Remove extra white spaces in C++ - c++

Question

I tried to write a script that removes extra white spaces but I didn't manage to finish it.
Basically I want to transform abc sssd g g sdg gg gf into abc sssd g g sdg gg gf.
In languages like PHP or C#, it would be very easy, but not in C++, I see. This is my code:
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#include <unistd.h>
#include <string.h>
char* trim3(char* s) {
int l = strlen(s);
while(isspace(s[l - 1])) --l;
while(* s && isspace(* s)) ++s, --l;
return strndup(s, l);
}
char *str_replace(char * t1, char * t2, char * t6)
{
char*t4;
char*t5=(char *)malloc(10);
memset(t5, 0, 10);
while(strstr(t6,t1))
{
t4=strstr(t6,t1);
strncpy(t5+strlen(t5),t6,t4-t6);
strcat(t5,t2);
t4+=strlen(t1);
t6=t4;
}
return strcat(t5,t4);
}
void remove_extra_whitespaces(char* input,char* output)
{
char* inputPtr = input; // init inputPtr always at the last moment.
int spacecount = 0;
while(*inputPtr != '\0')
{
char* substr;
strncpy(substr, inputPtr+0, 1);
if(substr == " ")
{
spacecount++;
}
else
{
spacecount = 0;
}
printf("[%p] -> %d\n",*substr,spacecount);
// Assume the string last with \0
// some code
inputPtr++; // After "some code" (instead of what you wrote).
}
}
int main(int argc, char **argv)
{
printf("testing 2 ..\n");
char input[0x255] = "asfa sas f f dgdgd dg ggg";
char output[0x255] = "NO_OUTPUT_YET";
remove_extra_whitespaces(input,output);
return 1;
}
It doesn't work. I tried several methods. What I am trying to do is to iterate the string letter by letter and dump it in another string as long as there is only one space in a row; if there are two spaces, don't write the second character to the new string.
How can I solve this?

Answer 1

There are already plenty of nice solutions. I propose you an alternative based on a dedicated <algorithm> meant to avoid consecutive duplicates: unique_copy():
void remove_extra_whitespaces(const string &input, string &output)
{
output.clear(); // unless you want to add at the end of existing sring...
unique_copy (input.begin(), input.end(), back_insert_iterator<string>(output),
[](char a,char b){ return isspace(a) && isspace(b);});
cout << output<<endl;
}
Here is a live demo. Note that I changed from c style strings to the safer and more powerful C++ strings.
Edit: if keeping c-style strings is required in your code, you could use almost the same code but with pointers instead of iterators. That's the magic of C++. Here is another live demo.

Answer 2

Here's a simple, non-C++11 solution, using the same remove_extra_whitespace() signature as in the question:
#include <cstdio>
void remove_extra_whitespaces(char* input, char* output)
{
int inputIndex = 0;
int outputIndex = 0;
while(input[inputIndex] != '\0')
{
output[outputIndex] = input[inputIndex];
if(input[inputIndex] == ' ')
{
while(input[inputIndex + 1] == ' ')
{
// skip over any extra spaces
inputIndex++;
}
}
outputIndex++;
inputIndex++;
}
// null-terminate output
output[outputIndex] = '\0';
}
int main(int argc, char **argv)
{
char input[0x255] = "asfa sas f f dgdgd dg ggg";
char output[0x255] = "NO_OUTPUT_YET";
remove_extra_whitespaces(input,output);
printf("input: %s\noutput: %s\n", input, output);
return 1;
}
Output:
input: asfa sas f f dgdgd dg ggg
output: asfa sas f f dgdgd dg ggg

Answer 3

Since you use C++, you can take advantage of standard-library features designed for that sort of work. You could use std::string (instead of char[0x255]) and std::istringstream, which will replace most of the pointer arithmetic.
First, make a string stream:
std::istringstream stream(input);
Then, read strings from it. It will remove the whitespace delimiters automatically:
std::string word;
while (stream >> word)
{
...
}
Inside the loop, build your output string:
if (!output.empty()) // special case: no space before first word
output += ' ';
output += word;
A disadvantage of this method is that it allocates memory dynamically (including several reallocations, performed when the output string grows).

