How to split an array of characters without using any basic function - c++

I have this function: int split(char* str, char s), so how to split the str without using strtok() or other functions?
E.g: str = "1,2,3,4,5", s = ','
After split(str, s), output will be:
1
2
3
4
5
Sorry guys, the int return -1 if str == NULL and return 1 if str != NULL.

How about this? I'm not sure what the int return type means in the function so I made it the count of splits.
#include <stdio.h>
int split(char* str, char s) {
int count = 0;
while (*str) {
if (s == *str) {
putchar('\n');
count++;
} else {
putchar(*str);
}
str++;
}
return count;
}

I didn't write code for years, but that should do?
while (*str) // as long as there are more chars coming...
{
if (*str == s) printf('\n'); // if it is a separator, print newline
else printf('%c',*str); // else print the char
str++; // next char
}

string split(const char* str, char s) {
std::string result = str;
std::replace(result.begin(), result.end(), s, '\n');
result.push_back('\n'); // if you want a trailing newline
return result;
}

Another approach...
#include <iostream>
using namespace std;
void split(char* str, char s){
while(*str){
if(*str==s){
cout << endl;
}else{
cout << *str;
}
str++;
}
cout << endl;
}
int main(){
split((char*)"herp,derp",',');
}

and another one with iterator
#include <iostream>
using namespace std;
int main() {
string s="1,2,3,4,5";
char cl=',';
for(string::iterator it=s.begin();it!=s.end();++it)
if (*it == cl)
cout << endl;
else cout << *it;
return 0;
}
http://ideone.com/RPrls7

Related

This is a program to recursively calculate the reverse of a string but it's not printing anything

What I have done is created a global array to store the reversed string.
#include <bits/stdc++.h>
using namespace std;
char arr[10];
int c = 1;
string Reverser(string z)
{
arr[c] = z[(z.size() - c)];
c++;
if (c == (z.size() + 1))
{
return 0;
}
else
{
Reverser(z);
}
return 0;
}
int main()
{
string z;
cin >> z;
string Reverser(z);
for (int i = 1; i <= z.size(); i++)
{
cout << arr[i];
}
return 0;
}
I have also tried to dry run it but I can't really find any error.
You can use a std::stringstream and pass it by reference in your recursive function. Also, you can pass the string by reference.
#include <iostream>
#include <string>
#include <sstream>
void reverse(const std::string& a, std::stringstream& ss, unsigned int pos)
{
ss << a[pos];
if (pos == 0) return;
reverse(a, ss, pos - 1);
}
void reverse(const std::string& a, std::stringstream& ss)
{
reverse(a, ss, a.length() - 1);
}
int main()
{
std::stringstream ss;
std::string input = "Hello";
reverse(input, ss);
std::cout << ss.str() << std::endl;
}

String function optimisation?

I'm new to C++ and i just wrote a function to tell me if certain characters in a string repeat or not:
bool repeats(string s)
{
int len = s.size(), c = 0;
for(int i = 0; i < len; i++){
for(int k = 0; k < len; k++){
if(i != k && s[i] == s[k]){
c++;
}
}
}
return c;
}
...but i can't help but think it's a bit congested for what it's supposed to do. Is there any way i could write such a function in less lines?
Is there any way i could write such a function in less lines?
With std, you might do:
bool repeats(const std::string& s)
{
return std::/*unordered_*/set<char>{s.begin(), s.end()}.size() != s.size();
}
#include <algorithm>
bool repeats(std::string s){
for (auto c : s){
if(std::count(s.begin(), s.end(), c) - 1)
return true;
}
return false;
}
Assuming you are not looking for repeated substrings :
#include <iostream>
#include <string>
#include <set>
std::set<char> ignore_characters{ ' ', '\n' };
bool has_repeated_characters(const std::string& input)
{
// std::set<char> is a collection of unique characters
std::set<char> seen_characters{};
// loop over all characters in the input string
for (const auto& c : input)
{
// skip characters to ignore, like spaces
if (ignore_characters.find(c) == ignore_characters.end())
{
// check if the set contains the character, in C++20 : seen_characters.contains(c)
// and maybe you need to do something with "std::tolower()" here too
if (seen_characters.find(c) != seen_characters.end())
{
return true;
}
// add the character to the set, we've now seen it
seen_characters.insert(c);
}
}
return false;
}
void show_has_repeated_characters(const std::string& input)
{
std::cout << "'" << input << "' ";
if (has_repeated_characters(input))
{
std::cout << "has repeated characters\n";
}
else
{
std::cout << "doesn't have repeated characters\n";
}
}
int main()
{
show_has_repeated_characters("Hello world");
show_has_repeated_characters("The fast boy");
return 0;
}
std::string str;
... fill your string here...
int counts[256]={0};
for(auto s:str)
counts[(unsigned char)s]++;
for(int i=0;i<256;i++)
if(counts[i]>1) return true;
return false;
6 lines instead of 9
O(n+256) instead of O(n^2)
This is your new compact function :
#include <iostream>
#include <algorithm>
using namespace std;
int occurrences(string s, char c) {
return count(s.begin(), s.end(), c); }
int main() {
//occurrences count how many times char is repetated.
//any number other than 0 is considered true.
occurrences("Hello World!",'x')?cout<<"repeats!":cout<<"no repeats!";
//It is equal write
//
// if(occurrences("Hello World!",'x'))
// cout<<"repeats!";
// else
// cout<<"no repeats!";
//So to count the occurrences
//
// int count = occurrences("Hello World!",'x');
}

How to insert an integer with leading zeros into a std::string?

