The function I am using is
vector<string> tokenise(string s){
}
Firstly, I intend to split the string into substrings, in which case the string is always arithmetic expression (e.g. "100+5") and there could be some whitespaces.
"100+5" is needed to convert to "100", "+", "5"
After conversion, the substrings will be stored in a vector and return it. I am struggling with the fist step and using the subscript to loop over a string. The type of the value returned is char, so there is no way to put it in the vector.
You could just call the string's substring method, after figuring out the range of characters that are either digits, arithmetic characters, or unwanted.
You mentioned - The type of the value returned is char, so there is no way to put it in the vector.
You have some function that returns a character. You want to then insert the equivalent string into the vector.
Assuming your vector is defined as
std::vector<std::string> broken_strings;
So you can do it as follows.
char ch = ...; // Here comes the character that you get from the function.
std::string str(1, ch);
broken_strings.push_back(str);
Then you can return broken_strings.
Edit:
OP mentions that he wants to tokenize algebraic expressions.
So it will have to be done it a different way.
Following is a simple approach.
std::vector<std::string> broken;
std::string temp;
for ( int i = 0; i<s.length() ;i++){
char ch = s[i];
if (ch == ' ')
continue;
else if (ch >= '0' && ch <='9')
temp += ch;
else{
if (temp.length() != 0)
broken.push_back(temp);
temp = "";
temp += ch;
broken.push_back(temp);
temp = "";
}
}
if (temp.length() != 0)
broken.push_back(token);
return broken;
You can see the demo of the same here
Ideone
vector<string> tokenise(string s)
{
vector<string> v;
string number;
for(int i = 0; i < s.length(); ++i)
{
if((s[i] >= '0') && (s[i] <= '9'))
{
number += string(1, s[i]);
}
else if(s[i] == '.')
{
number += string(1, s[i]);
}
else if((s[i] == '+') || (s[i] == '-') || (s[i] == '*') || (s[i] == '/'))
{
if(number.size())
{
v.push_back(number);
number.clear();
}
v.push_back(string(1, c));
}
}
if(number.size())
{
v.push_back(number);
number.clear();
}
return v;
}
Related
Basically I have to encode a name into a Soundex Code. The helper functions I implemented do the following:
Discard all non-letter characters from the surname: dashes, spaces, apostrophes, and so on.
Encode each letter as a digit
Coalesce adjacent duplicate digits from the code (e.g. 222025 becomes 2025).
Replace the first digit of the code with the first letter of the original name, converting to uppercase.
Remove all zeros from the code.
Make the code exactly length 4 by padding with zeros or truncating the excess.
Excuse the implementation of the helper functions, I know they could be implemented better. But when I manually pass the output from one function to another I see that the result is what I want. It's only when I combine them all into one function that I see that the output I pass is as if I didn't modify the input I passed at all. I believe my issue might have to do with passing by reference but doing that for all my functions made no difference or gave an incorrect output.
#include <iostream>
#include <string>
string removeNonLetters(string s) {
string result = "";
for (int i = 0; i < s.length(); i++) {
if (isalpha(s[i])) {
result += s[i];
}
}
return result;
}
string encode(string name) {
std::transform(name.begin(), name.end(), name.begin(), ::toupper);
string encoded = "";
for (int i = 0; i < name.size(); ++i) {
if (name[i] == 'A' || name[i] == 'E' || name[i] == 'I' || name[i] == 'O' || name[i] == 'U' || name[i] == 'H' || name[i] == 'W' || name[i] == 'Y')
encoded += '0';
else if (name[i] == 'B' || name[i] == 'F' || name[i] == 'P' || name[i] == 'V')
encoded += '1';
else if (name[i] == 'C' || name[i] == 'G' || name[i] == 'J' || name[i] == 'K' || name[i] == 'Q' || name[i] == 'S' || name[i] == 'X' || name[i] == 'Z')
encoded += '2';
else if (name[i] == 'D' || name[i] == 'T')
encoded += '3';
else if (name[i] == 'L')
encoded += '4';
else if (name[i] == 'M' || name[i] == 'N')
encoded += '5';
else if (name[i] == 'R')
encoded += '6';
}
return encoded;
}
string removeDuplicate(string encoded) {
for (int i = 0; i < encoded.size(); ++i) {
if (encoded[i] == encoded[i+1])
encoded[i] = '\0';
}
return encoded;
}
string removeZeros(string digits) {
for (int i = 0; i < digits.size(); ++i) {
if (digits[i] == '0')
digits[i] = '\0';
}
return digits;
}
string padding(string output) {
int size = output.size();
if (size < 4) {
for (int i = size; i < 4; ++i)
output += '0';
}
else if (size > 4) {
for (int j = size; j > 3; --j)
output[j] = '\0';
}
return output;
}
/* TODO: Replace this comment with a descriptive function
* header comment.
