How to replace all "pi" from a string by "3.14"? Example: INPUT = "xpix" ___ OUTPUT = "x3.14x" for a string, not character array.
This doesn't work:
#include<iostream>
using namespace std;
void replacePi(string str)
{
if(str.size() <=1)
return ;
replacePi(str.substr(1));
int l = str.length();
if(str[0]=='p' && str[1]=='i')
{
for(int i=l;i>1;i--)
str[i+2] = str[i];
str[0] = '3';
str[1] = '.';
str[2] = '1';
str[3] = '4';
}
}
int main()
{
string s;
cin>>s;
replacePi(s);
cout << s << endl;
}
There is a ready to use function in the C++ lib. It is called: std::regex_replace. You can read the documentation in the CPP Reference here.
Since it uses regexes it is very powerful. The disadvantage is that it may be a little bit too slow during runtime for some uses case. But for your example, this does not matter.
So, a common C++ solution would be:
#include <iostream>
#include <string>
#include <regex>
int main() {
// The test string
std::string input{ "Pi is a magical number. Pi is used in many places. Go for Pi" };
// Use simply the replace function
std::string output = std::regex_replace(input, std::regex("Pi"), "3.14");
// Show the output
std::cout << output << "\n";
}
But my guess is that you are learning C++ and the teacher gave you a task and expects a solution without using elements from the std C++ library. So, a hands on solution.
This can be implemented best with a temporary string. You check character by character from the original string. If the characters do not belong to Pi, then copy them as is to new new string. Else, copy 3.14 to the new string.
At the end, overwrite the original string with the temp string.
Example:
#include <iostream>
#include <string>
using namespace std;
void replacePi(string& str) {
// Our temporay
string temp = "";
// Sanity check
if (str.length() > 1) {
// Iterate over all chararcters in the source string
for (size_t i = 0; i < str.length() - 1; ++i) {
// Check for Pi in source string
if (str[i] == 'P' and str[i + 1] == 'i') {
// Add replacement string to temp
temp += "3.14";
// We consumed two characters, P and i, so increase index one more time
++i;
}
else {
// Take over normal character
temp += str[i];
}
}
str = temp;
}
}
// Test code
int main() {
// The test string
std::string str{ "Pi is a magical number. Pi is used in many places. Go for Pi" };
// Do the replacement
replacePi(str);
// Show result
std::cout << str << '\n';
}
What you need is string::find and string::replace. Here is an example
size_t replace_all(std::string& str, std::string from, std::string to)
{
size_t count = 0;
std::string::size_type pos;
while((pos=str.find(from)) != str.npos)
{
str.replace(pos, from.length(), to);
count++;
}
return count;
}
void replacePi(std::string& str)
{
replace_all(str, "pi", "3.14");
}
I get the same string back when I try to reverse it in C++.
I read that recursion is a good way to reverse things. I tried to implement a recursive algorithm by returning the first character of the string and calling the same function with the first character removed until the string has a size of 1. My first function removes the first character of the string and the second function reverses it:
string deleteFirstElement(string input) {
if (input.size() == 1) {
return input;
}
// This loop brings the first element to the last position
for (int i = 0; i < input.size()-1; i++) {
char temp;
temp = input.at(i);
input.at(i) = input.at(i+1);
input.at(i + 1) = temp;
}
input.pop_back(); // Delete last element of the string
return input;
}
string reverseit(string input) {
if (input.size() == 1) {
return input;
}
else {
return input.at(0) + reverseit(deleteFirstElement(input));
}
}
But why do I get the same string back and no reverse?
You get the same string back because you build the same string again.
Using the example of "ABC" you'll see what the function does:
reverseit("ABC") returns 'A' + reverseit("BC")
reverseit("BC") returns 'B' + reverseit("C")
reverseit("C") returns 'C'
You'd want
char firstChar = input.at(0);
return reverseit(deleteFirstElement(input)) + firstChar;
But really you should be looking into another solution. Recursion
reduces readability
is slow
uses lots of stack memory
easily creates hard to debug endless loops
in general it should be avoided if possible. Some solutions are really elegant with it, to be sure, but loops are almost always faster.
change the else part from this
string reverseit(string input) {
if (input.size() == 1) {
return input;
}
else {
return input.at(0) + reverseit(deleteFirstElement(input));
}
}
to this
string reverseit(string input) {
if (input.size() == 1) {
return input;
}
else {
return reverseit(deleteFirstElement(input))+ input.at(0);
}
}
Recursion is not the easy way to reverse things.
