A couple of questions

This is a discussion on A couple of questions within the C++ Programming forums, part of the General Programming Boards category; I am learning the STL and have some questions. Code: #include <iostream> #include <vector> #include <string> using namespace std; int ...

  1. #1
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    A couple of questions

    I am learning the STL and have some questions.
    Code:
    #include <iostream>
    #include <vector>
    #include <string>
    using namespace std;
    
    int main()
    {
    	vector<string> vec;
    	vec.push_back("test");
    
    	string s(vec[0]);	//turn vec[0] into a string named s
    	std::string c( s.size(), char() );
    	cout << c << endl;
    	copy( s.begin(), s.end(), c.begin() );
    	cout << c << endl;
    
    	
    	return 0;
    }
    1. In the first cout, why doesn't c print out?
    2. If I eliminate std::string and define string c above, this doesn't work. I get an error: no match for call to ‘(std::string) (size_t, char)’. What is it that std::string is doing in this case?

  2. #2
    The larch
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    Because at first c contains only characters '\0' which are non-printable.

    Well, for me it outputs spaces, which you could see when you printed something around it:

    Code:
    cout << "'" << c << "'" << endl;
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  3. #3
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    Because at first c contains only characters '\0' which are non-printable.
    So is this similar to how C puts '\0' at the end of an array only at the beginning of a string? And why do I need std::string in front of c( s.size(), char() );? If I declare c as a string and remove std::string, I get an error.

  4. #4
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by swappo
    And why do I need std::string in front of c( s.size(), char() );? If I declare c as a string and remove std::string, I get an error.
    I am not sure what you mean. Please post the smallest and simplest program that demonstrates that error.
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  5. #5
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    Here is the code with the changes as compared to the original in the first post.
    Code:
    #include <iostream>
    #include <vector>
    #include <string>
    using namespace std;
    
    int main()
    {
    	vector<string> vec;
    	vec.push_back("test");
    	string c;//added
    
    	string s(vec[0]);	
    	c( s.size(), char() );//deleted std::string
    	cout << c << endl;
    	copy( s.begin(), s.end(), c.begin() );
    	cout << c << endl;
    
      	
    	return 0;
    }
    Code:
    string_copy1.cpp: In function ‘int main()’:
    string_copy1.cpp:13: error: no match for call to ‘(std::string) (size_t, char)’

    Also, If I change some of the code to the following:
    Code:
    std::string c( s.size(), char() );
    	for(int i = 0; i < c.size(); i++)
    		cout << c[i];
    	cout << endl << c.size();
    I find that, cout << c[i];, produces all spaces yet c.size() prints out the number 4. Is c( s.size(), char() ) an integer? I hope I've explained this well enough. I am really confused on this.

  6. #6
    and the Hat of Guessing tabstop's Avatar
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    You can only call the constructor when you define the variable. Just like you had
    Code:
    std::string c( s.size(), char() );
    you can have
    Code:
    string c( s.size(), char() );

  7. #7
    C++まいる!Cをこわせ! Elysia's Avatar
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    The former calls the constructor and the latter calls operator () (of which there is no defined for std::string).
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

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