Difficulties with string declaration/reference parameters (c++) - c++

Last week I got an homework to write a function: the function gets a string and a char value and should divide the string in two parts, before and after the first occurrence of the existing char.
The code worked but my teacher told me to do it again, because it is not well written code. But I don't understand how to make it better. I understand so far that defining two strings with white spaces is not good, but i get out of bounds exceptions otherwise. Since the string input changes, the string size changes everytime.
#include <iostream>
#include <string>
using namespace std;
void divide(char search, string text, string& first_part, string& sec_part)
{
bool firstc = true;
int counter = 0;
for (int i = 0; i < text.size(); i++) {
if (text.at(i) != search && firstc) {
first_part.at(i) = text.at(i);
}
else if (text.at(i) == search&& firstc == true) {
firstc = false;
sec_part.at(counter) = text.at(i);
}
else {
sec_part.at(counter) = text.at(i);
counter++;
}
}
}
int main() {
string text;
string part1=" ";
string part2=" ";
char search_char;
cout << "Please enter text? ";
getline(cin, text);
cout << "Please enter a char: ? ";
cin >> search_char;
divide(search_char,text,aprt1,part2);
cout << "First string: " << part1 <<endl;
cout << "Second string: " << part2 << endl;
system("PAUSE");
return 0;
}

I would suggest you, learn to use c++ standard functions. there are plenty utility function that can help you in programming.
void divide(const std::string& text, char search, std::string& first_part, std::string& sec_part)
{
std::string::const_iterator pos = std::find(text.begin(), text.end(), search);
first_part.append(text, 0, pos - text.begin());
sec_part.append(text, pos - text.begin());
}
int main()
{
std::string text = "thisisfirst";
char search = 'f';
std::string first;
std::string second;
divide(text, search, first, second);
}
Here I used std::find that you can read about it from here and also Iterators.
You have some other mistakes. you are passing your text by value that will do a copy every time you call your function. pass it by reference but qualify it with const that will indicate it is an input parameter not an output.

Why is your teacher right ?
The fact that you need to initialize your destination strings with empty space is terrible:
If the input string is longer, you'll get out of bound errors.
If it's shorter, you got wrong answer, because in IT and programming, "It works " is not the same as "It works".
In addition, your code does not fit the specifications. It should work all the time, independently of the current value which is stored in your output strings.
Alternative 1: your code but working
Just clear the destination strings at the beginning. Then iterate as you did, but use += or push_back() to add chars at the end of the string.
void divide(char search, string text, string& first_part, string& sec_part)
{
bool firstc = true;
first_part.clear(); // make destinations strings empty
sec_part.clear();
for (int i = 0; i < text.size(); i++) {
char c = text.at(i);
if (firstc && c != search) {
first_part += c;
}
else if (firstc && c == search) {
firstc = false;
sec_part += c;
}
else {
sec_part += c;
}
}
}
I used a temporary c instead of text.at(i) or text\[i\], in order to avoid multiple indexing But this is not really required: nowadays, optimizing compilers should produce equivalent code, whatever variant you use here.
Alternative 2: use string member functions
This alternative uses the find() function, and then constructs a string from the start until that position, and another from that position. There is a special case when the character was not found.
void divide(char search, string text, string& first_part, string& sec_part)
{
auto pos = text.find(search);
first_part = string(text, 0, pos);
if (pos== string::npos)
sec_part.clear();
else sec_part = string(text, pos, string::npos);
}

