line.find won't compile, line is not declared - c++

I am a very novice programmer, and I am trying to understand the find functions for strings. At uni we are told to use c-strings, which is why I think that it isn't working. The problem comes when I compile, there is a compile error that line was not declared. This is my code:
#include <iostream>
#include <fstream>
#include <cstring>
#include <string>
using namespace std;
int main()
{
char test[256];
char ID[256];
cout << "\nenter ID: ";
cin.getline(ID, 256);
int index = line.find(ID);
cout << index << endl;
return 0;
}
Please help, it has become really frustrating as I need to understand this function to complete my assignment :/

You're trying to use C-style strings. But find is a member of the C++ string class. If you want to use C-style strings, use functions that operate on C style strings like strcmp, strchr, strstr, and so on.

Supposing you actually input some data into test also, then one way to do it would be:
char *found = strstr(test, ID);
if ( !found )
cout << "The ID was not found.\n";
else
cout << "The index was " << (found - test) << '\n';

Because find fuction a member function string class,You should declare a string class's object. I think you will do that like this:
string test = "This is test string";
string::size_type position;
position = test.find(ID);
if (position != test.npos){
cout << "Found: " << position << endl;
}
else{
cout << "not found ID << endl;
}

Related

Cout does not work on my other turbo c++ complier

Hello guys i am beginner on the language c++
i was trying to run this code below on my ide"codeblocks" and it works
https://www.youtube.com/watch?v=vLnPwxZdW4Y (link for the tutorial that following )
#include <iostream>
using namespace std;
int main()
{
string charactername = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is" << charactername<< endl;
cout << "i am " << characterage << endl;
return 0;
}
this code does not work on my other compiler running on dosbox ? any ideas why ?
I suggest you stop using Turbo C++ as it is a very outdated and a discontinued compiler. However, if you don't have the option of using new compilers (I had the same issue as I studied C++ at school), you will have to make the following changes:
using namespace std; cannot be used in Turbo C++. You will have to remove that and replace #include<iostream> with #include<iostream.h>
Data-type string cannot be used in Turbo C++. You will have to declare a character array instead.
You will have to use #include<stdio.h> and the function puts(); to display the character array in case of Turbo C++. Alternatively you can use a loop-statement.
This will be your final code:
#include <iostream.h>
#include <stdio.h>
int main()
{
char charactername[] = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is ";
puts(charactername);
cout << "i am " << characterage << endl;
return 0;
}
Note: The puts(); function automatically puts the cursor on the next line. So you don't need to use endl;
Or, if you want to use a loop-statement to display the character array
#include <iostream.h>
int main()
{
char charactername[] = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is ";
int i=0;
while(charactername[i]!='\0') {
cout<<charactername[i];
i++;
}
cout<<endl;
cout << "i am " << characterage << endl;
return 0;
}
'\0' is the last element of the character array. So as long as the loop does not reach the last element, it will print the character array.
a[] = "arnold"; basically means an array is created like this: a[0]='a', a[1]='r', a[2]='n',.... a[5]='d', a[6]='\0'.
Alternatively, if you use cout << charactername; instead of the while loop, it will print the whole name. (This is only in the case of a string variable (character array), for an integer array or any other array you will need the while loop)

C++ replace words in a string (text file)