Answer 4

There are plenty of ways of doing this (e.g., using regular expressions), but one way you could do this is using std::copy_if with a stateful functor remembering whether the last character was a space:
#include <algorithm>
#include <string>
#include <iostream>
struct if_not_prev_space
{
// Is last encountered character space.
bool m_is = false;
bool operator()(const char c)
{
// Copy if last was not space, or current is not space.
const bool ret = !m_is || c != ' ';
m_is = c == ' ';
return ret;
}
};
int main()
{
const std::string s("abc sssd g g sdg gg gf into abc sssd g g sdg gg gf");
std::string o;
std::copy_if(std::begin(s), std::end(s), std::back_inserter(o), if_not_prev_space());
std::cout << o << std::endl;
}

Answer 5

You can use std::unique which reduces adjacent duplicates to a single instance according to how you define what makes two elements equal is.
Here I have defined elements as equal if they are both whitespace characters:
inline std::string& remove_extra_ws_mute(std::string& s)
{
s.erase(std::unique(std::begin(s), std::end(s), [](unsigned char a, unsigned char b){
return std::isspace(a) && std::isspace(b);
}), std::end(s));
return s;
}
inline std::string remove_extra_ws_copy(std::string s)
{
return remove_extra_ws_mute(s);
}
std::unique moves the duplicates to the end of the string and returns an iterator to the beginning of them so they can be erased.
Additionally, if you must work with low level strings then you can still use std::unique on the pointers:
char* remove_extra_ws(char const* s)
{
std::size_t len = std::strlen(s);
char* buf = new char[len + 1];
std::strcpy(buf, s);
// Note that std::unique will also retain the null terminator
// in its correct position at the end of the valid portion
// of the string
std::unique(buf, buf + len + 1, [](unsigned char a, unsigned char b){
return (a && std::isspace(a)) && (b && std::isspace(b));
});
return buf;
}

Answer 6

for in-place modification you can apply erase-remove technic:
#include <string>
#include <iostream>
#include <algorithm>
#include <cctype>
int main()
{
std::string input {"asfa sas f f dgdgd dg ggg"};
bool prev_is_space = true;
input.erase(std::remove_if(input.begin(), input.end(), [&prev_is_space](unsigned char curr) {
bool r = std::isspace(curr) && prev_is_space;
prev_is_space = std::isspace(curr);
return r;
}), input.end());
std::cout << input << "\n";
}
So you first move all extra spaces to the end of the string and then truncate it.
The great advantage of C++ is that is universal enough to port your code to plain-c-static strings with only few modifications:
void erase(char * p) {
// note that this ony works good when initial array is allocated in the static array
// so we do not need to rearrange memory
*p = 0;
}
int main()
{
char input [] {"asfa sas f f dgdgd dg ggg"};
bool prev_is_space = true;
erase(std::remove_if(std::begin(input), std::end(input), [&prev_is_space](unsigned char curr) {
bool r = std::isspace(curr) && prev_is_space;
prev_is_space = std::isspace(curr);
return r;
}));
std::cout << input << "\n";
}
Interesting enough remove step here is string-representation independent. It will work with std::string without modifications at all.

Answer 7

I have the sinking feeling that good ol' scanf will do (in fact, this is the C school equivalent to Anatoly's C++ solution):
void remove_extra_whitespaces(char* input, char* output)
{
int srcOffs = 0, destOffs = 0, numRead = 0;
while(sscanf(input + srcOffs, "%s%n", output + destOffs, &numRead) > 0)
{
srcOffs += numRead;
destOffs += strlen(output + destOffs);
output[destOffs++] = ' '; // overwrite 0, advance past that
}
output[destOffs > 0 ? destOffs-1 : 0] = '\0';
}
We exploit the fact that scanf has magical built-in space skipping capabilities. We then use the perhaps less known %n "conversion" specification which gives us the amount of chars consumed by scanf. This feature frequently comes in handy when reading from strings, like here. The bitter drop which makes this solution less-than-perfect is the strlen call on the output (there is no "how many bytes have I actually just written" conversion specifier, unfortunately).
Last not least use of scanf is easy here because sufficient memory is guaranteed to exist at output; if that were not the case, the code would become more complex due to buffering and overflow handling.

