Debug my reimplementation of strtok and split - c++

I'm supposed to be writing code that takes a string of comma separated values without spaces (ex. my,name,is,jack). First we had to write a function
string nextstring(string str, int start_index)
that returns a single "value" from your initial string depending on the start index. The second part of the problem was to write a function
int split(string str, string a[], int max_size)
that will identify all the values in the initial string and put them in a string array and then return the total number of values stored in the array; i.e. if you had initially input my,name,is it would return 3.
My function never returns the correct value and whatever it returns changes depending on what the length of the words are.
#include <iostream>
#include <string>
using namespace std;
string nextstring(string str, int start_index);
int split(string str, string a[], int max_size);
int main()
{
string str;
int cnt;
string a[100];
cout<< "what is your string" << endl;
getline(cin, str);
cnt= split(str, a, 100);
cout << "There are " << cnt << " values in this string" << endl;
for(int i=0; i<cnt; i++)
{
cout << a[i] << endl;
}
return 0;
}
string nextstring(string str, int start_index)
{
string ans;
if(str[start_index] == ',' || str[start_index] == '\0')
{
ans=" ";
}
else{
ans=str[start_index]+nextstring(str, start_index+1);
}
return ans;
}
int split(string str, string a[], int max_size)
{
int j=0;
int ans=0;
double k=0;
while(j<max_size)
{
a[j]= nextstring(str,k);
string check=a[j];
if(isalpha(check[0])!= 0)
{
ans++;
}
k=k+a[j].length();
j++;
}
return ans;
}

It seems that your problem is that while(j<max_size){...} leads to j being incremented up to max_size. The line a[j]= nextstring(str,k); is at some points reading values that are outside your string which is really bad!
Replacing while(j<max_size){...} by while(j<max_size && k<str.length()){...} seems to be enough to make your code work!
Apart from that:
k has no reason to be a double! It should be an int (or something similar).
Since you are already using string, you should also learn to use vector. split is better written as:
int split(string str, vector<string> &a, int max_size)
{
int ans=0;
int k=0;
while(k<str.length())
{
string next = nextstring(str,k);
if(isalpha(next[0])!= 0)
{
ans++;
a.append(next);
}
k += next.length();
}
return ans;
}

The problem in your approach is to identify the end of the string, as there is no null terminator in a c++ string. Consider to update nextstring() to look for the end of string in a different manner:
string nextstring(string str, int start_index)
{
...
if(start_index == str.size() || str[start_index] == ',' ) //<===
{
ans=" ";
}
...
}
online demo
Additional recommendation
Note that it is not very nice to return a blank string when in reality it should be empty to reflect its real value (e.g. ",,"). You have no choice because otherwise you would have no mean in the calling function, to determine that the end of string was reached. But the consequence is thar all your strings have a trailing blank.
When you call recursively the function adding char to build the return string, you risk to have a considerable overhead. You could consider avoiding this, by replacing the else part:
ans=str.substr(start_index, str.find(',', start_index+1)-start_index);
However, as you have no trailing blank anymore, you need to adapt split() so to adapt its way to count the total number of chars parsed:
k=k+a[j].length()+1; // +1 because there's no longer a trailing blank.
Online demo

Related

Print a char string if the number of characters is even, or the middle character is removed

I have to make a console program that will print:
If the string is even, all the characters - let's say I have rain, it will print rain.
If the string is odd, it will remove the middle character, and then print the remaining ones - let's say I have telephone, it will remove p, and print telehone.
I've searched the StackOverflow forum for 3 hours but I couldn't solve this. This is my try:
#include <iostream>
#include <cstring>
#include <string>
using namespace std;
char s[21], x[21];
int j;
int main()
{
cin.get(s, 21);
if(strlen(s) % 2 == 0)
strcpy(x, s);
cout << x;
return 0;
}
string removeMiddle(const string)
{
int str = strlen(s);
bool isOdd;
if(str % 2 != 0)
isOdd == true;
if(isOdd == true)
{
j = str / 2;
}
s.erase (j+1, 1);
cout << s;
}
I've tried a lot of snippets and this is the best I can do. Thanks guys!
In your code you never called the removeMiddle function and removeMiddle functions is not returning anything even though its return type is string.
Here's the right code:
#include<bits/stdc++.h>
using namespace std;
int main(){
string str;
cin>>str;
if(str.length()%2==0){
cout<<str;
}
else{
int r = str.length()/2;
str.erase(r,1); //string& string ::erase (idx,len);
//Erases at most, len characters starting from index idx;
cout<<str;
}
return 0;
}
We first checked if the length of the string is even, then print the string, else we used the erase function to remove the middle character.

what is the issue with the below code?

