I'm trying to learn strings and I've figured out how to replace as well as insert into an existing string. I have 3 strings at the moment which I've declared as constants, I've merged them into one string variable which puts them all one after eachother.
I've also changed every single occurance of "Hi" to "Bye" in those strings. My 3 strings bundled into a single one are as following:
"Hi! My name is xxxx! I would like to be on my own but I don't know how to, could you help me?"
I want it to display as:
Hi!
My name is xxxx!
I would like to be on my own but I don't know how to, could you help me?
As soon as a puncutation occurs I'd like to insert a line break "\n", using replace works but that means the punctuation will disappear, using insert will first insert the line break before the punctuation, and it won't continue to the next one which results in:
"Hi!
My name is xxxx! I would like to be on my own but I don't know how to, could you help me?"
I changed the code to only include dots to simplify it, once solved the same solution can be applied to any other part such as question marks or exclamation marks.
Any tips on how to fix this?
#include <iostream>
#include <string>
using namespace std;
string const Text0 = "Hi.";
string const Text1 = "My name is xxxx.";
string const Text2 = "I would like to be on my own but I don't know how to, could you help me.";
string const Text3 = "I would, but I don't know how to.";
string text = Text0 + Text1 + Text2 + Text3;
int main() {
while (text.find("I") != string::npos) {
text.replace(text.find("I"), 1, "J");
}
while (text.find("like") != string::npos) {
text.replace(text.find("like"), 4, "milk");
}
text.insert(text.find("."), "\n");
cout << text;
return 0;
}
You can create your own short function that will add newline after every punctuation sign.
For example:
void addNewLines(std::string *text)
{
for (int i = 0; i < text->length(); i++)
{
if ((*text)[i] == '!' || (*text)[i] == '?' || (*text)[i] == '.')
{
(*text)[i + 1] = '\n';
}
}
}
As you can see in this piece of code, in the for loop you are going from the first to the last character of the string, and after every punctuation sign you replace empty space with \n character.
I'm using pointers here to prevent copying of the string to the function, in case it is a huge string, but you could do it without pointers, that way syntax is a little bit cleaner.
Related
I'm trying to add a space behind capital letters of a string for example. "ILikeBananas" would turn into "I like bananas"
I've tried using a while loop, isupper, and .insert((i-1), " ").
I noticed that it would work if it were i+1 however that would give me the wrong output.
void fixedInput(string &userInput) {
int i = 1;
while (userInput[i]) {
if (isupper(userInput[i])) {
userInput.insert((i-1)," ");
tolower(userInput[i]);
}
i++;
}
}
with (i-1) there is no output
Take a look at https://stackoverflow.com/a/14494432/11397643
Using that concept you should be able to append a space as well when doing a conversion
Given the code:
procedure example {
x=3;
y = z +c ;
while p {
b = a+c ;
}
}
I would like to split the code by using the delimiters {, ;, and }.
After splitting, I would like to get the information before it together with the delimiter.
So for example, I would like to get procedure example {, x=3;, y=z+c;, }. Then I would like to push it into a list<pair<int, string>> sList. Could someone explain how this can be done in c++?
I tried following this example: Parse (split) a string in C++ using string delimiter (standard C++), but I could only get one token. I want the entire line. I am new to c++, and the list, splitting, etc. is confusing.
Edit: So I have implemented it, and this is the code:
size_t openCurlyBracket = lines.find("{");
size_t closeCurlyBracket = lines.find("}");
size_t semiColon = lines.find(";");
if (semiColon != string::npos) {
cout << lines.substr(0, semiColon + 1) + "\n";
}
However, it seems that it can't separate based on semicolon separately, openBracket and closeBracket. Anyone knows how to separate based on these characters individually?
2nd Edit:
I have done this (codes below). It is separating correctly, I have one for open curly bracket. I was planning on adding the value to the list in the commented area below. However, when i think about it, if i do that, then the order of information in the list will be messed up. As i have another while loop which separates based on open curly bracket. Any idea on how i can add the information in an order?
Example:
1. procedure example {
2. x=3;
3. y = z+c
4. while p{
and so on.
while (semiColon != string::npos) {
semiColon++;
//add list here
semiColon = lines.find(';',semiColon);
}
I think that you should read about std::string::find_first_of function.
Searches the string for the first character that matches any of the characters specified in its arguments.
I have a problem to understand what you really want to achieve. Let's say this is an example of the find_first_of function use.
list<string> split(string lines)
{
list<string> result;
size_t position = 0;
while((position = lines.find_first_of("{};\n")) != string::npos)
{
if(lines[position] != '\n')
{
result.push_back(lines.substr(0, position+1));
}
lines = lines.substr(position+1);
}
return result;
}
I am making a roman numeral converter. I have everything figured out except there is one problem at the end.