Here is the easy way:
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
int main()
{
string s { "hello" };
reverse(s.begin(), s.end());
cout << s << endl;
}
Here is a recusive solution. It works but is much less reasonable and efficient.
#include <iostream>
#include <string>
using namespace std;
string recursiveReverse(string value, string accum = "")
{
if(value.size() == 0)
{
return accum;
}
else
{
accum += value.back();
value.pop_back();
return recursiveReverse(value, accum);
}
}
int main()
{
string s { "hello" };
s = recursiveReverse(s);
cout << s << endl;
}
If you have boost, this problem basicly boils down:
string reverseit(string input) {
string result(input);
boost::range::reverse(result);
return result;
}
Without boost, you may instead use:
std::reverse(result.begin(), result.end());
As noted, recursion reduced readability of your program and should be reserved for rare cases (usually where another solution is just more complicated).
I'm trying to figure out how to can fold a word from a string. For example "code" after the folding would become "ceod". Basically start from the first character and then get the last one, then the second character. I know the first step is to start from a loop, but I have no idea how to get the last character after that. Any help would be great. Heres my code.
#include <iostream>
using namespace std;
int main () {
string fold;
cout << "Enter a word: ";
cin >> fold;
string temp;
string backwards;
string wrap;
for (unsigned int i = 0; i < fold.length(); i++){
temp = temp + fold[i];
}
backwards= string(temp.rbegin(),temp.rend());
for(unsigned int i = 0; i < temp.length(); i++) {
wrap = fold.replace(backwards[i]);
}
cout << wrap;
}
Thanks
#Supreme, there are number of ways to do your task and I'm going to post one of them. But as #John had pointed you must try your own to get it done because real programming is all about practicing a lot. Use this solution just as a reference of one possibility and find many others.
int main()
{
string in;
cout <<"enter: "; cin >> in;
string fold;
for (int i=0, j=in.length()-1; i<in.length()/2; i++, j--)
{
fold += in[i];
fold += in[j];
}
if( in.length()%2 != 0) // if string lenght is odd, pick the middle
fold += in[in.length()/2];
cout << endl << fold ;
return 0;
}
good luck !
There are two approaches to this form of problem, a mathematically exact method would be to create a generator function which returns the number in the correct order.
An easier plan would be to modify the string to solve practically the problem.
Mathematical solution
We want a function which returns the index in the string to add. We have 2 sequences - increasing and decreasing and they are interleaved.
sequence 1 :
0, 1 , 2, 3.
sequence 2
len-1, len-2, len-3, len-4.
Given they are interleaved, we want even values to be from sequence 1 and odd values from sequence 2.
So our solution would be to for a given new index, choose which sequence to use, and then return the next value from that sequence.
int generator( int idx, int len )
{
ASSERT( idx < len );
if( idx %2 == 0 ) { // even - first sequence
return idx/2;
} else {
return (len- (1 + idx/2);
}
}
This can then be called from a function fold...
std::string fold(const char * src)
{
std::string result;
std::string source(src);
for (size_t i = 0; i < source.length(); i++) {
result += source.at(generator(i, source.length()));
}
return result;
}
Pratical solution
Although less efficient, this can be easier to think about. We are taking either the first or the last character of a string. This we will do using string manipulation to get the right result.
std::string fold2(const char * src)
{
std::string source = src;
enum whereToTake { fromStart, fromEnd };
std::string result;
enum whereToTake next = fromStart;
while (source.length() > 0) {
if (next == fromStart) {
result += source.at(0);
source = source.substr(1);
next = fromEnd;
}
else {
result += source.at(source.length() - 1); // last char
source = source.substr(0, source.length() - 1); // eat last char
next = fromStart;
}
}
return result;
}
You can take advantage of the concept of reverse iterators to write a generic algorithm based on the solution presented in Usman Riaz answer.
Compose your string picking chars from both the ends of the original string. When you reach the center, add the char in the middle if the number of chars is odd.