The question is to replace the spaces contained in a string with a "%20". So basically need to insert this in a string wherever there is a space. Therefore, I want to replace all spaces with %20 but only partial string is getting replaced. I can see the correct o/p in the replace function
#include<iostream>
#include<string>
using namespace std;
int spaces(char* s,int size) /*calculate number of spaces*/
{
int nspace=0;
for(int i=0;i<size;i++)
{
if(s[i]==' ')
{
nspace++;
}
}
return nspace;
}
int len_new_string(char* inp,int l) /*calculate the length of the new string*/
{
int new_length=l+spaces(inp,l)*2;
return new_length;
}
char* replace(char* s,int length) /*function to replace the spaces within a string*/
{
int len=len_new_string(s,length);
char new_string[len];
int j=0;
for(int i=0;i<length;i++)
{
if(s[i]==' ') /*code to insert %20 if space is found*/
{
new_string[j]='%';
new_string[j+1]='2';
new_string[j+2]='0';
j=j+3;
}
else /*copy the original string if no space*/
{
new_string[j]=s[i];
j++;
}
}
cout<<"Replaced String: "<<new_string<<endl;
return s=new_string;
}
int main()
{
char str[]="abc def ghi ";
int length=sizeof(str)/sizeof(str[0]);
cout<<"String is: "<<str<<endl;
char *new_str=replace(str,length);
cout<<"Replaced String is: "<<new_str<<endl;
}
The char array should go out of scope and be released. The only reason you don't get a segfault is that apparently no other program has reserved the memory in that spot yet. To avoid this, try using a char array with padding, handing it over by reference or pointer and filling it in place:
void replace(char *in, char *out, size_t length)
{
/* copy as-is for non-spaces, insert replacement for spaces */
}
int main()
{
char str[]="abc def ghi";
size_t buflen(strlen(str)+2*spaces(str, strlen(str)));
char output[buflen+1];
memset(output, 0, buflen+1);
replace(str, output, strlen(str));
}
Another option is to new[] the return array (remember to delete[] it afterwards, then!) or, which I think you left out for a reason, use std::string all along to avoid the array issue.

