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vector::front

Visual Studio 2005

Returns a reference to the first element in a vector.

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reference front( ); 
const_reference front( ) const;

Return Value

A reference to the first element in the vector object. If the vector is empty, the return is undefined.

Remarks

If the return value of front is assigned to a const_reference, the vector object cannot be modified. If the return value of front is assigned to a reference, the vector object can be modified.

When compiling with _SECURE_SCL 1, a runtime error will occur if you attempt to access an element in an empty vector. See Checked Iterators for more information.

Example

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// vector_front.cpp
// compile with: /EHsc
#include <vector>
#include <iostream>

int main( )
{
   using namespace std;   
   vector <int> v1;
   
   v1.push_back( 10 );
   v1.push_back( 11 );

   int& i = v1.front( );
   const int& ii = v1.front( );

   cout << "The first integer of v1 is "<< i << endl;
   i++;
   cout << "The second integer of v1 is "<< ii << endl;
}

Output

The first integer of v1 is 10
The second integer of v1 is 11

Requirements

Header: <vector>

Reference

vector Class
vector::front and vector::back
Standard Template Library

Other Resources

vector Members

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FAQ

Juanito's comment is very helpful

History

12/14/2011
Sumith

Confusing sample.

That's a confusing sample -- both vars i and ii refer to the same storage: the first item in the vector. The var i is mutable, and the change to it is then seen via the immutable ii, which also happens to match the content of the 2nd item pushed to the back of the vector. The comment exacerbates this, asserting "by incrementing i, we move the the[sic] front reference to the second element." No, instead the first item in the vector is modified from 10 to 11. Evidence for the doubter can be gathered by changing the push_back(11) to push_back(14), and noting that the program still produces 10 and 11.

Juanito

History

1/8/2011
JuanchoPanza