CODEWITHSUNDEEP
c++iterator
j l
j l

Reputation: 61

String Char Iterator

Just returned to programming in C++. the errors i get:

request for member begin in sent is of non-class type char[30]

request for member end in sent is of non-class type char[30]

char sent[] = "need to break this shiat down";
    for(vector<string>::iterator it=sent.begin(); it!=sent.end(); ++it){
        if(*it == " ")
            cout << "\n";
        else
            cout << *it << endl;
    }

should i change the char to string or define the vector differently?

Upvotes: 0

Views: 11408

Answers (5)

WhozCraig
WhozCraig

Reputation: 66254

You can also use streaming to throw out the whitespace and throw in newlines.

#include <iostream>
#include <sstream>
#include <string>
using namespace std;

int main(int argc, char *argv[])
{
    stringstream ss("need to break this shiat down.", ios_base::in);

    string s;
    while (ss >> s)
        cout << s << endl;

    return EXIT_SUCCESS;
}

Result:

need
to
break
this
shiat
down.

Upvotes: 3

PiotrNycz
PiotrNycz

Reputation: 24420

char sent[] is not std::string but string literal - but in this very case you can iterate over it:

int main() {
char sent[] = "need to break this shiat down";
    for(auto it = std::begin(sent); it!=std::end(sent) - 1; ++it){
        if(*it == ' ')
            cout << "\n";
        else
            cout << *it << endl;
    }
}

Note that I changed " " to ' ' - and skipped last null terminating char '\0'...

Live example: http://liveworkspace.org/code/55f826dfcf1903329c0f6f4e40682a12

For C++03 you can use this approach:

int main() {
char sent[] = "need to break this shiat down";
    for(char* it = sent; it!=sent+sizeof(sent) - 1; ++it){
        if(*it == ' ')
            cout << "\n";
        else
            cout << *it << endl;
    }
}

If this is string literal of size not known at that point - use strlen instead of sizeof...

Upvotes: 2

juanchopanza
juanchopanza

Reputation: 227558

It has been pointed out in other answers that you are iterating over the wrong type. You should define sent to be of std::string type and use std::string::begin() and std::string::end() to iterate, or, if you have C++11 support, you have some options for easily iterating over a fixed size array. You can iterate using std::begin and std::end`:

char sent[] = "need to break this shiat down";
for(char* it = std::begin(sent); it != std::end(sent); ++it){
    if(*it == ' ')
        std::cout << "\n";
    else
        std::cout << *it << "\n";
}

or you can use a range-based loop:

char sent[] = "need to break this shiat down";
for (const auto& c : sent)
{
  std::cout << c << "\n";
}

Upvotes: 4

Anirudha
Anirudha

Reputation: 32817

Use string instead of char[]

string sent = "need to break this shiat down";
for(string::iterator it=sent.begin(); it!=sent.end(); ++it){
    if(*it == ' ')
        cout << "\n";
    else
        cout << *it << endl;
}

char[] doesnt have begin and end methods..

Upvotes: 1

Has QUIT--Anony-Mousse
Has QUIT--Anony-Mousse

Reputation: 77495

Your variable sent is not of type vector<string>, but char[].

Your for loop however, tries to iterate over a vector of strings.

For a plain array, use C iteration:

 int len = strlen(sent);
 for (int i = 0; i < len; i++)

Upvotes: 1

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