Reverse String C++ using char array

I wrote a simple C++ program to reverse a string. I store a string in character array. To reverse a string I am using same character array and temp variable to swap the characters of an array.
#include<iostream>
#include<string>
using namespace std;
void reverseChar(char* str);
char str[50],rstr[50];
int i,n;
int main()
{
cout<<"Please Enter the String: ";
cin.getline(str,50);
reverseChar(str);
cout<<str;
return 0;
}
void reverseChar(char* str)
{
for(i=0;i<sizeof(str)/2;i++)
{
char temp=str[i];
str[i]=str[sizeof(str)-i-1];
str[sizeof(str)-i-1]=temp;
}
}
Now this method is not working and, I am getting the NULL String as result after the program execution.
So I want to know why I can't equate character array, why wouldn't this program work. And what is the solution or trick that I can use to make the same program work?
sizeof(str) does not do what you expect.
Given a char *str, sizeof(str) will not give you the length of that string. Instead, it will give you the number of bytes that a pointer occupies. You are probably looking for strlen() instead.
If we fixed that, we would have:
for(i=0;i<strlen(str)/2;i++)
{
char temp=str[i];
str[i]=str[strlen(str)-i-1];
str[strlen(str)-i-1]=temp;
}
This is C++, use std::swap()
In C++, if you want to swap the contents of two variables, use std::swap instead of the temporary variable.
So instead of:
char temp=str[i];
str[i]=str[strlen(str)-i-1];
str[strlen(str)-i-1]=temp;
You would just write:
swap(str[i], str[sizeof(str) - i - 1]);
Note how much clearer that is.
You're using C++, just use std::reverse()
std::reverse(str, str + strlen(str));
Global variables
It's extremely poor practice to make variables global if they don't need to be. In particular, I'm referring to i about this.
Executive Summary
If I was to write this function, it would look like one of the two following implementations:
void reverseChar(char* str) {
const size_t len = strlen(str);
for(size_t i=0; i<len/2; i++)
swap(str[i], str[len-i-1]);
}
void reverseChar(char* str) {
std::reverse(str, str + strlen(str));
}
When tested, both of these produce dlrow olleh on an input of hello world.
The problem is that within your function, str is not an array but a pointer. So sizeof will get you the size of the pointer, not the length of the array it points to. Also, even if it gave you the size of the array, that is not the length of the string. For this, better use strlen.
To avoid multiple calls to strlen, give the function another parameter, which tells the length:
void reverseChar(char* str, int len)
{
for(i=0; i<len/2; i++)
{
char temp=str[i];
str[i]=str[len-i-1];
str[len-i-1]=temp;
}
}
and call it with
reverseChar(str, strlen(str))
Another improvement, as mentioned in the comments, is to use std::swap in the loop body:
void reverseChar(char* str, int len)
{
for(i=0; i<len/2; i++)
{
std::swap(str[i], str[len-i-1]);
}
}
Also, there is std::reverse which does almost exactly that.
//reverse a string
#include<iostream>
using namespace std;
int strlen(char * str) {
int len = 0;
while (*str != '\0') {
len++;
str++;
}
return len;
}
void reverse(char* str, int len) {
for(int i=0; i<len/2; i++) {
char temp=str[i];
str[i]=str[len-i-1];
str[len-i-1]=temp;
}
}
int main() {
char str[100];
cin.getline(str,100);
reverse(str, strlen(str));
cout<<str<<endl;
getchar();
return 0;
}
If I were you, I would just write it like so:
int main()
{
string str;
cout << "Enter a string: " << endl;
getline(cin, str);
for (int x = str.length() - 1; x > -1; x--)
{
cout << str[x];
}
return 0;
}
This is a very simple way to do it and works great.
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
char str[80];
cout << "Enter a string bro: \n";
gets_s(str);
for (int i = strlen(str) - 1; i > -1; i--)
{
cout << str[i];
}
}

Value returned by strtok() for tokens of length 0?