As you understand yourself these declarations
string part1=" ";
string part2=" ";
do not make sense because the entered string in the object text can essentially exceed the both initialized strings. In this case using the string method at can result in throwing an exception or the strings will have trailing spaces.
From the description of the assignment it is not clear whether the searched character should be included in one of the strings. You suppose that the character should be included in the second string.
Take into account that the parameter text should be declared as a constant reference.
Also instead of using loops it is better to use methods of the class std::string such as for example find.
The function can look the following way
#include <iostream>
#include <string>
void divide(const std::string &text, char search, std::string &first_part, std::string &sec_part)
{
std::string::size_type pos = text.find(search);
first_part = text.substr(0, pos);
if (pos == std::string::npos)
{
sec_part.clear();
}
else
{
sec_part = text.substr(pos);
}
}
int main()
{
std::string text("Hello World");
std::string first_part;
std::string sec_part;
divide(text, ' ', first_part, sec_part);
std::cout << "\"" << text << "\"\n";
std::cout << "\"" << first_part << "\"\n";
std::cout << "\"" << sec_part << "\"\n";
}
The program output is
"Hello World"
"Hello"
" World"
As you can see the separating character is included in the second string though I think that maybe it would be better to exclude it from the both strings.
An alternative and in my opinion more clear approach can look the following way
#include <iostream>
#include <string>
#include <utility>
std::pair<std::string, std::string> divide(const std::string &s, char c)
{
std::string::size_type pos = s.find(c);
return { s.substr(0, pos), pos == std::string::npos ? "" : s.substr(pos) };
}
int main()
{
std::string text("Hello World");
auto p = divide(text, ' ');
std::cout << "\"" << text << "\"\n";
std::cout << "\"" << p.first << "\"\n";
std::cout << "\"" << p.second << "\"\n";
}

Your code will only work as long the character is found within part1.length(). You need something similar to this:
void string_split_once(const char s, const string & text, string & first, string & second) {
first.clear();
second.clear();
std::size_t pos = str.find(s);
if (pos != string::npos) {
first = text.substr(0, pos);
second = text.substr(pos);
}
}

The biggest problem I see is that you are using at where you should be using push_back. See std::basic_string::push_back. at is designed to access an existing character to read or modify it. push_back appends a new character to the string.
divide could look like this :
void divide(char search, string text, string& first_part,
string& sec_part)
{
bool firstc = true;
for (int i = 0; i < text.size(); i++) {
if (text.at(i) != search && firstc) {
first_part.push_back(text.at(i));
}
else if (text.at(i) == search&& firstc == true) {
firstc = false;
sec_part.push_back(text.at(i));
}
else {
sec_part.push_back(text.at(i));
}
}
}
Since you aren't handling exceptions, consider using text[i] rather than text.at(i).

Related

How can i split adjacent numbers and letters in c++?