so I worked on my program and now I am on a point where I can not find a solution. I need to replace some more signs in the fext file, for now the program only replaces "TIT" with the code number "*245$a", if I want to replace other letters the same way, the program does not change. Does anybody know how I can implement some more replacements in the text file? Let me know if there is a better option to replace more than 5 signs with another ones.
Thank you
#include <fstream>
#include <fstream>
#include <iostream>
#include <string>
using namespace std;
int main()
{
char dateiname[64], kommando[64];
ifstream iStream;
cout << "Choose an activity" << endl <<
" s - search " << endl <<
" c - convert" << endl <<
" * - end program" << endl;
cin.getline(kommando,64,'\n');
switch(kommando[0])
{
case 'c':
cout << "Enter a text file!" << endl;
cin.getline(dateiname,64,'\n');
iStream.open("C://users//silita//desktop//schwarz.txt");
case 's':
break;
case '*':
return 0;
default:
cout << "I can not read " << kommando << endl;
}
if (!iStream)
{
cout << "The File" << dateiname << "does not exist." << endl;
}
string s;
char o[] = "TIT";
while (getline(iStream, s))
{
while(s.find(o, 0) < s.length())
s.replace(s.find(o, 0), s.length() - s.find(o, 3),"*245$a");
cout << s << endl;
}
iStream.close();
}
You can use map in C++ STL to store multiple convert rules:
#include<map>
#include<algorithm>
using namespace std;
map<string,string> convertRules;
typedef map<string,string>::iterator MIT;
void setConvertRules(int numOfRules){
string word,code;
for(int i = 0 ; i < numOfRules; ++i){
cin>>word>>code;
//Use code as search key in order to decrypt
//If you want to encrypt, use convertrules[word] = code;
convertRules[code] = word;
}
}
To convert a file, just do as follows (some functions and classes need to be declared and implemented first, but here we mainly focus on top-level design):
/* Detailed class implementations are omitted for simplicity */
//a class to store contents of a file
class File;
//a processor to read, insert and overwrite certain file
class FileProcessor;
void FileProcessor::convert(const string &code, const string &word){
cursor == file.begin();
while(cursor != fp.end()){
_fp.convertNextLine(code,word);
}
}
File file;
FileProcessor filePcr;
int main()
const string sourceDir = "C://users//silita//desktop//schwarz.txt";
const string destDir = "C://users//silita//desktop//schwarz_decrypted.txt";
//Open a .txt file and read in its contents
if (!file.openAndReadIn(sourceDir)){
cerr << "The File" << sourceDir << "does not exist." << endl;
abort();
}
//Try to link processor to open file
if(!fp.linkTo(file)){
cerr << "Access to file" << sourceDir << "is denied." << endl;
abort();
}
//iterator is like a more safe version of C-style pointer
//the object type is a string-string pair
for(MIT it = convertRules.begin(); it != convertRules.end(); ++it){
fp.convert(it->first, it->second);
}
file.saveAs(destDir);
return 0;
}
Finally, if I would suggest you use C-style strstr function for efficiency when dealing with large files or batch processing. string::find adopts a naive sequential search startegy while strstr is implemented with the famous KMP algorithm for fast pattern match in strings, which is both efficient and thorough(can replace all matchs in one go instead of another for-loop).

Update: program shows adress of fstream instead of the text file

I am about to write a program which asks the user if they want to "search or convert" a file, if they choose convert, they need to provide the location of the file.
I do not know why the program shows the address of the file instead of opening it.
Here is my first approach:
#include <fstream>
#include <fstream>
#include <iostream>
#include <string>
using namespace std;
int main()
{
char dateiname[64], kommando[64];
ifstream iStream;
cout << "Choose an action: " << endl <<
" s - search " << endl <<
" c - convert" << endl <<
" * - end program" << endl;
cin.getline(kommando,64,'\n');
switch(kommando[0])
{
case 'c':
cout << "Enter a text file: " << endl;
cin.getline(dateiname,64,'\n');
iStream.open("C://users//silita//desktop//schwarz.txt");
case 's': break;
case '*': return 0;
default:
cout << "Invalid command: " << kommando << endl;
}
if (!iStream)
{
cout << "The file " << dateiname << " does not exist." << endl;
}
string s;
while (getline(iStream, s)) {
while(s.find("TIT", 0) < s.length())
s.replace(s.find("TIT", 0), s.length() - s.find("TIT", 3),"*245$a");
cout << iStream << endl;
}
iStream.close();
}
At first you can't compare c-strings using ==. You must use strcmp(const char*, const char*). More info about it you can find there: http://www.cplusplus.com/reference/cstring/strcmp/
For example: if (i == "Konvertieren") must become if(!strcmp(i,"Konvertieren"))
As mentioned in Lassie's answer, you can't compare strings in this way using c or c++; just to flesh it out, however, I'll explain why.
char MyCharArr[] = "My Character Array"
// MyCharArr is now a pointer to MyCharArr[0],
// meaning it's a memory address, which will vary per run
// but we'll assume to be 0x00325dafa
if( MyCharArr == "My Character Array" ) {
cout << "This will never be run" << endl;
}
Here the if compares a pointer (MyCharArr) which will be a memory address, ie an integer, to a character array literal. Obviously 0x00325dafa != "My Character Array".
Using cstrings (character arrays), you need to use the strcmp() function which you will find in the cstring library, which will give you a number telling you "how different" the strings are, essentially giving the difference a numerical value. In this instance we are only interested in no difference, which is 0, so what we need is this:
#include <cstring>
using namespace std;
char MyCharArr[] = "My Character Array"
if( strcmp(MyCharArr,"My Character Array")==0 ) {
// If there is 0 difference between the two strings...
cout << "This will now be run!" << endl;
}
While you are not doing so in your question, If we were using c++ strings rather than character arrays, we would use the compare() method to similar affect:
#include <string>
using namespace std;
string MyString = "My C++ String"
if( MyString.compare("My C++ String")==0 ) {
// If there is 0 difference between the two strings...
cout << "This will now be run!" << endl;
}