In a C++14 program, I am given a string like
std::string s = "MyFile####.mp4";
and an integer 0 to a few hundred. (It'll never be a thousand or more, but four digits just in case.) I want to replace the "####" with the integer value, with leading zeros as needed to match the number of '#' characters. What is the slick C++11/14 way to modify s or produce a new string like that?
Normally I would use char* strings and snprintf(), strchr() to find the "#", but figure I should get with modern times and use std::string more often, but know only the simplest uses of it.
What is the slick C++11/14 way to modify s or produce a new string like that?
I don't know if it's slick enough but I propose the use of std::transform(), a lambda function and reverse iterators.
Something like
#include <string>
#include <iostream>
#include <algorithm>
int main ()
{
std::string str { "MyFile####.mp4" };
int num { 742 };
std::transform(str.rbegin(), str.rend(), str.rbegin(),
[&](auto ch)
{
if ( '#' == ch )
{
ch = "0123456789"[num % 10]; // or '0' + num % 10;
num /= 10;
}
return ch;
} // end of lambda function passed in as a parameter
); // end of std::transform()
std::cout << str << std::endl; // print MyFile0742.mp4
}
I would use regex since you're using C++14:
#include <iostream>
#include <regex>
#include <string>
#include <iterator>
int main()
{
std::string text = "Myfile####.mp4";
std::regex re("####");
int num = 252;
//convert int to string and add appropriate number of 0's
std::string nu = std::to_string(num);
while(nu.length() < 4) {
nu = "0" + nu;
}
//let regex_replace do it's work
std::regex_replace(std::ostreambuf_iterator<char>(std::cout),
text.begin(), text.end(), re, nu);
std::cout << std::endl;
return 0;
}
WHy not use std::stringstream and than convert it to string.
std::string inputNumber (std::string s, int n) {
std::stringstream sstream;
bool numberIsSet = false;
for (int i = 0; i < s; ++i) {
if (s[i] == '#' && numberIsSet == true)
continue;
else if (s[i] == '#' && numberIsSet == false) {
sstream << setfill('0') << setw(5) << n;
numberIsSet = true;
} else
sstream << s[i];
}
return sstream.str();
}
I would probably use something like this
#include <iostream>
using namespace std;
int main()
{
int SomeNumber = 42;
std:string num = std::to_string(SomeNumber);
string padding = "";
while(padding.length()+num.length()<4){
padding += "0";
}
string result = "MyFile"+padding+num+".mp4";
cout << result << endl;
return 0;
}
Mine got out of control while I was playing with it, heh.
Pass it patterns on its command line, like:
./cpp-string-fill file########.jpg '####' test###this### and#this
#include <string>
#include <iostream>
#include <sstream>
std::string fill_pattern(std::string p, int num) {
size_t start_i, end_i;
for(
start_i = p.find_first_of('#'), end_i = start_i;
end_i < p.length() && p[end_i] == '#';
++end_i
) {
// Nothing special here.
}
if(end_i <= p.length()) {
std::ostringstream os;
os << num;
const std::string &ns = os.str();
size_t n_i = ns.length();
while(end_i > start_i && n_i > 0) {
end_i--;
n_i--;
p[end_i] = ns[n_i];
}
while(end_i > start_i) {
end_i--;
p[end_i] = '0';
}
}
return p;
}
int main(int argc, char *argv[]) {
if(argc<2) {
exit(1);
}
for(int i = 1; i < argc; i++) {
std::cout << fill_pattern(argv[i], 1283) << std::endl;
}
return 0;
}
I would probably do something like this:
using namespace std;
#include <iostream>
#include <string>
int main()
{
int SomeNumber = 42;
string num = std::to_string(SomeNumber);
string guide = "myfile####.mp3";
int start = static_cast<int>(guide.find_first_of("#"));
int end = static_cast<int>(guide.find_last_of("#"));
int used = 1;
int place = end;
char padding = '0';
while(place >= start){
if(used>num.length()){
guide.begin()[place]=padding;
}else{
guide.begin()[place]=num[num.length()-used];
}
place--;
used++;
}
cout << guide << endl;
return 0;
}

Return a pointer to the last appearance of a character in a C-Style string (C++)