I've got a large text document that including adjacent numbers and letters.
Just like that,
JACK1940383DAVID30284HAROLD68372TROY4392 etc.
How can i split this like below in C++
List: Jack / 1940383 , David/30284, ...
You can use std::string::find_first_of() and std::string::find_first_not_of() in a loop, using std::string::substr() to extract each piece, eg:
std::string s = "JACK1940383DAVID30284HAROLD68372TROY4392";
std::string::size_type start = 0, end;
while ((end = s.find_first_of("0123456789", start)) != std::string::npos) {
std::string name = s.substr(start, end-start);
start = end;
int number;
if ((end = s.find_first_not_of("0123456789", start)) != std::string::npos) {
number = std::stoi(s.substr(start, end-start));
}
else {
number = std::stoi(s.substr(start));
}
start = end;
// use name and number as needed...
}
Online Demo
You can use regex like this:
#include <iostream>
#include <string>
#include <regex>
#include <vector>
// create a struct to group your data
// this makes it easy to store it in a vector.
struct person_t
{
std::string name;
std::string number;
};
// overloaded output operator for printing one person's details
std::ostream& operator<<(std::ostream& os, const person_t& person)
{
std::cout << person.name << ": " << person.number << std::endl;
return os;
}
// get a vector of person_t based on the input
auto get_persons(const std::string& input)
{
// make a regex in this case a regex that will match one or more capital letters
// and groups them using the ()
// then match one or more digits and group them too.
static const std::regex rx{ "([A-Z]+)([0-9]+)" };
std::smatch match;
// a vector to hold all the persons
std::vector<person_t> persons;
// start at begin of string and look for first part of the string
// that matches the regex.
auto cbegin = input.cbegin();
while (std::regex_search(cbegin, input.cend(), match, rx))
{
// match[0] will contain the whole match,
// match[1]-match[n] will contain the groups from the regular expressions
// match[1] will contain the match with characters and thus the name
// match[2] will contain the match with the numbers and thus the number.
// create a person_t struct with this info
person_t person{ match[1], match[2] };
// and add it to the vector
persons.push_back(person);
cbegin = match.suffix().first;
}
return persons;
}
int main()
{
// parse and split the string
auto persons = get_persons("JACK1940383DAVID30284HAROLD68372TROY4392");
// show the output
for (const auto& person : persons)
{
std::cout << person;
}
}
As pointed in other good answers you can use
find_first_of(), find_first_not_of() and substr() from std::string in a loop
regex
But it may be too much. I will add 3 more examples that you may find
simpler.
The first 2 programs expects the file name on the command line for (my) convenience here, and the test file is in.txt. Contents are the same as posted
JACK1940383DAVID30284HAROLD68372TROY4392
The last example just parses the string data declared as a char[]
1. Using fscanf()
Since the target is to consume formatted data, fscanf() is an option. As the data structure is very simple, the program is just a one line loop:
char mask[] = "%50[^0-9]%50[0-9]";
while ( 2 == fscanf(F, mask, tk_key, tk_value))
std::cout << tk_key << "/" << tk_value << "\n";
program output
output is the same for all examples
JACK/1940383
DAVID/30284
HAROLD/68372
TROY/4392
code for ex. 1
#include <errno.h>
#include <iostream>
int main(int argc,char** argv)
{
if (argc < 2)
{ std::cerr << "Use: pgm FileName\n";
return -1;
}
FILE* F = fopen(argv[1], "r");
if (F == NULL)
{
perror("Could not open file");
return -1;
}
std::cerr << "File: \"" << argv[1] << "\"\n";
char tk_key[50], tk_value[50];
char mask[] = "%50[^0-9]%50[0-9]";
while ( 2 == fscanf(F, mask, tk_key, tk_value))
std::cout << tk_key << "/" << tk_value << "\n";
fclose(F);
return 0;
}
using a state machine
There are just 2 states so it is not a fancy FSA ;) State machines are good for representing this kind of stuff, albeit here this seems to be overkill.
#define S_LETTER 0
#define S_DIGIT 1
#include <algorithm>
#include <iostream>
#include <fstream>
using iich = std::istream_iterator<char>;
int main(int argc,char** argv)
{
std::ifstream in_file{argv[1]};
if ( not in_file.good()) return -1;
iich p {in_file}, eofile{};
std::string token{}; // string to build values
char st = S_LETTER; // state value for FSA
std::for_each(p, eofile,
[&token,&st](char ch)
{
char temp = 0;
switch (st)
{
case S_LETTER:
if ((ch >= '0') && (ch <= '9'))
{
std::cout << token << "/";
token = ch;
st = S_DIGIT; // now in number
}
else token += ch; // concat in string
break;
case S_DIGIT:
default:
if ((ch < '0') || (ch > '9'))
{ // is a letter
std::cout << token << "\n";
token = ch;
st = S_LETTER; // now in name
}
else token += ch; // concat in string
break;
}; // switch()
});
std::cout << token << "\n"; // print last token
}
Here we have no loop. for_each gets the data from an iterator and passes it to a function that builds the name and the value as strings and couts them
Output is the same
3. a simple FSA to consume the data
#define S_LETTER 0
#define S_DIGIT 1
#include <iostream>
int main(void)
{
char one[] = "JACK1940383DAVID30284HAROLD68372TROY4392";
char* p = (char*)&one;
char* token = p;
char st = S_LETTER;
char temp = 0;
while (*p != 0)
{
switch (st)
{
case S_LETTER:
if ((*p >= '0') && (*p <= '9'))
{
temp = *p;
*p = 0;
std::cout << token << "/";
*p = temp;
token = p;
st = S_DIGIT; // now in number
}
break;
case S_DIGIT:
default:
if ( (*p < '0') || (*p > '9'))
{ // letter
temp = *p;
*p = 0;
std::cout << token << "\n";
*p = temp;
token = p;
st = S_LETTER; // now in name
}
break;
}; // switch()
p += 1; // next symbol
}; // while()
std::cout << token << "\n"; // print last token
}
This code just uses a C-style loop to parse the input data

