Remove character from array where spaces and punctuation marks are found [duplicate] - c++

Question

This question already has answers here:
C++ Remove punctuation from String
(12 answers)
Closed 9 years ago.
In my program, I am checking whole cstring, if any spaces or punctuation marks are found, just add empty character to that location but the complilor is giving me an error: empty character constant.
Please help me out, in my loop i am checking like this
if(ispunct(str1[start])) {
str1[start]=''; // << empty character constant.
}
if(isspace(str1[start])) {
str1[start]=''; // << empty character constant.
}
This is where my errors are please correct me.
for eg the word is str,, ing, output should be string.

Answer 1

There is no such thing as an empty character.
If you mean a space then change '' to ' ' (with a space in it).
If you mean NUL then change it to '\0'.

Answer 2

Edit: the answer is no longer relevant now that the OP has edited the question. Leaving up for posterity's sake.
If you're wanting to add a null character, use '\0'. If you're wanting to use a different character, using the appropriate character for that. You can't assign it nothing. That's meaningless. That's like saying
int myHexInt = 0x;
or
long long myIndeger = L;
The compiler will error. Put in the value you wanted. In the char case, that's a value from 0 to 255.

Answer 3

UPDATE:
From the edit to OP's question, it's apparent that he/she wanted to trim a string of punctuation and space characters.
As detailed in the flagged possible duplicate, one way is to use remove_copy_if:
string test = "THisisa test;;';';';";
string temp, finalresult;
remove_copy_if(test.begin(), test.end(), std::back_inserter(temp), ptr_fun<int, int>(&ispunct));
remove_copy_if(temp.begin(), temp.end(), std::back_inserter(finalresult), ptr_fun<int, int>(&isspace));
ORIGINAL
Examining your question, replacing spaces with spaces is redundant, so you really need to figure out how to replace punctuation characters with spaces. You can do so using a comparison function (by wrapping std::ispunct) in tandem with std::replace_if from the STL:
#include <string>
#include <algorithm>
#include <iostream>
#include <cctype>
using namespace std;
bool is_punct(const char& c) {
return ispunct(c);
}
int main() {
string test = "THisisa test;;';';';";
char test2[] = "THisisa test;;';';'; another";
size_t size = sizeof(test2)/sizeof(test2[0]);
replace_if(test.begin(), test.end(), is_punct, ' ');//for C++ strings
replace_if(&test2[0], &test2[size-1], is_punct, ' ');//for c-strings
cout << test << endl;
cout << test2 << endl;
}
This outputs:
THisisa test
THisisa test another

Answer 4

Try this (as you asked for cstring explicitly):
char str1[100] = "str,, ing";
if(ispunct(str1[start]) || isspace(str1[start])) {
strncpy(str1 + start, str1 + start + 1, strlen(str1) - start + 1);
}
Well, doing this just in pure c language, there are more efficient solutions (have a look at #MichaelPlotke's answer for details).
But as you also explicitly ask for c++, I'd recommend a solution as follows:
Note you can use the standard c++ algorithms for 'plain' c-style character arrays also. You just have to place your predicate conditions for removal into a small helper functor and use it with the std::remove_if() algorithm:
struct is_char_category_in_question {
bool operator()(const char& c) const;
};
And later use it like:
#include <string>
#include <algorithm>
#include <iostream>
#include <cctype>
#include <cstring>
// Best chance to have the predicate elided to be inlined, when writing
// the functor like this:
struct is_char_category_in_question {
bool operator()(const char& c) const {
return std::ispunct(c) || std::isspace(c);
}
};
int main() {
static char str1[100] = "str,, ing";
size_t size = strlen(str1);
// Using std::remove_if() is likely to provide the best balance from perfor-
// mance and code size efficiency you can expect from your compiler
// implementation.
std::remove_if(&str1[0], &str1[size + 1], is_char_category_in_question());
// Regarding specification of the range definitions end of the above state-
// ment, note we have to add 1 to the strlen() calculated size, to catch the
// closing `\0` character of the c-style string being copied correctly and
// terminate the result as well!
std::cout << str1 << endl; // Prints: string
}
See this compilable and working sample also here.

Answer 5

As I don't like the accepted answer, here's mine:
#include <stdio.h>
#include <string.h>
#include <cctype>
int main() {
char str[100] = "str,, ing";
int bad = 0;
int cur = 0;
while (str[cur] != '\0') {
if (bad < cur && !ispunct(str[cur]) && !isspace(str[cur])) {
str[bad] = str[cur];
}
if (ispunct(str[cur]) || isspace(str[cur])) {
cur++;
}
else {
cur++;
bad++;
}
}
str[bad] = '\0';
fprintf(stdout, "cur = %d; bad = %d; str = %s\n", cur, bad, str);
return 0;
}
Which outputs cur = 18; bad = 14; str = string
This has the advantage of being more efficient and more readable, hm, well, in a style I happen to like better (see comments for a lengthy debate / explanation).