Answer 8

Since you are writing c-style, here's a way to do what you want.
Note that you can remove '\r' and '\n' which are line breaks (but of course that's up to you if you consider those whitespaces or not).
This function should be as fast or faster than any other alternative and no memory allocation takes place even when it's called with std::strings (I've overloaded it).
char temp[] = " alsdasdl gasdasd ee";
remove_whitesaces(temp);
printf("%s\n", temp);
int remove_whitesaces(char *p)
{
int len = strlen(p);
int new_len = 0;
bool space = false;
for (int i = 0; i < len; i++)
{
switch (p[i])
{
case ' ': space = true; break;
case '\t': space = true; break;
case '\n': break; // you could set space true for \r and \n
case '\r': break; // if you consider them spaces, I just ignore them.
default:
if (space && new_len > 0)
p[new_len++] = ' ';
p[new_len++] = p[i];
space = false;
}
}
p[new_len] = '\0';
return new_len;
}
// and you can use it with strings too,
inline int remove_whitesaces(std::string &str)
{
int len = remove_whitesaces(&str[0]);
str.resize(len);
return len; // returning len for consistency with the primary function
// but u can return std::string instead.
}
// again no memory allocation is gonna take place,
// since resize does not not free memory because the length is either equal or lower
If you take a brief look at the C++ Standard library, you will notice that a lot C++ functions that return std::string, or other std::objects are basically a wrapper to a well written extern "C" function. So don't be afraid to use C functions in C++ applications, if they are well written and you can overload them to support std::strings and such.
For example, in Visual Studio 2015, std::to_string is written exactly like this:
inline string to_string(int _Val)
{ // convert int to string
return (_Integral_to_string("%d", _Val));
}
inline string to_string(unsigned int _Val)
{ // convert unsigned int to string
return (_Integral_to_string("%u", _Val));
}
and _Integral_to_string is a wrapper to a C function sprintf_s
template<class _Ty> inline
string _Integral_to_string(const char *_Fmt, _Ty _Val)
{ // convert _Ty to string
static_assert(is_integral<_Ty>::value,
"_Ty must be integral");
char _Buf[_TO_STRING_BUF_SIZE];
int _Len = _CSTD sprintf_s(_Buf, _TO_STRING_BUF_SIZE, _Fmt, _Val);
return (string(_Buf, _Len));
}

Answer 9

Well here is a longish(but easy) solution that does not use pointers.
It can be optimized further but hey it works.
#include <iostream>
#include <string>
using namespace std;
void removeExtraSpace(string str);
int main(){
string s;
cout << "Enter a string with extra spaces: ";
getline(cin, s);
removeExtraSpace(s);
return 0;
}
void removeExtraSpace(string str){
int len = str.size();
if(len==0){
cout << "Simplified String: " << endl;
cout << "I would appreciate it if you could enter more than 0 characters. " << endl;
return;
}
char ch1[len];
char ch2[len];
//Placing characters of str in ch1[]
for(int i=0; i<len; i++){
ch1[i]=str[i];
}
//Computing index of 1st non-space character
int pos=0;
for(int i=0; i<len; i++){
if(ch1[i] != ' '){
pos = i;
break;
}
}
int cons_arr = 1;
ch2[0] = ch1[pos];
for(int i=(pos+1); i<len; i++){
char x = ch1[i];
if(x==char(32)){
//Checking whether character at ch2[i]==' '
if(ch2[cons_arr-1] == ' '){
continue;
}
else{
ch2[cons_arr] = ' ';
cons_arr++;
continue;
}
}
ch2[cons_arr] = x;
cons_arr++;
}
//Printing the char array
cout << "Simplified string: " << endl;
for(int i=0; i<cons_arr; i++){
cout << ch2[i];
}
cout << endl;
}

Answer 10

I don't know if this helps but this is how I did it on my homework. The only case where it might break a bit is when there is spaces at the beginning of the string EX " wor ds " In that case, it will change it to " wor ds"
void ShortenSpace(string &usrStr){
char cha1;
char cha2;
for (int i = 0; i < usrStr.size() - 1; ++i) {
cha1 = usrStr.at(i);
cha2 = usrStr.at(i + 1);
if ((cha1 == ' ') && (cha2 == ' ')) {
usrStr.erase(usrStr.begin() + 1 + i);
--i;//edit: was ++i instead of --i, made code not work properly
}
}
}

Answer 11

I ended up here for a slighly different problem. Since I don't know where else to put it, and I found out what was wrong, I share it here. Don't be cross with me, please.
I had some strings that would print additional spaces at their ends, while showing up without spaces in debugging. The strings where formed in windows calls like VerQueryValue(), which besides other stuff outputs a string length, as e.g. iProductNameLen in the following line converting the result to a string named strProductName:
strProductName = string((LPCSTR)pvProductName, iProductNameLen)
then produced a string with a \0 byte at the end, which did not show easily in de debugger, but printed on screen as a space. I'll leave the solution of this as an excercise, since it is not hard at all, once you are aware of this.