#include <iostream>
#include <conio.h>
using namespace std;
int main(){
char command[1024];
char newchar;
cout << "Command Line Interface Test with Intellisense" << endl;
cout << endl;
newchar = _getch();
command = command + newchar;
}
Why does this not work?
Why command = command + newchar is wrong?
You should use std::string and append the char to it.
http://en.cppreference.com/w/cpp/string/basic_string/append
Or with C++11, you can use the += operator with std::string
(You will have to #include string header)
It doesn't work because C++ is statically typed. A char[1024] object will stay the same type for its entire life, and it can't change to a char[1025]. The same applies to a std::string, but the size of a string is not part of its type and can therefore change:
std::string command = "abc";
comamnd += "d";
In command + newchar, command becomes a (const) pointer and newchar is an integer value, so you are making a pointer to a "bigger" address, but when assigning the result to command, you are trying to change the (const) pointer to an array, which luckily is not allowed.
char* pNew = command + newchar;
This would have worked, but not doing as expected. As others already answered: use std::string.
Are You trying to do something like this?
#include <iostream>
#include <conio.h>
using namespace std;
#define BUFFER_SIZE 1024
int main(){
char command[BUFFER_SIZE];
cout << "Command Line Interface Test with Intellisense" << endl;
cout << endl;
for(unsigned int i = 0; i < BUFFER_SIZE; ++i)
char newchar = _getch();
if(newchar == '\n') break;
// do some magic with newchar if you wish
command[i] = newchar;
}
}
Related
I have the following code:
#include <fstream>
#include <iostream>
using namespace std;
int main() {
ofstream os;
char fileName[] = "0.txt";
for(int i = '1'; i <= '5'; i++)
{
fileName[0] = i;
os.open(fileName);
os << "Hello" << "\n";
os.close();
}
return 0;
}
The aim is to write my code output into multiple .txt files up to, say 64 different times. When I change this loop to run more than 10, that is
for(int i = '1'; i <= '10'; i++)
I get the following error:
warning: character constant too long for its type
Any ideas how to write to more than 10 files? Moreover, how do I write a number after each "Hello", e.g. "Hello1 ... Hello10"?
Cheers.
I believe the reason you receive that warning is because you're attempting to assign two chars into one slot of the char array:
fileName[0] = i;
because when i = 10;, it's no longer a single character.
#include <fstream>
#include <iostream>
#include <string>//I included string so that we can use std::to_string
using namespace std;
int main() {
ofstream os;
string filename;//instead of using a char array, we'll use a string
for (int i = 1; i <= 10; i++)
{
filename = to_string(i) + ".txt";//now, for each i value, we can represent a unique filename
os.open(filename);
os << "Hello" << std::to_string(i) << "\n";//as for writing a number that differs in each file, we can simply convert i to a string
os.close();
}
return 0;
}
Hopefully this resolved the issues in a manner that you're satisfied with; let me know if you need any further clarification! (:
Is there any way to tell a stringstream to ignore a null terminating char and read a certain amount of chars anyway?
As you can see from this minimum example, even though the char array consists of 3 chars, the stringstream terminates at the second position:
#include <sstream>
#include <iostream>
using namespace std;
int main(int argc, char* argv[]) {
char test[3];
test[0] = '1';
test[1] = '\0';
test[2] = '2';
stringstream ss(test);
char c;
cout << "start" << endl;
while (ss.get(c)) {
cout << c << endl;
}
if (ss.eof()) {
cout << "eof" << endl;
}
}
$ ./a.out
start
1
eof
This question is not about stringstreams. The problem is you are implicitly constructing a std::string from a const char* for that stringstream constructor argument, and doing so using the overload that expects a C-string. So, naturally, you should expect C-string-like behaviour.
Instead you can form the argument using the std::string(const char*, std::size_t) constructor, or send the data to a default-constructed stringstream using .write.
In addition to the other answer explaining the underlying problem (creating a std::string from a char*), here's one (of many) ways around the problem, using std::string_literals:
#include <iostream>
#include <string>
#include <sstream>
int main(){
using namespace std::string_literals;
const std::string str_with_null = "1\0002"s;
std::stringstream ss(str_with_null);
char c;
while (ss.get(c)) {
std::cout << static_cast<int>(c) << '\n';
}
}
When run, this should print out:
49
0
50
As a topic says. I try to combine few chars (from array) into one string.
i tried
char test[]={'A'};
char testt[]={'a'};
string testtt= test[0]+testt[0];
But it doesn't work.
char test[]={'A'};
char testt[]={'a'};
string testtt="";
testtt+=test[0];
testtt+=testt[0];
You can either use the constructor that takes char*, length.
#include<iostream>
#include<string>
int main(){
char test[]={'A'};
char testt[]={'a','w'};
std::string testtt= std::string(&test[0], 1) + std::string(&testt[1],1);/
std::cout<<testtt<<std::endl;
return 0;
}
or constructor that takes a range:
std::string testtt= std::string(&test[0], &test[1]) + std::string(&testt[0],&testt[1]);
It will do the trick ;)
#include <sstream>
string toStr(char* tab, int length)
{
stringstream ss;
for (int i=0 ; i<length; i++)
ss << tab[i];
return ss.str();
}
if you look inside class string you'll find that it doesn't overload copy constructor to take one character parameter.
you may think that you implement your own class containing a string object:
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
class MyStr
{
public:
MyStr(char c){ itsString = c;}
string GetStr()const {return itsString;}
private:
string itsString;
};
int main(int argc, char* argv[])
{
char test[]={'A'};
char testt[]={'a'};
MyStr testtt = test[0] + testt[0];
cout << testtt.GetStr() << endl;
cout << endl << endl << endl;
return 0;
}
this program runs and the compiler never complains BUT the result won't be as you may think: it won't be "Aa" but just: 'ت' because in fact the above statement: string testtt= test[0]+testt[0]; is equivalent to write: cout << (char)('A' + 'a');
which means summing two characters and the result is the integer ASCII value then this value will be converted back again to charater because you are really ivoking string(char); which will result to (charater of 162)
i have a question about a simple c++ function.