How to use Arrays, with Strings

I'm a C++ programmer, who's still in the nest, and not yet found my wings. I was writing a Calendar program, and I discovered, that C++ does not support a string type. How do I make an Array, that will be able to store strings of characters?
I've thought of creating an enumerated data type, as the array type. While, it will work, for my Calendar, it won't work if say I was creating a database of the names of students in my class.
http://prntscr.com/7m074w I got; "error, 'string' does not name a type."
that C++ does not support a string type.
Wrong info, you can create an character array as follows
char array[length];
//Where length should be a constant integer
Otherwise you can depend on standard template library container, std::string
If you have C++11 compiler you can depend on std::array
The C++ Standard Library includes a string type, std::string. See http://en.cppreference.com/w/cpp/string/basic_string
The Standard Library also provides a fixed-size array type, std::array. See http://en.cppreference.com/w/cpp/container/array
But you may also want to learn about the dynamically-sized array type, std::vector. See http://en.cppreference.com/w/cpp/container/vector
The language also includes legacy support for c-strings and c-arrays, which you can find in a good C++ or C book. See The Definitive C++ Book Guide and List
An example of how to use an array/vector of strings:
#include <string>
#include <array>
#include <vector>
#include <iostream>
int main() {
std::array<std::string, 3> stringarray;
stringarray[0] = "hello";
stringarray[1] = "world";
// stringarray[2] contains an empty string.
for (size_t i = 0; i < stringarray.size(); ++i) {
std::cout << "stringarray[" << i << "] = " << stringarray[i] << "\n";
}
// Using a vector, which has a variable size.
std::vector<std::string> stringvec;
stringvec.push_back("world");
stringvec.insert(stringvec.begin(), "hello");
stringvec.push_back("greetings");
stringvec.push_back("little bird");
std::cout << "size " << stringvec.size()
<< "capacity " << stringvec.capacity()
<< "empty? " << (stringvec.empty() ? "yes" : "no")
<< "\n";
// remove the last element
stringvec.pop_back();
std::cout << "size " << stringvec.size()
<< "capacity " << stringvec.capacity()
<< "empty? " << (stringvec.empty() ? "yes" : "no")
<< "\n";
std::cout << "stringvec: ";
for (auto& str : stringvec) {
std::cout << "'" << str << "' ";
}
std::cout << "\n";
// iterators and string concatenation
std::string greeting = "";
for (auto it = stringvec.begin(); it != stringvec.end(); ++it) {
if (!greeting.empty()) // add a space between words
greeting += ' ';
greeting += *it;
}
std::cout << "stringvec combined :- " << greeting << "\n";
}
Live demo: http://ideone.com/LWYevW
You can create an array of characters by char name[length];.
C++ also has a data type string. You can create an array of strings and store what values you'd like. here .
So
use array of characters
use string data type
For Example -
#include <iostream>
#include <string>
int main ()
{
//To Create a String
std::string s0 ("Initial string");
return 0;
}
C++ does have a string type: string from #include <string>
If you don't want to use that, you can also use char* name = "YourTextHere..." or `char[length+1] name = "YourTextHere"

How do you append an int to a string in C++? [duplicate]

