithm of finding the longest word in the string. It is similar to finding the longest integer in an array of integers.
Return the longest word.
My original solution contained a bug: If you were to input 2 words of length n and 1 word of length n + k, then it would output those three words.
You should make a separate if condition to check whether the word length is the same as before, if yes, then you can append "; " and the other word.
This is what I would do:
Change if(counter > max_word) to if(counter >= max_word) so words of the same length are also taken into account.
Make the max string by default (so "" instead of " "). (See next point)
Add an if condition in the if(counter >= max_word)second if condition to see if the max string is not empty, and if it's not empty append "; "
Changing max = to max += so that it appends the words (in the second condition)
Wouldn't it be easier for you to split the whole line into a vector of string ?
Then you could ask the length of each element of the string and then print them. Because right now you still have all the words in a single string making each individual word hard to analyse.
It would also be hard, as you requested, to print all the words with the same length if you use a single string.
EDIT :
Start by looping through your whole input
Keep the greater length of word between the current one and the previously saved
Make a substring for each word and push_back it into a vector
Print the length of the bigger word
Loop through the vector and print each word of that size.
Look the following website for all references about vectors. dont forget to #include
www.cplusplus.com/reference/vector/vector/
#include <iostream>
#include <vector>
#include <string>
void LongestWord(std::string &str){
std::string workingWord = "";
std::string maxWord = "";
for (int i = 0; i < str.size(); i++){
if(str[i] != ' ')
workingWord += str[i];
else
workingWord = "";
if (workingWord.size() > maxWord.size())
maxWord = workingWord;
}
std::cout << maxWord;
}
int main(){
std::string str;
std::cout << "Enter a string:";
getline(std::cin, str);
LongestWord(str);
std::cout << std::endl;
return 0;
}
Source: http://www.cplusplus.com/forum/beginner/31169/
#include<bits/stdc++.h>
using namespace std;
int main()
{
string s,a;
char ch;
int len,mlen=0;
getline(cin,s);
char* token=strtok(&s[0]," ");
string r;
while(token!=NULL)
{
r=token;
len=r.size();
if(mlen<len)
{
mlen=len;
a=token;
}
token = strtok(NULL, " ");
}
cout<<a;
return 0;
}
#include<iostream>
using namespace std;
int main()
{
string str;
getline(cin,str);
cin.ignore();
int len =str.length();`
int current_len=0,max_len=0;
int initial=0,start=0;
int i=0;
while(1)
{
if(i==len+2)
{break;}
if(str[i]==' '|| i==len+1)
{
if(current_len>max_len)
{
initial=start;
max_len=current_len;
}
current_len=0;
start=i+1;
}
else
{
current_len++;
}
i++;
}
for (int i = 0; i < max_len; i++)
{
cout<<str[i+initial];
}
cout<<endl<<max_len<<endl;
return 0 ;
}

Reverse words of a sentence in c++, Asked on Google Code jam

How to solve this? https://code.google.com/codejam/contest/351101/dashboard#s=p1?
The code I ended up with is below, but it can only reverse strings up to one space because it was code keeping the literal logic in mind, Reverse the whole string, reverse the words, and done. The spaces are slightly messed up and when I tried a loop to detect number of spaces and act accordingly it failed. Please help! Code:
#include <iostream>
#include <string>
using namespace std;
int main()
{
char revwrd[100];
char revstr[100];
string str;
getline(cin, str);
cout<<str;
int sps[10];
int len,y=0;
len = str.length();
cout<<"\n"<<"The Length of the string is:"<<len;
for(int x=len-1;x>-1;x--)
{
revstr[x] = str[y];
y++;
}
cout<<"\n"<<"The inverse of the string is:"<<"\n";
for(int z = 0;z<len;z++)
{
cout<<revstr[z];
}
cout<<"\n";
int no=0;
int spaces=0;
for(int a=0;a<len;a++)
{
if(revstr[a]== ' ')
{
sps[no]=a;
no++;
spaces++;
}
}
int rinc=0;
int spinc;
cout<<"\n";
spinc=sps[0];
int spinc2 = sps[0]+1;
int lend;
for(rinc=0;rinc<sps[0]+1;rinc++)
{
revwrd[rinc] = revstr[spinc];
spinc--;
}
for(lend=len;lend>sps[0];lend--)
{
revwrd[spinc2] = revstr[lend];
spinc2++;
}
cout<<"Spaces in the string:"<<spaces<<"\n";
cout<<"The words inversed are:"<<"\n";
for(int inc=1;inc<len+1;inc++)
{
cout<<revwrd[inc];
}
return 0;
}
The conditions of the challenge are that there is only a single space between words, and that spaces don't appear at the beginning or the end of the line, so for this particular exercise you don't have to worry about preserving spacing; as long as you write the output with a single space between each word, you're good.