This is my cpp file:
#include <string>
#include <iostream>
#include <stdio.h>
#include <ros/ros.h>
#include <json_prolog/prolog.h>
using namespace std;
using namespace json_prolog;
int main(int argc, char *argv[])
{
ros::init(argc, argv, "Knives");
Prolog pl;
char M[]="drawer";
cout<<M<<endl;
if(strcmp(M,"knife")==0)
{
string q= "rdfs_individual_of(M, 'http://knowrob.org/kb/knowrob.owl#TableKnife')";
PrologQueryProxy bdgs = pl.query(q);
cout<< endl;
for(PrologQueryProxy::iterator it=bdgs.begin(); it != bdgs.end(); it++)
{
PrologBindings bdg = *it;
cout << "Knives individuals= "<< bdg["M"] << endl;
}
cout<< endl<< endl;
}
if(strcmp(M,"drawer")==0)
{
string q= "rdfs_individual_of(M, 'http://knowrob.org/kb/knowrob.owl#Drawer')";
PrologQueryProxy bdgs = pl.query(q);
cout<< endl;
for(PrologQueryProxy::iterator it=bdgs.begin(); it != bdgs.end(); it++)
{
PrologBindings bdg = *it;
cout << "Drawer individuals= "<< bdg["M"] << endl;
}
cout<< endl<< endl;
}
return 0;
}
this code is connect with an xml file to parse it.
if i compile it works and i have no problems.
Now i have to change it,because i don't want to define the variable char M but i want to give it in input. the problem is that i change :
char M[]=....
with:
char M;
cin>>M;
i have a problem about the conversion from char to const char [-fpermissive]
how i can solve it?
try this:
std::string M;
cin >> M;
replace lines like this:
if(strcmp(M,"drawer")==0)
with this:
if (M == "drawer" )
Why did you change an array of chars (char M[] = ...;) to a single char (char M;)? Aside from the other answer suggesting using a std::string instead (which you should do), if you must use an array of char variables, you need to use std::unique_ptr<char> M(new char[100]) before use in std::cin (I've assumed 100 chars are enough to hold the input; change as necessary).
std::string M = "drawer";
cout<<M<<endl;
if(M.compare("knife") == 0)
{
// do whatever
}
You can also do:
if(M == "knife")
{
// do whatever
}
Unless you need to use C strings, why would you not use the standard library string? No reason to make life more difficult.
I want to ask for word from the user and then convert the word from string to char using 'strcpy'. Then I want to determine the sum of the ascii codes for all of the letters in the word.
However, I am having difficulties. I don't understand exactly how I can do that. This is what I have been able to do so far.
#include <iostream>
#include <time.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
int main()
{
string word;
cout << "Enter word: ";
getline(cin, word);
/*
char w[word];
strcpy(w,word.c_str());
int ('A');
cout<<char(65);
*/
return 0;
}
The commented part is where I have been trying to do the converting. I copied the code from a worksheet. Even if it did work, I don't know how, and what it all means.
Thanks for your help.
char w[word];
strcpy(w, word.c_str());
char w[word] is incorrect. The square brackets is for the size, which must be a constant integral expression. word is of type std::string, so this makes neither logical nor practical sense. Maybe you meant it as:
char w = word;
But that still won't work because word is a string, not a character. The correct code in this case is:
char* w = new char[word.size() + 1];
That is, you allocate the memory for w using a char*. Then you use word.size() + 1 to initialize heap-allocated memory amounting to those bytes. Don't forget for the obligatory delete[] when you're finished using w:
delete[] w;
However, note that using raw pointers and explicit new is not needed in this case. Your code can easily be cleaned up into the following:
#include <numeric>
int main ()
{
std::string word;
std::getline(std::cin, word);
int sum = std::accumulate(word.begin(), word.end(), 0); /*
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ */
std::cout << "The sum is: " << sum << std::endl;
}
You don't need to use strcpy() (or use a char * at all, for that matter), but this'll do your counting using a char pointer:
#include <iostream>
#include <string>
int main() {
std::string word;
std::cout << "Enter word: ";
std::cin >> word;
const char * cword = word.c_str();
int ascii_total = 0;
while ( *cword ) {
ascii_total += *cword++;
}
std::cout << "Sum of ASCII values of characters is: ";
std::cout << ascii_total << std::endl;
return 0;
}
Output:
paul#local:~/src/cpp/scratch$ ./asccount
Enter word: ABC
Sum of ASCII values of characters is: 198
paul#local:~/src/cpp/scratch$
If you really do want to use strcpy(), I'll leave it as an exercise to you to modify the above code.
Here's a better way to do it, just using std::string (and C++11, and obviously presuming your system uses the ASCII character set in the first place):
#include <iostream>
#include <string>
int main() {
std::string word;
std::cout << "Enter word: ";
std::cin >> word;
int ascii_total = 0;
for ( auto s : word ) {
ascii_total += s;
}
std::cout << "Sum of ASCII values of characters is: ";
std::cout << ascii_total << std::endl;
return 0;
}