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up vote 14 down vote favorite
1

I referred to http://en.cppreference.com/w/cpp/language/typeid to write a code which does different things for different types.

Code is as below and explanation is given in the comments 

#include <iostream>
#include <typeinfo>

using namespace std;

template <typename T>
void test_template(const T &t) 
{
    if (typeid(t) == typeid(double))
        cout <<"double\n";
    if (typeid(t) == typeid(string))
        cout <<"string\n";
    if (typeid(t) == typeid(int))
        cout <<"int\n";
}
int main()
{
    auto a = -1; 
    string str = "ok"; 
    test_template(a); // prints int
    test_template("Helloworld"); // Does not print string
    test_template(str); // prints string
    test_template(10.00); // prints double

    return 0;
}

I am wondering why test_template(str) prints "string" whereas test_template("Helloworld") does not. BTW my g++ version is g++ (Ubuntu 5.4.0-6ubuntu1~16.04.4) 5.4.0 20160609

share|improve this question
11  
Why the downvotes? Sure, it's at a basic level, but the question is complete, clear, and OP even includes a compiler version! – TartanLlama 7 hours ago
1  
"Helloworld" is not a std::string. But "Helloworld"s would be. – Jarod42 6 hours ago
up vote 13 down vote accepted

In this call

test_template("Helloworld"); // Does not print string

the argument "Helloworld" is a string literal that has type const char[11].

Because the function parameter is a referenced type

void test_template(const T &t) 
                          ^^^

then within the function the argument (more precisely the parameter) has the type const char ( &t )[11].

String literals in C++ have types of constant character arrays with the number of elements equal to the number of characters in string literal plus the terminating zero.

In this call

test_template(str);

the argument has type std::string because the variable str is declared like

string str = "ok";
^^^^^^

It was initialized by the string literal "ok" nevertheless the object itself is of the type std::string.

share|improve this answer
1  
Should conat char be const char? (I can't submit an edit with just one change.) – Christophe Strobbe 5 hours ago
    
@ChristopheStrobbe What do you mean? – Vlad from Moscow 5 hours ago
    
Well, conat in conat char ( &t )[11] in your answer. – Christophe Strobbe 4 hours ago
1  
I mean const is a type qualifier. But what is conat? A typo or something else? – Christophe Strobbe 4 hours ago
1  
@ChristopheStrobbe Oh, I am sorry. It is a typo.:) – Vlad from Moscow 4 hours ago
up vote 14 down vote

String literals in C++ are of type const char[N+1], where N is the number of characters in the string. std::string is a standard library class which owns a string and provides a number of operations over it. A std::string can be constructed from a const char[N], but they are not the same thing.

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up vote 7 down vote

String literals like "Helloworld" are constants arrays of characters.

The std::string class have a constructor that can take pointers to string literals, but a string literal is in itself not a std::string object.

As a side-note, using a function like your is considered a code-smell and bad design. Use overloaded functions taking different arguments instead. That will also solve your problem with the strings.

share|improve this answer
1  
Agreed, there's no point in having a template at all here. There's no common functionality. – AndyG 7 hours ago
1  
Thanks! I will consider using overloaded functions instead. – Meehatpa 7 hours ago

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