With that in mind, you can read each word using regular formatted input:
std::string word;
...
while ( stream >> word )
// do something with word
You don't have to worry about buffer size, you don't have to worry about detecting spaces, etc. You do have to worry about detecting the newline character, but that's easily done using the peek method:
while ( stream >> word )
{
// do something with word;
if ( stream.peek() == '\n' )
break;
}
The above loop will read individual words from the input stream stream until it sees a newline character (there's probably a better way to do that, but it works).
Now, in order to reverse each line of input, you obviously need to store the strings somewhere as you read them. The easiest thing to do is store them to a vector:
std::vector< std::string > strings;
...
while ( stream >> word )
{
strings.push_back( word );
if ( stream.peek() == '\n' )
break;
}
So now you have a vector containing all the strings in the line, you just have to print them out in reverse order. You can use a reverse iterator to walk through the vector:
std::vector< std::string >::reverse_iterator it;
for ( it = strings.rbegin(); it != strings.rend(); ++it )
{
std::cout << *it << " ";
}
std::cout << std::endl;
The rbegin() method returns an iterator that points to the last element in the vector; the rend() method returns an iterator that points to an element before the first element of the vector; ++it advances the iterator to point to the next item in the vector, going back to front; and *it gives the string that the iterator points to. You can get a little more esoteric and use the copy template function:
std::copy( strings.rbegin(),
strings.rend(),
std::ostream_iterator<std::string>( std::cout, " " )
);
That single method call replaces the loop above. It creates a new ostream_iterator that will write strings to cout, separated by a single space character.
For the conditions of this particular exercise, this is more than adequate. If you were required to preserve spacing, or to account for punctuation or capitalization, then you'd have to do something a bit lower-level.
Just few loops and if's:
// Reverse Words
#include <iostream>
#include <string>
using namespace std;
int main() {
int tc; cin >> tc; cin.get();
for(int t = 0; t < tc; t++) {
string s, k; getline(cin, s);
for(int i = (s.length()- 1); i >= 0; i--) {
if(s[i] == ' ' || (i == 0)) {
if(i == 0) k += ' ';
for(int j = i; j < s.length(); j++) {
k += s[j];
if(s[j+1] == ' ' ) break;
}
}
}
cout << "Case #" << t + 1 << " " << k << endl;
}
return 0;
}
This might handle multi spaces:
std::string ReverseSentence(std::string in)
{
std::vector<string> words;
std::string temp = "";
bool isSpace = false;
for(int i=0; in.size(); i++)
{
if(in[i]!=' ')
{
if(isSpace)
{
words.push_back(temp);
temp = "";
isSpace = false;
}
temp+=in[i];
}
else
{
if(!isSpace)
{
words.push_back(temp);
temp = "";
isSpace = true;
}
temp += " ";
}
}
std::reverse(words.begin(),words.end());
std::string out = "";
for(int i=0; i<words.size(); i++)
{
out+=words[i];
}
return out;
}
you can follow this method :
step 1 : just check for spaces in input array store their index number in an integer array.
step 2 : now iterate through this integer array from end
step a : make a string by copying characters from this index to previous index .
note : since for first element there is no previous element in that case you will copy from this index to end of the input string .
step b : step a will give you a word from end of input string now add these word with a space to make your output string .
I hope this will help you .