This question already has answers here:
How to concatenate a std::string and an int
(25 answers)
Closed 6 years ago.
int i = 4;
string text = "Player ";
cout << (text + i);
I'd like it to print Player 4.
The above is obviously wrong but it shows what I'm trying to do here. Is there an easy way to do this or do I have to start adding new includes?
With C++11, you can write:
#include <string> // to use std::string, std::to_string() and "+" operator acting on strings
int i = 4;
std::string text = "Player ";
text += std::to_string(i);
Well, if you use cout you can just write the integer directly to it, as in
std::cout << text << i;
The C++ way of converting all kinds of objects to strings is through string streams. If you don't have one handy, just create one.
#include <sstream>
std::ostringstream oss;
oss << text << i;
std::cout << oss.str();
Alternatively, you can just convert the integer and append it to the string.
oss << i;
text += oss.str();
Finally, the Boost libraries provide boost::lexical_cast, which wraps around the stringstream conversion with a syntax like the built-in type casts.
#include <boost/lexical_cast.hpp>
text += boost::lexical_cast<std::string>(i);
This also works the other way around, i.e. to parse strings.
printf("Player %d", i);
(Downvote my answer all you like; I still hate the C++ I/O operators.)
:-P
These work for general strings (in case you do not want to output to file/console, but store for later use or something).
boost.lexical_cast
MyStr += boost::lexical_cast<std::string>(MyInt);
String streams
//sstream.h
std::stringstream Stream;
Stream.str(MyStr);
Stream << MyInt;
MyStr = Stream.str();
// If you're using a stream (for example, cout), rather than std::string
someStream << MyInt;
For the record, you can also use a std::stringstream if you want to create the string before it's actually output.
cout << text << " " << i << endl;
Your example seems to indicate that you would like to display the a string followed by an integer, in which case:
string text = "Player: ";
int i = 4;
cout << text << i << endl;
would work fine.
But, if you're going to be storing the string places or passing it around, and doing this frequently, you may benefit from overloading the addition operator. I demonstrate this below:
#include <sstream>
#include <iostream>
using namespace std;
std::string operator+(std::string const &a, int b) {
std::ostringstream oss;
oss << a << b;
return oss.str();
}
int main() {
int i = 4;
string text = "Player: ";
cout << (text + i) << endl;
}
In fact, you can use templates to make this approach more powerful:
template <class T>
std::string operator+(std::string const &a, const T &b){
std::ostringstream oss;
oss << a << b;
return oss.str();
}
Now, as long as object b has a defined stream output, you can append it to your string (or, at least, a copy thereof).
Another possibility is Boost.Format:
#include <boost/format.hpp>
#include <iostream>
#include <string>
int main() {
int i = 4;
std::string text = "Player";
std::cout << boost::format("%1% %2%\n") % text % i;
}
Here a small working conversion/appending example, with some code I needed before.
#include <string>
#include <sstream>
#include <iostream>
using namespace std;
int main(){
string str;
int i = 321;
std::stringstream ss;
ss << 123;
str = "/dev/video";
cout << str << endl;
cout << str << 456 << endl;
cout << str << i << endl;
str += ss.str();
cout << str << endl;
}
the output will be:
/dev/video
/dev/video456
/dev/video321
/dev/video123
Note that in the last two lines you save the modified string before it's actually printed out, and you could use it later if needed.
For the record, you could also use Qt's QString class:
#include <QtCore/QString>
int i = 4;
QString qs = QString("Player %1").arg(i);
std::cout << qs.toLocal8bit().constData(); // prints "Player 4"
cout << text << i;
One method here is directly printing the output if its required in your problem.
cout << text << i;
Else, one of the safest method is to use
sprintf(count, "%d", i);
And then copy it to your "text" string .
for(k = 0; *(count + k); k++)
{
text += count[k];
}
Thus, you have your required output string
For more info on sprintf, follow:
http://www.cplusplus.com/reference/cstdio/sprintf
cout << text << i;
The << operator for ostream returns a reference to the ostream, so you can just keep chaining the << operations. That is, the above is basically the same as:
cout << text;
cout << i;
cout << "Player" << i ;
cout << text << " " << i << endl;
The easiest way I could figure this out is the following..
It will work as a single string and string array.
I am considering a string array, as it is complicated (little bit same will be followed with string).
I create a array of names and append some integer and char with it to show how easy it is to append some int and chars to string, hope it helps.
length is just to measure the size of array. If you are familiar with programming then size_t is a unsigned int
#include<iostream>
#include<string>
using namespace std;
int main() {
string names[] = { "amz","Waq","Mon","Sam","Has","Shak","GBy" }; //simple array
int length = sizeof(names) / sizeof(names[0]); //give you size of array
int id;
string append[7]; //as length is 7 just for sake of storing and printing output
for (size_t i = 0; i < length; i++) {
id = rand() % 20000 + 2;
append[i] = names[i] + to_string(id);
}
for (size_t i = 0; i < length; i++) {
cout << append[i] << endl;
}
}
There are a few options, and which one you want depends on the context.
The simplest way is
std::cout << text << i;
or if you want this on a single line
std::cout << text << i << endl;
If you are writing a single threaded program and if you aren't calling this code a lot (where "a lot" is thousands of times per second) then you are done.
If you are writing a multi threaded program and more than one thread is writing to cout, then this simple code can get you into trouble. Let's assume that the library that came with your compiler made cout thread safe enough than any single call to it won't be interrupted. Now let's say that one thread is using this code to write "Player 1" and another is writing "Player 2". If you are lucky you will get the following:
Player 1
Player 2
If you are unlucky you might get something like the following
Player Player 2
1
The problem is that std::cout << text << i << endl; turns into 3 function calls. The code is equivalent to the following:
std::cout << text;
std::cout << i;
std::cout << endl;
If instead you used the C-style printf, and again your compiler provided a runtime library with reasonable thread safety (each function call is atomic) then the following code would work better:
printf("Player %d\n", i);
Being able to do something in a single function call lets the io library provide synchronization under the covers, and now your whole line of text will be atomically written.
For simple programs, std::cout is great. Throw in multithreading or other complications and the less stylish printf starts to look more attractive.
You also try concatenate player's number with std::string::push_back :
Example with your code:
int i = 4;
string text = "Player ";
text.push_back(i + '0');
cout << text;
You will see in console:
Player 4
You can use the following
int i = 4;
string text = "Player ";
text+=(i+'0');
cout << (text);
If using Windows/MFC, and need the string for more than immediate output try:
int i = 4;
CString strOutput;
strOutput.Format("Player %d", i);