*/
string soundex(string s) {
/* TODO: Fill in this function. */
string copy = s;
removeNonLetters(s);
encode(s);
removeDuplicate(s);
s[0]= copy[0];
removeZeros(s);
padding(s);
return s;
}
int main() {
string s = "Curie";
cout << soundex(s) << '\n';
// Output should be C600 but I keep getting "Curie."
}
Your functions return the adjusted strings, that's good. But your calling code doesn't use the returned values!
Something like this is what you want.
string soundex(string s) {
/* TODO: Fill in this function. */
string copy = s;
s = removeNonLetters(s);
s = encode(s);
s = removeDuplicate(s);
s[0] = copy[0];
s = removeZeros(s);
s = padding(s);
return s;
}
If you want to change the value of a variable you normally use =. I'm sure you know that but for some reason you forgot because functions are involved.
I have the debug assertion error in the if statement when i = 7:
Expression: c>= -1 && c <= 255
This is my code:
#include <iostream>
#include <string>
const char* clearString(std::string str)
{
for (int i = str.length() - 1; i >= 0; i--)
{
if ( !isdigit(str[i])
&& str[i] != ',')
{
str.erase(i, 1);
}
}
return str.c_str();
}
int main()
{
std::string str = "688,13 €";
std::cout << clearString(str);
}
I try to delete all characters in the string that are not numbers and ','.
For std::isdigit, see the Notes section as to why you are getting the assertion.
The fix is to cast to an unsigned char:
if (!isdigit(static_cast<unsigned char>(str[i]))
Second, your function returns the address of a local temporary, thus exhibits undefined behavior. Return a std::string instead.
std::string clearString(std::string str)
{
//…
return str;
}
Third, you could rewrite your function using std::remove_if and std::string::erase, instead of writing a loop that removes a character at a time.
#include <algorithm>
//...
std::string clearString(std::string str)
{
auto iter = std::remove_if(str.begin(), str.end(),
[&](char ch)
{ return !isdigit(static_cast<unsigned char>(ch)) && ch != ',';});
str.erase(iter, str.end());
return str;
}
The function isdigit() works only with chars that their decimal value is between -1 and 255.
The decimal value of the character € is -128 which the function doesn't support.
I would suggest to change the comparison instead of using isdigit(), compare the decimal values of the chars.
Change your function to this:
const char* clearString(std::string& str)
{
for (int i = str.length() - 1; i >= 0; i--)
{
if ((str[i] < '0' || str[i] > '9') && str[i] != ',')
{
str.erase(i, 1);
}
}
return str.c_str();
}
A little out of topic, about your algorithm.
It would be better if you don't erase every non-digit character, but shift your characters left, skipping all non-digits (except ',') and resize string.
About isdigit I would do how 0xBlackMirror suggested, compare to '0' and '9'.
Here is the code:
const char* clearString(std::string str)
{
int j = 0;
for (uint i = 0; i < str.size(); i++)
{
if ((str[i] >= '0' && str[i] <= '9') || str[i] == ',')
{
str[j++] = str[i];
}
}
str.resize(j);
return str.c_str();
}
Using the C++ language, the function LetterChanges(str) takes the str parameter being passed and modifies it using the following algorithm.
Replace every letter in the string with the letter following it in the
alphabet (ie. c becomes d, z becomes a). Then capitalize every vowel
in this new string (a, e, i, o, u) and finally return this modified
string.
