#include <iostream>
#include <stack>
using namespace std;
void reverseSentence(string s)
{
stack<int> st;
for (int i = 0; i < s.length(); i++)
{
string word = "";
while (s[i] != ' ' && i < s.length())
{
word += s[i];
i++;
}
st.push(word);
}
while (!st.empty())
{
cout << st.top() << " ";
st.pop();
}
cout << endl;
}
int main()
{
string s;
cin >> s;
reverseSentence(s);
return 0;
}
The image shows the error displayed in the push(), pop(), empty() functions
I am currently using MacOS Ventura 13.1 , it would be great if anyone could help.
Since the functions aren't working, I cannot use the STL templates for my codes.
This is not working because you are creating stack<int> and you are pushing string, templates are compile time evaluated and compiler generate code based on the type you provided. so you are creating stack with int and you are pushing string so stack<int> does not have push for string. Changestack<int> to stack<string>.
Related
When you use a specific keyword, it is a matter of printing out all the words that contain that keyword. You shouldn't use arrays in solving problems.
I used the find() function to get the code that prints from the starting index of the keyword until ',' appears, but there is an error.
Here is the code:
#include <iostream>
#include <string>
#include <limits.h>
using namespace std;
int main() {
string data = "사랑,프로그래머,의자,사랑의바보,영통역,천년의사랑,냉장고,객체지향";
string keyword;
cout << "키워드 : ";
cin >> keyword;
int i = 0;
while (1) {
i = data.find(keyword, i);
if (i == INT_MAX) return 0;
while (data[i] != ',') {
cout << data[i] << " ";
i++;
}
}
}
Even if it compiles successfully, if the keyword is after the word, it cannot be printed.
In summary, my questions are:
Problems in the above code, assuming the keyword is in front of the word.
Idea to print the word even when the keyword is behind it.
std::string::find returns std::string::npos if it doesn't find anything.
The result is of type std::size_t and not int: std::size_t i = 0;
The inner while loop goes out of bounds after the last word: while (i < data.length() && data[i] != ',')
#include <iostream>
#include <string>
using namespace std;
int main() {
string data = "사랑,프로그래머,의자,사랑의바보,영통역,천년의사랑,냉장고,객체지향";
string keyword;
cout << "키워드 : ";
cin >> keyword;
std::size_t i = 0;
while (1) {
i = data.find(keyword, i);
if (i == std::string::npos) return 0;
while (i < data.length() && data[i] != ',') {
cout << data[i] << " ";
i++;
}
}
}
beginner here
i wrote the below in C++, it's a short program that currently takes 2 words as inputs, and outputs the same words back but the words are split into even and odd instead. I would like to be able to do this for 'T' words instead, but I can't figure it out. I would like to be able to first input the number of words that will follow, for example 10. Then to input the words and get T results back. So instead of just 2 words, an unlimited amount with the user specifying.
I need to put the below into a function and go from there sometime, but I want to learn the best technique to do so - any advice please?
Thanks!
Alex
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main() {
int T;
cin >> T;
string FirstWord;
cin >> FirstWord;
int LengthFirst;
LengthFirst = FirstWord.length();
string EvenFirst;
string OddFirst;
for (int i = 0; i < LengthFirst; i += 2){
EvenFirst = EvenFirst + FirstWord[i];
}
for (int i = 1; i < LengthFirst; i += 2){
OddFirst = OddFirst + FirstWord[i];
}
string SecondWord;
cin >> SecondWord;
int LengthSecond;
LengthSecond = SecondWord.length();
string EvenSecond;
string OddSecond;
for (int i = 0; i < LengthSecond; i += 2){
EvenSecond += SecondWord[i];
}
for (int i = 1; i < LengthSecond; i += 2){
OddSecond += SecondWord[i];
}
cout << EvenFirst << " " << OddFirst << endl;
cout << EvenSecond << " " << OddSecond << endl;
return 0;
}
Think I got it here, I was over-thinking this one
I put it in a for loop, as below - so any number of words can be input, user has to input the number of test cases at the
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main() {
int T;
cin >> T;
for (int i = 0; i < T; i++){
string FirstWord;
cin >> FirstWord;
int LengthFirst;
LengthFirst = FirstWord.length();
string EvenFirst;
string OddFirst;
for (int i = 0; i < LengthFirst; i += 2){
EvenFirst = EvenFirst + FirstWord[i];
}
for (int i = 1; i < LengthFirst; i += 2){
OddFirst = OddFirst + FirstWord[i];
}
cout << EvenFirst << " " << OddFirst << endl;
}
return 0;
}
Ultimately, you are performing the same task N times.