I have the following piece of C++ code:
string dots="...";
char *points=(char *)malloc(sizeof(char)*20);
strcpy(points,dots.c_str());
points=strtok(points,".");
while(points!=NULL)
{
cout<<points<<endl;
points=strtok(NULL,".");
}
The cout statement prints nothing. What is this character that cout returns for 0 length token match? I have tried to check for '\0' but does not work. Please Help.
EDIT: Complete Program to Validate IP Addresses
#include<iostream>
#include<cstring>
#include<stdlib.h>
using namespace std;
int validateIP(string);
int main()
{
string IP;
cin>>IP;
int result=validateIP(IP);
if(result==0)
cout<<"Invalid IP"<<endl;
if(result==1)
cout<<"Valid IP"<<endl;
return 0;
}
//function definition validateIP(string)
int validateIP(string IP)
{
char ip[16];
int dotCount=0;
strcpy(ip,IP.c_str());
//check number of dots
for(int i=0;i<strlen(ip);++i)
{
if(ip[i]=='.')
{
dotCount++;
}
}
if(dotCount!=3)
return 0;
//check range
char *numToken;
numToken = strtok (ip,".");
while (numToken!= NULL)
{
int number;
if(numToken!=NULL) //check for token of length 0(e.g. case: ...)
number=atoi(numToken); //i also checked for (numToken[0]!='\O')
else return 0;
if(number<0 or number>255)
return 0;
numToken=strtok (NULL,".");
}
return 1;
}
The program prints ValidIP for input: ...
Your code has undefined behavior, you haven't allocate memory for points, accessing it invokes UB.
Update, I might write validateIP by using string and STL functions only if I could. Mix C/C++ is not good for maintenance.
#include <sstream>
int to_int(const std::string& s)
{
int i(0);
std::stringstream ss(s);
ss >> i;
return i;
}
bool isValidIp(const std::string& IP)
{
if (std::count(IP.begin(), IP.end(), '.') != 3)
{
return false;
}
std:stringstream ss(IP);
int token;
std::string s;
while(std::getline(ss, s, '.'))
{
int token = to_int(s);
if (token < 0 || token > 255)
{
return false;
}
}
return true;
}
Then you call it:
if (isValidIp(IP))
{
std::cout << "Valid IP" << std::endl;
}
else
{
std::cout << "Invalid IP" << std::endl;
}
The strtok function returns sub-string of the given string delimited by the given character. IMO (to be tested) if your string only contains delimiting characters, the strtok function will return NULL (no more tokens) at the first call.
Moreover in your code snippet, you copy the string to an uninitialized pointer. Replace your call to strcpy by a call to strdup for the underlying memory to be allocated before copying. Edit: you modified your question as I were answering
strtok is used to tokenize the string. Say, i have a string "abc.def.ghi.jkl" then we can use strtok to get the tokens besed on the delimiter.
char a[]="abc.def.ghi.jkl";
char tmp=strtok(a, ".");
if (tmp != NULL) //Required because strtok will return null if it failes find the delimiter
printf("\n value is [%s]", tmp); //out put is abc
So, in your case "..." is the string and '.' is the delimiter which result in empty string because there is no characters between first character and the delimiter '.'
your code will return empty string say "" as an output. for all the sttok function call.
Second you have to allocate memory to the points variable like
char points[dots.length()+1];
If the string only contains delimiting characters, strok return NULL
You probably want this:
int main()
{
string dots=". . ."; //Notice space
char *points=(char *)malloc(sizeof(char)*20);
char *p; // Use a char pointer
strcpy(points,dots.c_str());
p=strtok(points,".");
while(p!=NULL)
{
cout<<points<<endl;
p=strtok(NULL,".");
}
/* Free Memory */
free(points);
}

