a programm which removes excess spaces from a string [duplicate] - c++

Question

This question already has answers here:
Replace multiple spaces with one space in a string
(5 answers)
Closed 9 years ago.
I wrote a program which is supposed to remove excess spaces from a string. But it only shows characters before spaces. It finds a space and checks the character after that whether it is a space. Depending on excess spaces it shifts other characters over excess spaces. But output is very confusing.
input: "qwe(2 spaces)rt(one space)y"
output: "qwe(one space)rt(one space)y"
#include <iostream>
#include <stdlib.h>
#include <string>
using namespace std;
int main(){
string a;
cin >> a;
int len = a.length();
int new_len=len;
int z,s=0;
for(int i=0; i<new_len; i++){
if(a[i]==' '){
z=i+1;
s=0;
//Assigning the number of excess spaces to s.
while(a[z]==' '){
s++;
z++;
}
//doing the shifting here.
if(s>0){
for(int l=i+1; l<new_len-s; l++){
a[l]=a[s+l];
}
}
new_len-=s;
}
}
cout << a << endl;
cout << a.length();
system("pause");
return 0;
}

Answer 1

Most of your code is semi-pointless -- when you use the normal string extractor (stream >> string) it automatically skips across all consecutive leading white-space, and stops reading at the first whitespace character. As such, it's already doing almost everything the rest of your code is intended to accomplish. That leaves a much simpler approach to accomplishing the same task:
std::copy(std::istream_iterator<std::string>(std::cin),
std::istream_iterator<std::string>(),
std::ostream_iterator<std::string>(std::cout, " "));
This does have one problem: it'll leave one extra space at the end of the output. If you don't want that, you can use the infix_ostream_iterator I've posted previously. With that, you'd change the above to something like this:
std::copy(std::istream_iterator<std::string>(std::cin),
std::istream_iterator<std::string>(),
infix_ostream_iterator<std::string>(std::cout, " "));

Answer 2

If you're using C++11 doing this your way is overkill - you can just use a regex. Something like the following should do it (untested):
#include <regex>
#include <iostream>
#include <string>
using namespace::std;
int main(){
string a;
cin >> a;
regex r(" +");
a = regex_replace(a,r," ");
cout << a << endl;
cout << a.length();
system("pause");
return 0;
}

Answer 3

Your code is highly ineffective. Imagine a following string with 1,000,000 characters:
a a a a a a a...
Each time your algorithm encounters a second space, it goes through the whole string to shift it one char left. I would attempt another approach:
Create two iterators, like realPos and charPos. Set them to 0 at the beginning.
Create a variable, which stores a number of spaces encountered so far, like spacesSeen. Set it to 0.
Now, while realPos is lower than length of the whole string:
If string[realPos] != ' ' and charPos != realPos, make an assignment: string[charPos] = string[realPos]. Then increase both realPos and charPos by one. Set spacesSeen to 0.
If string[realPos] == ' ' and spacesSeen == 0, increase spacesSeen by one, copy characters and advance both iterators.
If string[realPos] == ' ' and spacesSeen > 0, then increase spacesSeen and then increase only realPos.
Now charPos marks the place where your final string ends, adjust strings size, such that it ends there.
In simpler words: copy the characters one by one and skip multiple spaces on the way.