The string looks like IVV
I need to make it IX
I have split the string at each new letter, then appended them back on, then using an if statement to see if it contains 2 "V"s. I want to know if there is a simpler way to do this.
Using std::string should help you tremendously as you can leverage its search and replace functionality. You'll want to start with the find function which allows you to search for a character or a string and returns an index where what you are searching for exists or npos if the search fails.
You can then call replace passing it the index returned by find, the number of characters you want to replace and what replace the range with.
The code below should help you get started.
#include <string>
#include <iostream>
int main()
{
std::string roman("IVV");
// Search for the string you want to replace
std::string::size_type loc = roman.find("VV");
// If the substring is found replace it.
if (loc != std::string::npos)
{
// replace 2 characters staring at position loc with the string "X"
roman.replace(loc, 2, "X");
}
std::cout << roman << std::endl;
return 0;
}
You could use std string find and rfind operations, these find the position of the first and the last occurrence of the entered parameter, check if these are not equal and you will know
Answer updated
#include <string>
int main()
{
std::string x1 = "IVV";
if (x1.find('V') !=x1.rfind('V'))
{
x1.replace(x1.find('V'), 2, 'X');
}
return 0;
}
I'm making a function that removes elements from a string. However, I cant seem to get both of my loops to work together. The first while loop works flawlessly. I looked into it and I believe it might be because when "find_last_of" isn't found, it still returns a value (which is throwing off my loop). I haven't been able to figure out how I can fix it. Thank you.
#include <iostream>
#include <string>
using namespace std;
string foo(string word) {
string compare = "!##$";
string alphabet = "abcdefghijklmnopqrstuvxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
while(word.find_first_of(compare) < word.find_first_of(alphabet)) {
int position = word.find_first_of(compare);
word = word.substr(++position);
}
while(word.find_last_of(compare) > word.find_last_of(alphabet)){
int size = word.length();
word = word.substr(0, --size);
}
return word;
}
int main() {
cout << foo("!!hi!!");
return 0;
}
I wrote it like this so compound words would not be affected. Desired result: "hi"
It's not entirely clear what you're trying to do, but how about replacing the second loop with this:
string::size_type p = word.find_last_not_of(compare);
if(p != string::npos)
word = word.substr(0, ++p);
It's not clear if you just want to trim certain characters from the front and back of word or if you want to remove every one of a certain set of characters from word no matter where they are. Based on the first sentence of your question, I'll assume you want to do the latter: remove all characters in compare from word.
A better strategy would be to more directly examine each character to see if it needs to be removed, and if so, do so, all in one pass through word. Since compare is quite short, something like this is probably good enough:
// Rewrite word by removing all characters in compare (and then erasing the
// leftover space, if any, at the end). See std::remove_if() docs.
word.erase(std::remove_if(word.begin(),
word.end(),
// Returns true if a character is to be removed.
[&](const char ch) {
return compare.find(ch) != compare.npos;
}),
word.end());
BTW, I'm not sure why there is both a compare and alphabet string in your example. It seems you would only need to define one or the other, and not both. A character is either one to keep or one to remove.
So basically, I might have some string that looks like: "hey this is a string * this string is awesome 97 * 3 = 27 * this string is cool".
However, this string might be huge. I'm trying to remove all the asterisks from the string, unless that asterisk appears to represent multiplication. Efficiency is somewhat important here, and I'm having trouble coming up with a good algorithm to remove all the non-multiplication asterisks from this.
In order to determine whether an asterisk is for multiplication, I can obviously just check whether it's sandwiched in between two numbers.
Thus, I was thinking I could do something like (pseudocode):
wasNumber = false
Loop through string
if number
set wasNumber = true
else
set wasNumber = false
if asterisk
if wasNumber
if the next word is a number
do nothing
else
remove asterisk
else
remove asterisk
However, that^ is ugly and inefficient on a huge string. Can you think of a better way to accomplish this in C++?
Also, how could I actually check whether a word is a number? It's allowed to be a decimal. I know there's a function to check if a character is a number...