Here is a possible implementation:
#include <iostream>
#include <string>
#include <vector>
#include <utility>
#include <algorithm>
#include <iterator>
template <class ForwardIt, class OutputIt>
OutputIt fold(ForwardIt source, ForwardIt end, OutputIt output)
{
auto reverse_source = std::reverse_iterator<ForwardIt>(end);
auto reverse_source_end = std::reverse_iterator<ForwardIt>(source);
auto source_end = std::next(source, std::distance(source, end) / 2);
while ( source != source_end )
{
*output++ = *source++;
*output++ = *reverse_source++;
}
if ( source != reverse_source.base() )
{
*output++ = *source;
}
return output;
}
int main() {
std::vector<std::pair<std::string, std::string>> tests {
{"", ""}, {"a", "a"}, {"stack", "sktca"}, {"steack", "sktcea"}
};
for ( auto const &test : tests )
{
std::string result;
fold(
std::begin(test.first), std::end(test.first),
std::back_inserter(result)
);
std::cout << (result == test.second ? " OK " : "FAILED: ")
<< '\"' << test.first << "\" --> \"" << result << "\"\n";
}
}
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?
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.
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
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).
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;
}
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;
}
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.
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.
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));
}
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;
}
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
}
}
}
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.
My friend got an assignment and I can't help him. Basically, using recursion he needs to print the words of a sentence in the reverse order. For example:
Input - This is a sentence
Output - sentence a is This
Here is an example I wrote him for a normal print, and I could do the whole sentence reversal with no problem, but I can't get neither an idea of a starting point for reversing only the words recursively without a linear approach and using the string library or linked lists or any other way:
#include <iostream>
using namespace std;
void revSentence(char sentence[], int i)
{
if (sentence[i] == 0)
return;
cout<<sentence[i];
i++;
revSentence (sentence, i);
}
int main()
{
char sentence[100];
cin.getline (sentence, 100);
int i = 0;
revSentence(sentence, i);
return 0;
}
Thing is it probably is something very simple because he is doing a fast course with only the basics, so they didn't use anything but the iostream library, so it must be something simple or at least not too complicated. So I'm asking for at least an idea of approach, or a solution. I've got a feeling I'm missing something very simple here but just can't see it.
Thanks in advance
It's not as easy as you might think.
You need to change the location where the printing command is called and place it below revSentence() recursive call, this is called post-order traversal, while the one you are doing is called pre-order traversal.
You also need a stack to push the reversed words.
#include <iostream>
#include <string>
#include <stack>
void revSentence(std::string const &str, std::stack<char> &stk, int i) {
if(i == str.size()) return;
revSentence (str, stk, i + 1);
if((!stk.empty() && stk.top() != ' ' && str[i] == ' ') || i == 0) {
if(!i) std::cout << str[i];
while(!stk.empty()) {
std::cout << stk.top();
stk.pop();
}
if(i) std::cout << str[i];
}
stk.push(str[i]);
if(!i) std::cout << std::endl;
}
int main() {
std::string sentence;
std::stack<char> stk;
std::getline (std::cin, sentence);
int i = 0;
revSentence(sentence, stk, i);
return 0;
}
I can not tell if you are allowed to use std::string and it's methods ... but if so...
In the following approach, I emphasize the finding of the first and last words in the input sentence, then using recursion on the leftover (here called sMiddle).
No additional stack of stl used.
The input string is duplicated in automatic variables, so it uses a bit more stack than some other choices (not a problem on Linux).
Assumes - no padding before 1st word, or after last word.
std::string revSentence(std::string s)
{
std::string retVal;
do
{
size_t posEndW1 = s.find(' ');
if(posEndW1 == std::string::npos) break;
// from line start to end of 1st word
std::string wFirst = s.substr(0, (posEndW1));
size_t posBeginW2 = s.rfind(' '); // from end of line
if(posBeginW2 == std::string::npos) break;
std::string wLast = s.substr(posBeginW2+1, std::string::npos); // to line end
std::string sMiddle = s.substr(posEndW1+1, (posBeginW2-posEndW1-1));
retVal = wLast + " " + revSentence(sMiddle) + " " + wFirst;
}while(0);
return(retVal);
}