This problem is really made for recursion:
void reverse()
{
string str;
cin >> str;
if (cin.peek() != '\n' || cin.eof()) {
str = " " + str;
reverse();
}
cout << str;
}
int main(int argc, const char * argv[])
{
int count = 0;
cin >> count;
for (int i = 0; i < count; i++) {
cout << "Case #" << (i + 1) << ": ";
reverse();
cout << endl;
}
return 0;
}
So I read word by word and add one space in front of the word until end of line or file is reached. Once end of line is reached the recursion unwraps and prints the read strings in reversed order.
I went for the brute force, and I badly wanted to use pointers!
get the sentence
detect every word and put them in a container.
read backwards the container.
This is it:
#include <iostream>
#include <string>
#include <vector>
int main()
{
char *s1 = new char[100];
std::cin.getline(s1, 100);
std::vector<std::string> container;
char* temp = new char[100];
char *p1, *p0;
p1 =p0 = s1;
int i;
do{
if (*p1==' ' || *p1=='\0'){
//std::cout<<p1-p0<<' ';
for(i=0;i<p1-p0;++i) temp[i]=p0[i]; temp[i]='\0';
p0 = p1+1;
container.push_back(temp);
std::cout<<temp;
}
p1++;
}while(*(p1-1)!='\0');
std::cout<<std::endl;
for(int i=container.size()-1;i>=0;i--) std::cout<<container[i]<<' ';
return 0;
}

Concatenating 2 strings together

I have two strings:
string word;
string alphabet;
word has some input that I passed on to; let's say "XYZ".
alphabet has the alphabet except XYZ, so it's "ABCDEFGHIJKLMNOPQRSTUVW"
When I tried to concat them with "+=", all I get is word (i.e. "XYZ"). I want to make it look like this:
XYZABCDEFGHIJKLMNOPQRSTUVW
What am I doing wrong? Code is basicly this vvvv
===========================encryption.cpp=================================================
#include <cstdlib>
#include <iostream>
#include "encryption.h"
#include <string>
encryption::encryption()
{
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
}
string encryption::removeletter(string word,char letter)
{
//remove letter
string temp;
int indis=0;
for(int i = 0; word[i] != '\0'; i++)
{
if(word[i] != letter)
{
temp[indis]=word[i] ;
indis++;
}
}
word=temp;
return word;
}
This is the function i have proplems with :
void encryption::encrypt(string word)//main.cpp is just calling this atm
{
string temp;
int pos;
//getting rid of the repeating letters
for(int i = 0; i < word.length(); i++)
{
if( (pos = temp.find(kelime[i])) < 0)
temp += word[i];
}
word=temp;//that ends here
//taking words letters out of the alphabet
for(int i = 0; i < word.length(); i++)
{
alfabet=removeletter(alfabe,word[i]);
}
for(int i = 0; i < word.length()-1; i++)
{
for(int j=0;alfabet[j] !='\0'; j++)
if(alfabet[j+1] =='\0') alfabet[j]='\0';
}
/* I tried += here */
}
===========================encryption.h====================================================
#ifndef encryption_h
#define encryption_h
#include<string>
class encryption
{
public:
encryption();
void encrypt(string word);
string removeletter(string word,char letter);
//some other functions,i deleted them atm
public:
string alphabet;
string encryptedalphabet;
//staff that not in use atm(deleted)
};
#endif
===========================main.cpp======================================================
#include <cstdlib>
#include <iostream>
#include "encryption.h"
#include <string>
string word;
encryption encrypted;
cin>>word;
encrypted.encrypt( word);
+= is how you do it, you must be doing something else.
string word = "XYZ";
string alphabet = "ABCDEFGHIJKLMNOPQRSTUVW";
alphabet += word;
cout << alphabet << "\n";
output
ABCDEFGHIJKLMNOPQRSTUVWXYZ
Use
word.append(alphabet);
Note that when you want to concatenate many strings, it's more efficient to use a stringstream to avoid unnecessary string copying.
string word = "XYZ";
string alphabet = "ABCDEFGHIJKLMNOPQRSTUVW";
word += alphabet;
Now you can print 'word', and you will find your answer.