Write a recursive function that reverses the input string

I've been reading the book C++ For Everyone and one of the exercises said to write a function string reverse(string str) where the return value is the reverse of str.
Can somebody write some basic code and explain it to me? I've been staring at this question since yesterday and can't figure it out. The furthest I've gotten is having the function return the first letter of str (Which I still don't know how it happened)
This is as far as I got (An hour after posting this question):
string reverse(string str)
{
string word = "";
if (str.length() <= 1)
{
return str;
}
else
{
string str_copy = str;
int n = str_copy.length() - 1;
string last_letter = str_copy.substr(n, 1);
str_copy = str_copy.substr(0, n);
word += reverse(str_copy);
return str_copy;
}
return word;
}
If I enter "Wolf", it returns Wol. Somebody help me out here
If I return word instead of return str_copy then I get a w
If I return last_letter then I get an l
I'll instead explain the recursive algorithm itself. Take the example "input" which should produce "tupni". You can reverse the string recursively by
If the string is empty or a single character, return it unchanged.
Otherwise,
Remove the first character.
Reverse the remaining string.
Add the first character above to the reversed string.
Return the new string.
Try this one
string reverse(string &s)
{
if( s.length() == 0 ) // end condtion to stop recursion
return "";
string last(1,s[s.length()-1]); // create string with last character
string reversed = reverse(s.substr(0,s.length()-1));
return last+reversed; // Make he last character first
}
A recursive function must have the following properties
It must call itself again
It must have a condition when the recursion ends. Otherwise you have a function which
will cause a stack overflow.
This recursive function does basically create a string of the last character and then call itself again with the rest of the string excluding the last character. The real switching happens at the last line where last+reversed is returned. If it would be the other way around nothing would happen.
It is very inefficient but it works to show the concept.
Just to suggest a better way of handling recursion:
String reversal using recursion in C++:
#include <iostream>
#include <string>
using namespace std;
string reverseStringRecursively(string str){
if (str.length() == 1) {
return str;
}else{
return reverseStringRecursively(str.substr(1,str.length())) + str.at(0);
}
}
int main()
{
string str;
cout<<"Enter the string to reverse : ";
cin>>str;
cout<<"The reversed string is : "<<reverseStringRecursively(str);
return 0;
}
I won't write a full-blown algorithm for you, but here's a hint:
How about swapping the two outermost characters, and then apply the same to the characters in the middle?
Oh, and if that book really proposed string reverse(string str) as an appropriate function signature for this, throw it away and buy a good book instead.
Here is my version of a recursive function that reverses the input string:
void reverse(char *s, size_t len)
{
if ( len <= 1 || !s )
{
return;
}
std::swap(s[0], s[len-1]);// swap first and last simbols
s++; // move pointer to the following char
reverse(s, len-2); // shorten len of string
}
Shortest and easiest
class Solution {
public:
string reverseString(string s) {
string str;
if(s != "\0"){
str = reverseString(s.substr(1, s.length()));
str += s.substr(0,1);
}
return str;
}
};
1-line recursive solution:
string RecursiveReverse(string str, string prev = "") {
return (str.length() == 0 ? prev : RecursiveReverse(str.substr(0, str.length()-1), prev += str[str.length()-1]));
}
You call it like this:
cout << RecursiveReverse("String to Reverse");
I know I shouldn't give a solution, but since no one mentioned this easy solution I though I should share it. I think the code literally is the algorithm so there is no need for a pseudo-code.
void c_plusplus_recursive_swap_reverse(std::string::iterator start,
std::string::iterator end)
{
if(start >= end) {
return;
}
std::iter_swap(start, end);
c_plusplus_recursive_swap_reverse(++start, --end);
}
To call it use:
c_plusplus_recursive_swap_reverse(temp.begin(), temp.end());
All existing solutions had way too much code that didn't really do anything, so, here's my take at it:
#include <iostream>
#include <string>
std::string
r(std::string s)
{
if (s.empty())
return s;
return r(s.substr(1)) + s[0];
}
int
main()
{
std::cout << r("testing") << std::endl;
}
P.S. I stumbled upon this question trying to find a C++ way for std::string of what s+1 for a char * in C is; without going the whole route of s.substr(1, s.length()-1), which looks too ugly. Turns out, there's std::string::npos, which means until the end of the string, and it's already the default value for the second argument, so, s.substr(1) is enough (plus, it also looks more efficient and on par with the simple s + 1 in C).
Note, however, that recursion in general doesn't scale as the input grows larger, unless the compiler is able to do what is known as tail-recursion optimisation. (Recursion is rarely relied upon in imperative languages.)
However, in order for the tail recursion optimisation to get activated, it is generally required that, (0), the recursion only happens within the return statement, and that, (1), no further operations are performed with the result of the recursive call back in the parent function.