How to run a string search algorithm through whole body of text

I am using the brute force string search algorithm to search through a small sentence, however I want the algorithm to return every time it finds the certain string instead of finding it once and then stopping
//Declare and initialise variables
string pat, text;
text = "This is a test sentence, find test within this string";
cout << text << endl;
//User input for pat
cout << "Please enter the string you want to search for" << endl;
cin >> pat;
//Set the length of the pat and text
int patLength = pat.size();
int textLength = text.size();
//Algorithm
for (int i = 0; i < textLength - patLength; ++i)
{
//Do while loop to run through the whole text
do
{
int j;
for (j = 0; j < patLength; j++)
{
if (text[i + j] != pat[j])
break; // Doesn't match here.
}
if (j == patLength)
{
finds.push(i); // Matched here.
}
} while (i < textLength);
}
//Print output
cout << "String: " << pat << " was found at positions: " << finds.top();
The program stores each find in a queue. When I run this program, it asks for the 'pat', then does nothing. I have done a bit of debugging and found that it is probably the do while loop. However I can't find a fix
You could use the std::string::find function combined with a function that you call for each find.
#include <iostream>
#include <functional>
#include <vector>
#include <sstream>
void Algorithm(
const std::string& text, const std::string& pat,
std::function<void(const std::string&,size_t)> f, std::vector<size_t>& positions)
{
size_t pos=0;
while((pos=text.find(pat, pos)) != std::string::npos) {
// store the position
positions.push_back(pos);
// call the supplied function
f(text, pos++);
}
}
// function to call for each position in which the pattern is found
void gotit(const std::string& found_in, size_t pos) {
std::cout << "Found in \"" << found_in << "\" # " << pos << "\n";
}
int main(int argc, char* argv[]) {
std::vector<std::string> args(argv+1, argv+argc);
if(args.size()==0)
args.push_back("This is a test sentence, find test within this string");
for(const auto& text : args) {
std::vector<size_t> found_at;
std::cout << "Please enter the string you want to search for: ";
std::string pat;
std::cin >> pat;
Algorithm(text, pat, gotit, found_at);
std::cout << "collected positions:\n";
for(size_t pos : found_at) {
std::cout << pos << "\n";
}
}
}
My first bit of advice would be to structure your code into separate functions.
Let's say you have a function that returns the position of the pattern's first occurrence in a sequence of characters:
using position = typename std::string::const_iterator;
position first_occurrence(position text_begin, position text_end, const std::string& pattern);
If there is no more occurrence of the pattern, it returns text_end.
You can now write a very simple loop:
auto occurrence = first_occurrence(text_begin, pattern);
while (occurrence != text_end) {
occurrences.push_back(occurrence);
occurrence = first_occurence(occurrence + 1, text_end, pattern);
}
to accumulate all the occurrences of the pattern.
The first_occurrence function already exists in the standard library under the name of std::search. Since C++17, you can customize this function with pattern-searching specialized searchers, such as std::boyer_moore_searcher: it pre-processes the pattern to make it faster to look for in the string. Here's an example application to your problem:
#include <algorithm>
#include <string>
#include <vector>
#include <functional>
using occurrence = typename std::string::const_iterator;
std::vector<occurrence> find_occurrences(const std::string& input, const std::string& pattern) {
auto engine = std::boyer_moore_searcher(pattern.begin(), pattern.end());
std::vector<occurrence> occurrences;
auto it = std::search(input.begin(), input.end(), engine);
while (it != input.end()) {
occurrences.push_back(it);
it = std::search(std::next(it), input.end(), engine);
}
return occurrences;
}
#include <iostream>
int main() {
std::string text = "This is a test sentence, find test within this string";
std::string pattern = "st";
auto occs = find_occurrences(text, pattern);
for (auto occ: occs) std::cout << std::string(occ, std::next(occ, pattern.size())) << std::endl;
}

C++ remove punctuation marks and spaces from a string

How can I remove punctuation marks and spaces from a string in a simple way without using any library functions?
int main()
{
string s = "abc de.fghi..jkl,m no";
for (int i = 0; i < s.size(); i++)
{
if (s[i] == ' ' || s[i] == '.' || s[i] == ',')
{
s.erase(i, 1); // remove ith char from string
i--; // reduce i with one so you don't miss any char
}
}
cout << s << endl;
}
Assuming you can use library I/O like <iostream> and types like std::string and you just don't want to use the <cctype> functions like ispunct().
#include <iostream>
#include <string>
int main()
{
const std::string myString = "This. is a string with ,.] stuff in, it.";
const std::string puncts = " [];',./{}:\"?><`~!-_";
std::string output;
for (const auto& ch : myString)
{
bool found = false;
for (const auto& p : puncts)
{
if (ch == p)
{
found = true;
break;
}
}
if (!found)
output += ch;
}
std::cout << output << '\n';
return 0;
}
No idea about the performance, I'm sure it can be done in multiple better ways.