#include <iostream>
using namespace std;
string LetterChanges(string str) {
// code goes here
string str2 = "abcdefghijklmnopqrstuvwxyz";
int j;
for (int i = 0; str[i] != '\0'; i++) {
for (j = 0; str2[j] != '\0'; j++) {
if (str[i] == str2[j]) {
str[i] = str2[j + 1];
}
}
}
for (int i = 0; str[i] != '\0'; i++) {
if (str[i] == 'a') {
str[i] = 'A';
}
else if (str[i] == 'e') {
str[i] = 'E';
}
else if (str[i] == 'i') {
str[i] = 'I';
}
else if (str[i] == 'o') {
str[i] = 'O';
}
else if (str[i] == 'u') {
str[i] = 'U';
}
}
return str;
}
int main() {
// keep this function call here
cout << LetterChanges(gets(stdin));
return 0;
}
I am trying to run this code but it is not giving the desire output please help..
The Function
Let's start with your first loop:
for (int i = 0; str[i] != '\0'; i++) {
for (j = 0; str2[j] != '\0'; j++) {
if (str[i] == str2[j]) {
str[i] = str2[j + 1];
}
}
}
There are a couple things here. Firstly, we don't even need to deal with str2. Characters in C++ use ASCII encoding, meaning we can actually do something like str[i]++ to change an 'a' to a 'b' or an 'e' to an 'f', etc...
Also, I'd advise against using str[i] != '\0'. We're using the standard library strings instead of c-strings for a reason, so we might as well make our lives easier and use str.size(). Along these same lines, I'd suggest str.at(i) as opposed to str[i] as the former will do bounds checking for us.
Lastly, if you include cctype, then we can use the isalpha function to make sure we're only modifying alphabetic characters (no numbers or spaces, etc..).
Thus your first loop can become:
for (int i = 0; i < str.size(); i++) {
if(isalpha(str.at(i)){
if(str.at(i) == 'z') str.at(i) = 'a'; //special case
else str.at(i)++;
}
}
As far as your second loop, you don't even need it! We can actually incorporate everything straight into the first one. As long as we make sure to do the vowel modification after we've changed the individual letters.
A conversion from lowercase to uppercase can be done with some ASCII math as well. The difference between the lowercase and uppercase letters is 'A'-'a', so if we add that to any lowercase letter, it'll give us its uppercase version!
With all of this, we can modify your code to:
for (int i = 0; i < str.size(); i++) {
if(isalpha(str.at(i)){ //Make sure it's a letter!
if(str.at(i) == 'z') str.at(i) = 'a'; //special case
else str.at(i)++;
if(str.at(i) == 'a' | str.at(i) == 'e' | str.at(i) == 'i'
| str.at(i) == 'o' | str.at(i) == 'u') {
str.at(i) += 'A' - 'a';
}
}
Your Main
There's only one thing to fix here. Don't use gets for input. If you're looking for a single word, use the extraction operator, >>, or if you want a whole line, use getline.
string word, line;
getline(cin, line);
cin >> word;
cout << LetterChanges(line) << endl;
cout << LetterChanges(word) << endl;
This question already has answers here:
C++ Remove punctuation from String
(12 answers)
Closed 9 years ago.
I got a code. It should give me an output that will erase the middle character between 'z' and 'p'. for example: zipZap("zipXzap"): expected [zpXzp] but found [z pXz p]
std::string zipZap(const std::string& str){
string a = str;
string b = "";
size_t len = str.length();
for (size_t i = 0; i < len; i++){
if (str[i] == 'z')
if (str[i+2] == 'p')
a[i+1] = ' ';
}
return a;
}
When i replaced a[i+1] = ''; it gave me an error.
You are not removing the chars, you are replacing them with ' '.