First, let's discuss how to store the information. Functionally, we have one string as input which yields two strings as output. std::pair (from <utility>) lets us easily represent this. But for sake of even-odd, std::array might be a better representation for us. Since we have a variable number of words as input, a variable number of std::array will be output. std::vector (from <vector>) is our friend here.
Second, let's discuss how to process the information. Using named variables for each output component does not scale, so let's switch to a fixed array (noted below as array<string,2>. By switching to a fixed array for output, addressing each split becomes a function of the loop index (index % 2). Below is a solution that generalizes on a known split size at compile time.
#include <string>
#include <array>
#include <vector>
#include <iostream>
int main() {
int N;
std::cin >> N;
constexpr const int Split = 2;
using StringPack = std::array<std::string, Split>;
std::vector<StringPack> output;
for (int wordIndex = 0; wordIndex < N; ++wordIndex) {
std::string word;
std::cin >> word;
StringPack out;
{
int index = 0;
for (char c : word) {
out[index % Split] += c;
++index;
}
}
output.emplace_back(out);
}
for (const auto & out : output) {
for (const auto & word : out) {
std::cout << word << ' ';
}
std::cout << '\n';
}
}
This program is suppose to generate passwords and compare to what the user inputed, if they match it breaks the while loop and outputs the user's input, but for some reason, the generated passwords are just one characters. I am new to C++, I just started like last Friday.
#include <iostream>
#include <unistd.h>
using namespace std;
int main()
{
string Password, Passwords;
cout << "Enter a password: ";
getline(cin, Password);
sleep(.7);
system("clear");
while(Password.compare(Passwords)!= 0)
{
for (int x = 0; x <= Password.length(); x++)
{
for (char Alpha = 'a'; Alpha <= 'z'; Alpha++)
{
if(Alpha == 'z')
{
Alpha = 'a';
}
for(int I=0; I <= 10; I++)
{
Passwords = Alpha + I;
system("clear");
sleep(.7);
cout << Passwords <<endl;
}
}
}
}
cout << "Password found: " << Passwords <<endl;
return 0;
}
After a long back and forward in the comments, the OP explained what was his purpose. To generate random words of the same size as input and stop when it matched the input.
This code does what you want. It's in c++14 so you need a recent compiler and to set the c++14 option. Please note the actual use of random.
#include <iostream>
#include <string>
#include <random>
#include <algorithm>
using std::cin;
using std::cout;
using std::cerr;
using std::endl;
class RandomCharGenerator {
private:
static std::string s_chars_;
private:
std::random_device rd_{};
std::default_random_engine r_eng_{rd_()};
std::uniform_int_distribution<int> char_dist_{
0, static_cast<int>(s_chars_.size())};
public:
RandomCharGenerator() = default;
auto getRandomChar() -> char { return s_chars_[char_dist_(r_eng_)]; }
auto setRandomString(std::string &str) -> void {
std::generate(std::begin(str), std::end(str),
[this] { return this->getRandomChar(); });
}
};
std::string RandomCharGenerator::s_chars_ = "abcdefghijklmnopqrstuvwxyz";
auto main() -> int {
RandomCharGenerator rand_char;
auto input = std::string{};
cin >> input;
auto generated = std::string(input.size(), ' ');
do {
rand_char.setRandomString(generated);
cout << generated << endl;
} while (input != generated);
cout << "We generated what you input" << endl;
return 0;
}
For input longer than 4 characters it takes a long time to generate the input.
Ideone demo
To understand why you had only 1 char in your Passwords:
Passwords = Alpha + I;
Alpha is a char, I is an int. Their sum is an int. This is converted to char when assigning to Passwords which is a string. So Passwords is now a string composed of only one char.
It's not clear what that actual line of code was supposed to do, so can't tell you what would have been the fix. Maybe you meant to append to Passwords. Then you should have written Passwords += Alpha + I.
The code below is an example of what I am trying to make. I did not make the code below, am just giving you and example of what am trying to do in the code above
#include <iostream>
#include <string>
#include <unistd.h>
using namespace std;
int main()
{
string password;
string Generated;
cout << "Password to find: ";
cin >> password;
char Alpha[]={'a'-1,'a','a','a','a','a','a','a','a','a'};
while( password.compare(Generated) != 0 )
{
Alpha[0]++;
for(int x=0;x<password.length();x++)
{
if (Alpha[x] == 'z'+1)
{
Alpha[x] = 'a';
Alpha[x + 1]++;
}
}
Generated=Alpha[password.length()-1];
for(int i=password.length()-2; i>=0 ; i-- )
Generated+= Alpha[i];
system("clear");
cout << "Trying: "<< Generated << endl;
}
system("clear");
sleep(1);
cout <<"Access Granted: "<< Generated << endl;
return 0;
}
Im trying to find the number of times a substring repeats within a string input but for some reason when I call the function it gives me a weird number. I have already tested the function within main and it works fine but when I make a standalone function it doesn't work.