E.g., in the case above, the + s[0] is logically done by the parent after the child call completes (and it probably would be so even if you go the more uglier s[s.length()-1] + route), so, it might as well prevent most compilers from doing a tail-recursion-optimisation, thus making the function very inefficient on large inputs (if not outright broken due to heap exhaustion).
(For what it's worth, I've tried writing a more tail-recursion-friendly solution (making sure to grow the return result through an argument to the function itself), but disassembly of the resulting binary seems to suggest that it's more involved than that in the imperative languages like C++, see gcc: is there no tail recursion if I return std::string in C++?.)
you can implement your own reverse similar to std::reverse.
template <typename BidirIt>
void reverse(BidirIt first, BidirIt last)
{
if((first == last) || (first == --last))
return;
std::iter_swap(first, last);
reverse(++first, last);
}
I did something like this, it did the reversal in place. I took two variables that traverse the string from two extreme end to the centre of the string and when they overlap or equal to each other then reversal terminates.
Take an example: input string str = "abcd" and call the function as
ReverseString(str,0,str.length()-1);
and increment/decrement the variable pointers recursively.
First the pointers points to 'a' and 'd' and swap them, then they point to 'b' and 'c' and swap them. Eventually i >= j which calls for the base case to be true and hence the recursion terminates. The main take away for this question is to pass input string as reference.
string ReverseString(string& str,int i,int j){
if(str.length() < 1 || str == "" || i >= j){
return "";
}
else{
char temp = str[i];
str[i] = str[j];
str[j] = temp;
ReverseString(str,i+1,j-1);
}
return str;
}
String can be reversed in-place. If we start from smallest possible string i.e. one character string, we don't need to do anything. This is where we stop or return from our recursive call and it becomes our base case.
Next, we have to think of a generic way to swap the smallest string i.e. two characters or more. Simplest logic is to swap the current character str[current_index] with character on the opposite side str[str_length-1 - current_index].
In the end, call the reverse function again for next index.
#include <iostream>
using namespace std;
void reverse_string(std::string& str, int index, int length) {
// Base case: if its a single element, no need to swap
// stop swapping as soon as we reach the mid, hence index*2
// otherwise we will reverse the already reversed string
if( (length - index*2) <= 1 ) {
return;
}
// Reverse logic and recursion:
// swap current and opposite index
std::swap(str[index], str[length-1 - index]);
// do the same for next character (index+1)
reverse_string(str, index+1, length);
}
int main() {
std::string s = "World";
reverse_string(s, 0, s.length());
std::cout << s << endl;
}
There are already some good answer but I want to add my approach with full working Recursive reversing string.
#include <iostream>
#include <string>
using namespace std;
char * reverse_s(char *, char*, int,int);
int main(int argc, char** argv) {
if(argc != 2) {
cout << "\n ERROR! Input String";
cout << "\n\t " << argv[0] << "STRING" << endl;
return 1;
}
char* str = new char[strlen(argv[1])+1];
strcpy(str,argv[1]);
char* rev_str = new char[strlen(str)+1];
cout<<"\n\nFinal Reverse of '" << str << "' is --> "<< reverse_s(str, rev_str, 0, strlen(str)) << endl;
cin.ignore();
delete rev_str, str;
return 0;
}
char* reverse_s(char* str, char* rev_str, int str_index, int rev_index ) {
if(strlen(str) == 1)
return str;
if(str[str_index] == '\0' ) {
rev_str[str_index] = '\0';
return rev_str;
}
str_index += 1;
rev_index -=1;
rev_str = reverse_s(str, rev_str, str_index, rev_index);
if(rev_index >= 0) {
cout << "\n Now the str value is " << str[str_index-1] << " -- Index " << str_in
dex << " Rev Index: " << rev_index;
rev_str[rev_index] = str[str_index-1];
cout << "\nReversed Value: " << rev_str << endl;
}
return rev_str;
}
void reverse(string &s, int &m) {
if (m == s.size()-1)
return;
int going_to = s.size() - 1 - m;
string leader = s.substr(1,going_to);
string rest = s.substr(going_to+1,s.size());
s = leader + s.substr(0,1) + rest;
reverse(s,++m);
}
int main ()
{
string y = "oprah";
int sz = 0;
reverse(y,sz);
cout << y << endl;
return 0;
}
void ClassName::strgRevese(char *str)
{
if (*str=='\0')
return;
else
strgRevese(str+1);
cout <<*str;
}
here is my 3 line string revers
std::string stringRevers(std::string s)
{
if(s.length()<=1)return s;
string word=s.at(s.length()-1)+stringRevers(s.substr(0,s.length()-1));//copy the last one at the beginning and do the same with the rest
return word;
}
The question is to write a recursive function. Here is one approach. Not a neat code, but does what is required.
/* string reversal through recursion */
#include <stdio.h>
#include <string.h>
#define size 1000
char rev(char []);
char new_line[size];
int j = 0;
int i =0;
int main ()
{
char string[]="Game On";
rev(string);
printf("Reversed rev string is %s\n",new_line);
return 0;
}
char rev(char line[])
{
while(line[i]!='\0')
{
i++;
rev(line);
i--;
new_line[j] = line[i];
j++;
return line[i];
}
return line[i];
}
It will reverse Original string recursively
void swap(string &str1, string &str2)
{
string temp = str1;
str1 = str2;
str2 = str1;
}
void ReverseOriginalString(string &str, int p, int sizeOfStr)
{
static int i = 0;
if (p == sizeOfStr)
return;
ReverseOriginalString(str, s + 1, sizeOfStr);
if (i <= p)
swap(&str[i++], &str[p])
}
int main()
{
string st = "Rizwan Haider";
ReverseOriginalString(st, 0, st.length());
std::cout << "Original String is Reversed: " << st << std::endl;
return 0;
}