There are many ways to do this. One simple way is to build a new string, only adding chars when the proper conditions are met:
std::string zipZap(const std::string& str)
{
string a;
size_t len = str.length();
for (size_t i = 0; i < len; i++) {
// Always add first and last chars. As well as ones not between 'z' and 'p'
if (i == 0 || i == len-1 || (str[i-1] != 'z' && str[i+1] != 'p')) {
a += str[i];
}
}
return a;
}
Use string.erase() :
std::string zipZap(const std::string& str){
std::string a = str;
std::string b = "";
size_t len = str.length();
for (size_t i = 0; i < len; i++){
if (a[i] == 'z')
if (a[i+2] == 'p')
a.erase(i+1,1);
}
return a;
}
You're completely right that you cant replace an element of the string with ''.
A string is an array of chars, and '' is not a char at all. It is nothing.
If we look at the cplusplus page for a string
http://www.cplusplus.com/reference/string/string/
We see that we can use erase(iterator p) to "Erase characters from string (public member function)"
So if we change:
for (size_t i = 0; i < len; i++){
if (str[i] == 'z')
if (str[i+2] == 'p')
a.erase(a.begin() + i + 1);
We're closer now, but we can see that len is no longer the same as str.length(). the length of a is now actually 1 char shorter than len. To remedy this however we can simply add:
for (size_t i = 0; i < len; i++){
if (str[i] == 'z')
if (str[i+2] == 'p')
a.erase(a.begin() + i + 1);
len -= 1;
Hope that helps
If you #include <regex>, you can do a regular expression replacement.
std::string zipZap(const std::string& str){
regex exp("z.p");
string a = str;
a = regex_replace(a, exp "zp");
return a;
}
My input string is
\\?\bac#dos&ven_bb&prod_open-v&rev_5001#1&7f6ac24&0&353020304346333030363338#{53f56307-b6bf-11d0-94f2-00a0c91efb8b}
Required output is
bac\dos&ven_bb&prod_open-v&rev_5001\1&7f6ac24&0&353020304346333030363338_0
I have written a following code but is not working...need help is figuring out the problem.
Forgive my ignorance :) Also let me know if there is any better and efficient way to do it.
The rule for the output string is
In the second string i am removing all the "\" and "?" .And where is see the "#" i replace it with "\". and the second string is only till you see the charater "{" but does not include "#" at the end of it.
THanks
int main()
{
char s[] = "\\?\bac#dos&ven_bb&prod_open-v&rev_5001#1&7f6ac24&0&353020304346333030363338#{53f56307-b6bf-11d0-94f2-00a0c91efb8b}";
char s1[] = {0};
printf("OUtput string is : ");
for(int i = 0; s[i] != '{'; i++)
{
if(s[i] != '\\' && s[i] != '?')
{
int j = 0;
if(s[i] == '#')
{
s1[j] = '\\';
continue;
}
s1[j] = s[i];
j++;
}
}
for(int i = 0; s1[i] != '\0'; i++)
{
cout<<s1[i];
}
getch();
}
I would suggest looking into using the std::string::replace() function. There is plenty of online documentation on this. Take a look at some of the other functions that std::string has to offer as they might be of use too. If you are using c++, the use of std::string is usually preferable to tinkering with char arrays and indices.
Note the fixed scope of j. In your version you were always assigning to s1[0].
for(int i = 0, j = 0; s[i] != '{'; i++)
{
if(s[i] != '\\' && s[i] != '?')
{
// int j = 0;
if(s[i] == '#')
{
s1[j] = '\\';
}
else
{
s1[j] = s[i];
}
j++;
}
}
The other thing is to allocate enough space for the new string. Since you haven't specified the size char s1[] = {0}; declares an array of size 1. You need to do something like:
char s1[sizeof(s)] = { 0 }; // the size of the old array, since we don't know how long the new one will be
But since you tagged the Q C++, take advantage of of dynamically resizable std::string.
std::string s = ".......";
std::string s1;
for(int i = 0; s[i] != '{'; i++)
{
if(s[i] != '\\' && s[i] != '?')
{
if(s[i] == '#')
s1 += '\\';
else
s1 += s[i];
}
}
Your s1 buffer needs to be increased, as it stands now there is no room for the new string.
E.g.
char* s1 = calloc(strlen(s)+1,sizeof(char)); // same size should be enough, free(s1) later
the calloc ensures that it is \0 terminated, in your code you forgotten to add the \0 so the printout act erratically.