Return a pointer to the last appearance of c
appearing inside s and nullptr (0) if c does not appear inside s.
#include <string>
#include <iostream>
#include <cassert>
using namespace std;
const char* myStrRChr(const char* s, char c)
{
int curIdx = 0;
char last;
while (s[curIdx] != '\0')
{
if (s[curIdx] == c)
last = s[curIdx];
curIdx++;
}
if (s[curIdx] == c)
return last;
else
// return '\0', nullptr, NULL
return "";
}
int main()
{
char cstr[50] = "Abadabadoo!";
char buf[10];
const char * cat = "cat";
char dog[] = "Labradoodle";
cout << "\nmyStrRChr(cstr, 'a') expects adoo!" << endl;
cout << " -- " << myStrRChr(cstr, 'a') << endl;
return 0;
}
This code returns "adabadoo!". I can't wrap my mind around as to how to get the last instance of "char c."
You can do this by obtaining a pointer to the end of the string and decrementing down the string searching for the character c, and a pointer to the beginning of the string to know where to stop looping:
const char *mystrrchr(const char *str, char c)
{
int len = strlen(str);
char *p = const_cast<char *>(&str[len-1]);
char *stop = const_cast<char *>(&str[0]);
while(p>=stop)
{
if(*p==c)
{
return p;
}
p--;
}
return nullptr;
}

Palindrome finder: non-alphanumeric character deletion problems

So I'm having a substantial amount of trouble with this one bit of code. I've included the whole program for context, but my issue lies in the cleanUp function, wherein I (attempt to) remove all characters that are not 'A' through 'Z'.
Any tips?
#include <iostream>
#include <cstdlib>
#include <string>
#include <stdio.h>
#include <ctype.h>
using namespace std;
bool again(string title); // Checks if you want to run again.
void makeUpper(char word[]);
void getReverse(char word[], char reversed[]);
char * find(char *str, char what);
bool equal(char word[], char reversed[]);
int size(char word[]);
char * cleanUp(char *str);
int main()
{
char word[256] = "Hello?? There!", reversedWord[256];
do
{
cout<<"Please enter the string to check: ";
makeUpper(word);
cout << word;
cleanUp(word);
getReverse(word,reversedWord);
if(equal(word, reversedWord))
cout<<"You have a palindrome!"<<endl;
else
cout<<"You do not have a palindrome!"<<endl;
} while(again("Do you want to do this again? "));
return 0;
}
bool again(string title)
{
string answer;
cout<<endl<<title;
getline(cin,answer);
return toupper(answer[0]) == 'Y';
}
void makeUpper(char word[])
{
char *ptr = word;
while (*ptr) {
*ptr = toupper(*ptr);
ptr++;
}
cout << "In uppercase:: " << word << endl;
}
char * cleanUp(char * astrid)
{
char *new_astrid;
for (*astrid; *astrid != '\0'; astrid++)
{
cout << "First loop";
if (isalpha(*astrid))
{
*new_astrid = *astrid;
new_astrid = ++new_astrid;
cout << "Here!";
}
}
cout << *new_astrid;
return *new_astrid;
}
void getReverse(char word[], char reversed[])
{
char *ptr_ind = find(word, '\0'), *ptr_ind_2 = reversed;
while(ptr_ind != word-1)
{
*ptr_ind_2 = *ptr_ind;
ptr_ind--;
ptr_ind_2++;
}
*ptr_ind_2 = '\0';
}
char * find(char *str, char what)
{
char *ptr = str;
while(*ptr != what && *ptr != '\0')
ptr++;
return *ptr == what ? ptr : NULL;
}
bool equal(char word[], char reverse[])
{
int total;
char * ptr;
ptr = word;
if((total = size(word)) != size(reverse))
return false;
for(char * ptr2 = reverse; *ptr != '\0' && *ptr == *ptr2; ptr++, ptr2++);
return *ptr == '\0';
}
int size(char word[])
{
int total = 0;
char * ptr = word;
while(*ptr != '\0') //while(!ptr)
{
ptr++;
total++;
}
return total;
}
There are several errors in your code.
new_astrid is not initialized and when you call *new_astrid = *astrid you try to copy a character to uninitialized memory, which will crash the program.
You also return the dereferenced pointer to new_astrid but the function prototype of cleanUp says that you return a pointer to char.
You should initialize new_astrid with new char[strlen(astrid)]. But then your code will result in memory leaks, since you increase the pointer (new_astid = ++new_astrid). So you should store the pointer first, so you can delete it later.
Instead of using raw pointers, i would suggest you use std::strings.
My suggestion for a palindrome tester would be:
#include <iostream>
#include <string>
#include <locale>
bool isPalindrome(std::string word)
{
std::locale loc;
for (std::string::size_type i = 0; i < word.length() / 2 + 1; ++i)
{
if (std::toupper(word[i],loc) != std::toupper(word[word.length() - i - 1],loc))
{
return false;
}
}
return true;
}
int main(int , char **)
{
std::string str = "Abba";
//Remove all non alpha values from string
str.erase(std::remove_if(str.begin(), str.end(), [](char const c){return !std::isalpha(c);}), str.end());
if (isPalindrome(str) == false)
{
std::cout << str << " is no palindrome" << std::endl;
}
else
{
std::cout << str << " is a palindrome" << std::endl;
}
return 0;
}
The erasion of non alpha values in the string is from this question.