Thank you in advance
#include <iostream>
#include <cstring>
#include <string>
using namespace std;
int checkHope(string word1)
{
int answer;
int counter;
for(int i = 0; word1[i] != '\0'; i++)
{
answer = word1.find("h", i);
if ((word1.find("o", (answer+1)) == i+1) && (word1.find("e", (answer+3)) == i+3)) counter++;
}
return counter;
}
int main()
{
string word1;
cout << "Please enter a word to check how many times the word \"hope\" appears. You can also have any letter instead of p.: ";
getline(cin, word1);
cout << checkHope(word1);
return 0;
}
//This will work,
//Changes initialize counter to 0, add condition to check alphabet 'p' also in if condition
#include <iostream>
#include <cstring>
#include <string>
using namespace std;
int checkHope(string word1)
{
int answer;
int counter=0;
for(int i = 0; word1[i] != '\0'; i++)
{
answer = word1.find("h", i);
if ((word1.find("o", (answer+1)) == i+1)
&& (word1.find("p", (answer+2)) == i+2)
&& (word1.find("e", (answer+3)) == i+3))
counter++;
}
return counter;
}
int main()
{
string word1;
cout << "Please enter a word to check how many times the word \"hope\" appears. You can also have any letter instead of p.: ";
getline(cin, word1);
cout << checkHope(word1);
return 0;
}
Ok guy i had to make a program to split elements of a string. And after that print those words.
there are some problems i am facing:
1) the array prints more than the size of the words in string i want that it should end printing as soon as last word is printed. i tried to prevent that but it always gives runtime error when i try to break at the last word.
2)is there any other efficient way to split and print ???
#include <sstream>
#include <iostream>
#include<cstdio>
#include<cstdlib>
#include <string>
using namespace std;
int main()
{
std::string line;
std::getline(cin, line);
string arr[1000];
int i = 0;
int l=line.length();
stringstream ssin(line);
while (ssin.good() && i < l)
{
ssin >> arr[i];
++i;
}
int size = sizeof(arr) / sizeof(arr[0]);
for(i = 0; i <size; i++){
cout << arr[i] << endl;
}
return 0;
}
int size = sizeof(arr) / sizeof(arr[0]);
That is a compile time value, and it's always going to be the number of elements in your array (1000). It has no idea how many strings you assigned to in your loop. You stored the number of successfully read strings (plus 1) in the i variable, so you could do this instead:
int size = i - 1;
But if it were up to me, I would just use a growable structure, like vector (#include <vector>)
std::vector<std::string> arr;
std::string temp;
while (ssin >> temp)
{
arr.push_back(temp);
}
for (auto const & str : arr)
{
std::cout << str << std::endl;
}
/* If you're stuck in the past (can't use C++11)
for (std::vector<std::string>::iterator = arr.begin(); i != arr.end(); ++i)
{
std::cout << *i << std::endl;
}
*/
For general purpose character based splitting, I would much prefer boost::split (I know you can't use it, but for future reference)
std::vector<std::string> arr;
boost::split(arr, line, boost::is_any_of(".,;!? "));
Read up on the function strtok. It is old school but very easy to use.
1) there are a couple of changes you should make to your program:
#include <sstream>
#include <iostream>
#include <string>
using namespace std;
int main()
{
std::string line("hello string world\n");
string arr[1000];
int i = 0;
stringstream ssin(line);
while (ssin.good() && i < 1000)
{
ssin >> arr[i++];
}
int size = i-1;
for(i = 0; i < size; i++){
cout << i << ": " << arr[i] << endl;
}
return 0;
}
namely, you don't want to print sizeof(arr)/sizeof(arr[0]) (i.e. 1000) elements. There is no point in the condition i < l
2) stringstream is fine if you just want to separate the single strings; if more is needed, use boost/tokenizer for splitting strings. It's modern c++, once you try it you'll never come back!
this is the best method i think no worry now
#include <sstream>
#include <iostream>
#include<cstdio>
#include<cstdlib>
#include <cstring>
#include <string>
using namespace std;
int main ()
{
std::string str;
std::getline(cin, str);
string arr[100];
int l=0,i;
char * cstr = new char [str.length()+1];
std::strcpy (cstr, str.c_str());
// cstr now contains a c-string copy of str
char * p = std::strtok (cstr,".,;!? ");
while (p!=0)
{
//std::cout << p << '\n';
arr[l++]=p;
p = strtok(NULL,".,;!? ");
}
for(i = 0; i <l; i++)
{
cout << arr[i] << endl;
}
delete[] cstr;
return 0;
}