Print out the words of a line in reverse order through recursion

I'm trying not to use any storage containers. I don't know if it's even possible. Here is what I have so far. (I'm getting a segmentation fault).
#include <iostream>
#include <string>
using namespace std;
void foo(string s)
{
size_t pos;
pos = s.find(' ');
if(pos == string::npos)
return;
foo(s.erase(0, pos));
cout << s.substr(0, pos) << " ";
}
int main()
{
foo("hello world");
return 0;
}
I know there's probably many things wrong with this code. So rip away. I'm eager to learn. I'm trying to imitate a post order print as you would do in a reverse print of a singly linked list. Thanks.
EDIT:
An example:
"You are amazing" becomes "amazing are You"
The segfault is a stack overflow.
foo( "hello world" ) erases everything up to the first space (" world") and recurses.
foo( " world" ) erases everything up to the first space (" world") and recurses.
foo( " world" )... you get the idea.
Also, once you called foo( s.erase( 0, pos ) ), trying to print s.substr( 0, pos ) after the recursion returns does not make sense. You need to save the substring somewhere before you erase it, so you still have it to print afterwards.
void foo(string s)
{
size_t pos = s.find(' '); // declare-and-use in one line
string out = s.substr( 0, pos ); // saving the substring
if ( pos != string::npos )
{
foo( s.erase( 0, pos + 1 ) ); // recurse, skipping the space...
cout << " "; // ...but *print* the space
}
cout << out; // print the saved substring
}
The problem is that your recursion continues until you run out of memory.
Pay attention to this line:
if(pos == string::npos)
when your erase the substring you don't erase the white space so in the next recursion s.find returns pos = 0 which means that your recursion never ends.
Here is a code that works. Also note that I added a level variable to be able to control the behaviour on the first level (in this case add a endl)
#include <iostream>
#include <string>
using namespace std;
void foo(string s, int l)
{
size_t pos;
pos = s.find(' ');
if(pos == string::npos){
cout << s << " ";
return;
}
string temp = s.substr(0, pos);
foo(s.erase(0, pos+1),l+1);
cout << temp << " ";
if(l == 0)
cout << endl;
}
int main()
{
foo("hello world", 0);
return 0;
}
An approach to recursion, which may allow your compiler to transform automatically to iteration, is to accumulate the result in the function arguments. This will be familiar if you've written recursive functions in any of the Lisp family of languages:
#include <iostream>
#include <string>
std::string reverse_words(const std::string& s, const std::string& o = {})
{
using std::string;
const auto npos = string::npos;
static const string whitespace(" \n\r\t");
// find start and end of the first whitespace block
auto start = s.find_first_of(whitespace);
if (start == npos)
return s + o;
auto end = s.find_first_not_of(whitespace, start);
if (end == npos)
return s + o;
auto word = s.substr(0, start);
auto space = s.substr(start, end-start);
auto rest = s.substr(end);
return reverse_words(rest, space + word + o);
}
int main()
{
std::cout << reverse_words("hello to all the world") << std::endl;
std::cout << reverse_words(" a more difficult\n testcase ") << std::endl;
return 0;
}
I tried to make a brief example by using standard algorithms. I also handles more kinds of spaces than just standard whitespace (tabs for instance).
#include <cctype>
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
void print_reverse(string words) {
// Termination condition
if(words.empty())
return;
auto predicate = (int(*)(int))isspace;
auto sit = begin(words);
auto wit = find_if_not(sit, end(words), predicate);
auto nit = find_if (wit, end(words), predicate);
print_reverse(string(nit, end(words)));
// word spaces
cout << string(wit, nit) << string(sit, wit);
}
int main() {
string line;
getline(cin, line);
print_reverse(line);
cout << endl;
}
Here is an example run:
$ ./print-out-the-words-of-a-line-in-reverse-order-through-recursion
You are amazing
amazing are You
The key is in adding 1 to pos in the erase statement.
So try:
#include <iostream>
#include <string>
using namespace std;
void foo(string s)
{
size_t pos;
pos = s.find(' ');
if(pos == string::npos)
{
cout << s << " ";
return;
}
string out = s.substr(0, pos);
foo(s.erase(0, pos+1));
cout << out << " ";
}
int main()
{
foo("hello world");
cout << endl;
return 0;
}
EDIT
Alternatively you could use a char* instead of a std::string, then you do not need to make a temp variable. Try it online.
#include <iostream>
#include <cstring>
void foo(char* s)
{
char* next = std::strchr(s, ' ');
if(next != nullptr)
{
foo(next + 1);
*next = 0;
}
std::cout << s << " ";
}
int main()
{
char s[] = "You are amazing";
foo(s);
std::cout << std::endl;
}
The problem is that you're not doing anything with the last word and you're not doing anything with the remaining chunk.
If you have a recursive reverse printer, you'll want something like this (pseudocode):
def recursive-reverse(string) {
pos = string.find-last(" ");
if pos doesn't exist {
print string;
return;
} else {
print string.create-substring(pos+1, string.end);
recursive-reverse(string.create-substring(0, pos));
}
}
To implement this in C++:
#include <iostream>
#include <string>
void recursive_reverse(std::string &s) {
// find the last space
size_t pos = s.find_last_of(" ");
// base case - there's no space
if(pos == std::string::npos) {
// print only word in the string
std::cout << s << std::endl;
// end of recursion
return;
} else {
// grab everything after the space
std::string substring = s.substr(pos+1);
// print it
std::cout << substring << std::endl;
// grab everything before the space
std::string rest = s.substr(0, pos);
// recursive call on everything before the space
recursive_reverse(rest);
}
}
int main() {
std::string s("Hello World!");
recursive_reverse(s);
return 0;
}
ideone

