In xcode 4.3 I have set the project to use c++11: I changed the voice c++ language dialect to c++11, and c++ standard library to "libc++ ( LLVM c++ standard library with c++11 support)".
Then I tried to compile and execute this simple code:
#include <iostream>
using namespace std;
int main (int argc, char** argv)
{
char buffer[100];
cin.getline(buffer,100);
cout << buffer << endl;
return 0;
}
The problem is that it asks two times the input.For example I type "hello" and the stream remains opened, waiting for another string.If I type another string then it prints out "hello".
If I don't use c++11 this problem doesn't occur.
Does anyone know how to solve this problem? I want to take in input maximum 100 characters without using std::string.
This is a bug in libc++. My apologies. It is fixed on Mountain Lion. You can work around it by using getline(istream&, string&) instead:
#include <iostream>
#include <string>
using namespace std;
int main (int argc, char** argv)
{
std::string buffer;
getline(cin, buffer);
cout << buffer << endl;
return 0;
}
Related
i keep getting this error. i know this is a c++ 11 function but it still isnt working with code blocks c++ compiler. am i using this function correctly of is it a problem with the codeblocks compiler. i tried changing the compiler. using the "have g++ follow the c++11 iso standard" i still keep getting this error. or getting the "stoi() does not exist in the current scope" error
#include <iostream>
#include <string>
using namespace std;
int main()
{
int test = 34;
cout << stoi(test);
}
stoi means "String To Int". It will read an int from a std::string (or std::wstring). See also the reference.
You were probably looking for the reverse std::to_string (reference). But you don't need either, there is no need to convert to string before printing:
#include <iostream>
int main()
{
int test = 34;
std::cout << test;
}
stoi means string to int. So it takes a string as an input.
This should work:
string test = "34"; cout << stoi(test);
I am trying to take a string and parse it into an int. I have read the many answers out there, and it seems that using stoi is the most up-to-date way. It appears to me that stoi uses std, but I am getting Function 'stoi' could not be resolved despitre using namespace std;
#include <iostream>
#include <string>
#include <cstring>
#include <fstream>
#include<stdlib.h>
using namespace std;
int main(int argc, char* argv[]) {
string line = "";
string five = "5";
int number = stoi(five); //Error here with stoi
return 0;
}
Any ideas what is causing this?
Update:
I am using Eclipse. My flags are: -c -fmessage-length=0 -std=c++11
If you are using GCC or MINGW, then this is the answer:
std::stoi doesn't exist in g++ 4.6.1 on MinGW
This is a result of a non-standard declaration of vswprintf on
Windows. The GNU Standard Library defines
_GLIBCXX_HAVE_BROKEN_VSWPRINTF on this platform, which in turn disables the conversion functions you're attempting to use. You can
read more about this issue and macro here:
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=37522.
If you're willing to modify the header files distributed with MinGW,
you may be able to work around this by removing the
!defined(_GLIBCXX_HAVE_BROKEN_VSWPRINTF) macro on line 2754 of
.../lib/gcc/mingw32/4.6.1/include/c++/bits/basic_string.h, and adding
it back around lines 2905 to 2965 (the lines that reference
std::vswprintf). You won't be able to use the std::to_wstring
functions, but many of the other conversion functions should be
available.
Please always provide platform and compiler information.
Toggle on C++11 support in your compiler flags. -std=c++11 for a recent gcc. For Eclipse, please refer to the corresponding question in the FAQ and this answer explains how to get rid of the remaining Eclipse warning.
If you are amenable to parsing an int another way, how about using an STL algorithm and a C++11 lambda expression?
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
int main() {
string str = "12345";
int num = 0;
for_each(str.begin(), str.end(), [&num](char c){ num = 10 * num + (c - '0'); });
cout << str << " = " << num << endl;
}
When I include the required library, the "#include.." line doesn't show any warning. But when I use the functions in that library, I find the Vim shows that "..use of undeclared function...". It seems that the library is not correctly included. So I want to know how to figure out this problem?