Fully functioning code:
#include <iostream>
#include <string>
using namespace std;
string RemoveAllAstericks(string);
void RemoveSingleAsterick(string&, int);
bool IsDigit(char);
int main()
{
string myString = "hey this is a string * this string is awesome 97 * 3 = 27 * this string is cool";
string newString = RemoveAllAstericks(myString);
cout << "Original: " << myString << "\n";
cout << "Modified: " << newString << endl;
system("pause");
return 0;
}
string RemoveAllAstericks(string s)
{
int len = s.size();
int pos;
for(int i = 0; i < len; i++)
{
if(s[i] != '*')
continue;
pos = i - 1;
char cBefore = s[pos];
while(cBefore == ' ')
{
pos--;
cBefore = s[pos];
}
pos = i + 1;
char cAfter = s[pos];
while(cAfter == ' ')
{
pos++;
cAfter = s[pos];
}
if( IsDigit(cBefore) && IsDigit(cAfter) )
RemoveSingleAsterick(s, i);
}
return s;
}
void RemoveSingleAsterick(string& s, int i)
{
s[i] = ' '; // Replaces * with a space, but you can do whatever you want
}
bool IsDigit(char c)
{
return (c <= 57 && c >= 48);
}
Top level overview:
Code searches the string until it encounters an *. Then, it looks at the first non-whitespace character before AND after the *. If both characters are numeric, the code decides that this is a multiplication operation, and removes the asterick. Otherwise, it is ignored.
See the revision history of this post if you'd like other details.
Important Notes:
You should seriously consider adding boundary checks on the string (i.e. don't try to access an index that is less than 0 or greater than len
If you are worried about parentheses, then change the condition that checks for whitespaces to also check for parentheses.
Checking whether every single character is a number is a bad idea. At the very least, it will require two logical checks (see my IsDigit() function). (My code checks for '*', which is one logical operation.) However, some of the suggestions posted were very poorly thought out. Do not use regular expressions to check if a character is numeric.
Since you mentioned efficiency in your question, and I don't have sufficient rep points to comment on other answers:
A switch statement that checks for '0' '1' '2' ..., means that every character that is NOT a digit, must go through 10 logical operations. With all due respect, please, since chars map to ints, just check the boundaries (char <= '9' && char >= '0')
You can start by implementing the slow version, it could be much faster than you think. But let's say it's too slow. It then is an optimization problem. Where does the inefficiency lies?
"if number" is easy, you can use a regex or anything that stops when it finds something that is not a digit
"if the next word is a number" is just as easy to implement efficiently.
Now, it's the "remove asterisk" part that is an issue to you. The key point to notice here is that you don't need to duplicate the string: you can actually modify it in place since you are only removing elements.
Try to run through this visually before trying to implement it.
Keep two integers or iterators, the first one saying where you are currently reading your string, and the second one saying where you are currently writing your string. Since you only erase stuff, the read one will always be ahead of the writing one.
If you decide to keep the current string, you just need to advance each of your integers/iterators one by one, and copying accordingly. If you don't want to keep it, just advance the reading string! Then you only have to cut the string by the amount of asterisks you removed. The complexity is simply O(n), without any additional buffer used.
Also note that your algorithm would be simpler (but equivalent) if written like this:
wasNumber = false
Loop through string
if number
set wasNumber = true
else
set wasNumber = false
if asterisk and wasNumber and next word is a number
do nothing // using my algorithm, "do nothing" actually copies what you intend to keep
else
remove asterisk
I found your little problem interesting and I wrote (and tested) a small and simple function that would do just that on a std::string. Here u go:
// TestStringsCpp.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <string>
#include <iostream>
using namespace std;
string& ClearAsterisk(string& iString)
{
bool bLastCharNumeric = false;
string lString = "0123456789";
for (string::iterator it = iString.begin(); it != iString.end() ; ++it) {
switch (*it) {
case ' ': break;//ignore whitespace characters
case '*':
if (bLastCharNumeric) {
//asterisk is preceded by numeric character. we have to check if
//the following non space character is numeric also
for (string::iterator it2 = it + 1; it2 != iString.end() ; ++it2) {
if (*it2 != ' ') {
if (*it2 <= '9' && *it2 >= '0') break;
else iString.erase(it);
break; //exit current for
}
}
}
else iString.erase(it);;
break;
default:
if (*it <= '9' && *it >= '0') bLastCharNumeric= true;
else bLastCharNumeric = false; //reset flag
}
}
return iString;
}
int _tmain(int argc, _TCHAR* argv[])
{
string testString = "hey this is a string * this string is awesome 97 * 3 = 27 * this string is cool";
cout<<ClearAsterisk(testString).c_str();
cin >> testString; //this is just for the app to pause a bit :)
return 0;
}
It will work perfectly with your sample string but it will fail if you have a text like this: "this is a happy 5 * 3day menu" because it checks only for the first nonspace character after the '*'. But frankly I can't immagine a lot of cases you would have this kind of construct in a sentence.
HTH,JP.
A regular expression wouldn't necessarily be any more efficient, but it would let you rely on somebody else to do your string parsing and manipulation.
Personally, if I were worried about efficiency, I would implement your pseudocode version while limiting needless memory allocations. I might even mmap the input file. I highly doubt that you'll get much faster than that.