String.erase giving out_of_range exception

I was meant to write some program which will read text from text file and erase given words.
Unfortunately, something's wrong with this particular part of code, I get the following exception notification:
This text is just a sample, based on other textterminate called after throwing
an instance of 'std::out_of_range' what<>: Basic_string_erase
I guess that there is something wrong with the way I use erase, I'm trying to to use do while loop, determine the beginning of word which is meant to be erased every time the loop is done and eventually erase text which begins at the beginning of word which is meant to be erased and the end of it - I'm using its length.
#include <iostream>
#include <string>
using namespace std;
void eraseString(string &str1, string &str2) // str1 - text, str2 - phrase
{
size_t positionOfPhrase = str1.find(str2);
if(positionOfPhrase == string::npos)
{
cout <<"Phrase hasn't been found... at all"<< endl;
}
else
{
do{
positionOfPhrase = str1.find(str2, positionOfPhrase + str2.size());
str1.erase(positionOfPhrase, str2.size());//**IT's PROBABLY THE SOURCE OF PROBLEM**
}while(positionOfPhrase != string::npos);
}
}
int main(void)
{
string str("This text is just a sample text, based on other text");
string str0("text");
cout << str;
eraseString(str, str0);
cout << str;
}
Your function is wrong. It is entirely unclear why you call method find twice after each other.
Try the following code.
#include <iostream>
#include <string>
std::string & eraseString( std::string &s1, const std::string &s2 )
{
std::string::size_type pos = 0;
while ( ( pos = s1.find( s2, pos ) ) != std::string::npos )
{
s1.erase( pos, s2.size() );
}
return s1;
}
int main()
{
std::string s1( "This text is just a sample text, based on other text" );
std::string s2( "text" );
std::cout << s1 << std::endl;
std::cout << eraseString( s1, s2 ) << std::endl;
return 0;
}
The program output is
This text is just a sample text, based on other text
This is just a sample , based on other
I think your trouble is that positionOfPhrase inside do loop can be string::npos, in which case erase will throw an exception. This can be fixed by changing logic to:
while (true) {
positionOfPhrase = str1.find(str2, positionOfPhrase + str2.size());
if (positionOfPhrase == string::npos) break;
str1.erase(positionOfPhrase, str2.size());
}