The screenshots for this question are attached as follows:
Try including it as follows:
#include <stdlib.h> //use <> instead of ""
Also, the "printf" function comes from the "cstdio" library so try implementing that library as well,
#include <stdio.h>
UPDATED
The easiest way to fix that problem is;
Include the stdio.h library
#include <stdio.h>
Then, instead of typing;
printf('s');
you do,
printf("s");
Now, if you really want to print a character 's', then use,
printf("%c", 's'); // Tells the printf function that 's' is a character
The final code would look like;
#include <stdio.h>
int main(int argc, char** argv) {
printf("s");
printf("%c", 's');
return 0;
}
Now, your comment was that "cout" does not work. In order for "cout" to work you need to include the iostream library:
#include <iostream>
Then, you can use "cout" in your code;
std::cout << 's';
std::cout << "s";
Or you can include "namespace std" and the "iostream" library to avoid using std:: before "cout"
include <iostream>
using namespace std;
Thereafter, use cout without std::
cout << 's';
cout << "s";
The final code would be;
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
cout << 's';
cout << "s";
return 0;
}
If you want to learn more about what is in the iostream library and how to use it I recommend using this site:
http://www.cplusplus.com/reference/iostream/
Also, for the stdio.h,
http://www.cplusplus.com/reference/cstdio/
I am trying to take a string and parse it into an int. I have read the many answers out there, and it seems that using stoi is the most up-to-date way. It appears to me that stoi uses std, but I am getting Function 'stoi' could not be resolved despitre using namespace std;
#include <iostream>
#include <string>
#include <cstring>
#include <fstream>
#include<stdlib.h>
using namespace std;
int main(int argc, char* argv[]) {
string line = "";
string five = "5";
int number = stoi(five); //Error here with stoi
return 0;
}
Any ideas what is causing this?
Update:
I am using Eclipse. My flags are: -c -fmessage-length=0 -std=c++11
If you are using GCC or MINGW, then this is the answer:
std::stoi doesn't exist in g++ 4.6.1 on MinGW
This is a result of a non-standard declaration of vswprintf on
Windows. The GNU Standard Library defines
_GLIBCXX_HAVE_BROKEN_VSWPRINTF on this platform, which in turn disables the conversion functions you're attempting to use. You can
read more about this issue and macro here:
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=37522.
If you're willing to modify the header files distributed with MinGW,
you may be able to work around this by removing the
!defined(_GLIBCXX_HAVE_BROKEN_VSWPRINTF) macro on line 2754 of
.../lib/gcc/mingw32/4.6.1/include/c++/bits/basic_string.h, and adding
it back around lines 2905 to 2965 (the lines that reference
std::vswprintf). You won't be able to use the std::to_wstring
functions, but many of the other conversion functions should be
available.
Please always provide platform and compiler information.
Toggle on C++11 support in your compiler flags. -std=c++11 for a recent gcc. For Eclipse, please refer to the corresponding question in the FAQ and this answer explains how to get rid of the remaining Eclipse warning.
If you are amenable to parsing an int another way, how about using an STL algorithm and a C++11 lambda expression?
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;
int main() {
string str = "12345";
int num = 0;
for_each(str.begin(), str.end(), [&num](char c){ num = 10 * num + (c - '0'); });
cout << str << " = " << num << endl;
}
When using ifstream class to read words from an input file, I have used the following expression:
#include <iostream>
#include <fstream>
int main(int argc, char *argv[])
{
std::ifstream inputStream(myFile.txt);
std::string myString;
myFile.open()
while(myFile.good())
{
myFile >> myString;
printf("%s \n", myString);
}
return 0;
}
The contents of myFile.txt are:
" This is a simple program. "
The compiles and executes as expected using g++ compiler.
However, the same code when compiled using msvc 2008, returns error at the extraction operator (>>) requiring me to replace the std::string with either an initialized character array or any of the supported native types.
This threw me off as I was expecting the usage of the standard library to be same across implementations.
I understand the compile error and know the way to fix it via using c_str().
But, it would help me a great deal, if someone could clarify why the usage for the standard library is different across platforms.
To me it is not starndard anymore !!
EDIT: Code updated to be complete. Content of myFile.txt updated.
Chances are that you forgot to #include <string>. Without it, Microsoft's version of <iostream> (and such) include enough of a declaration of std::string for some things to work, but other parts are missing, so you get strange, seemingly inexplicable failures.
One of the things that's missing is most of the operator overloads for std::string, which is exactly what you seem to be missing.
As an aside, while (myfile.good()) ... is pretty much a guaranteed bug -- you probably want:
while (myfile>>myString)
std::cout << myString << " \n";
Alternatively, you could do the job with a standard algorithm:
#include <string>
#include <algorithm>
#include <iterator>
#include <fstream>
#include <iostream>
int main() {
std::ifstream myfile("input.txt");
std::copy(std::istream_iterator<std::string>(myfile),
std::istream_iterator<std::string>(),
std::ostream_iterator<std::string>(std::cout, " \n"));
return 0;
}
The following compiles fine for me on MSVC 2010:
std::ifstream inputStream;
std::string myString;
inputStream.open("myFile.txt", std::ifstream::in);
while(inputStream.good())
{
inputStream >> myString;
}
Note: without using std::ifstream::in as my open mode, I got the same error as you. I suggest you check what value you have for this parameter.