bool isPresent=false;
vector<T>::iterator p;
In the code above the compiler does not know that vector<T>::iterator is a type
unless you use typename.
Example:
template <typename T>
bool isNumberInVector( vector<T>&v, T numToCheck)
{
bool isPresent=false;
typedef typename vector<T>::iterator Iter;
Iter p;
for(p=v.begin( );p!=v.end( );p++)
if((*p)==numToCheck ) isPresent=true;
return isPresent;
}
That should work.
Best Regards,
Jim Smith
E-mail: [email protected]
--- On Tue, 7/7/09, wolexzo2 <[email protected]> wrote:
From: wolexzo2 <[email protected]>
Subject: [c-prog] problem with vector iterator
To: [email protected]
Date: Tuesday, July 7, 2009, 12:28 AM
Below is a template of a function isNumberInVector.
I get errors about the iterator when I run this program.
errors such as
error: expected `;' before `p'
`p' was not declared in this scope
error: dependent-name `std::vector: :iterator' is parsed as a non-type, but
instantiation yields a type
I don't get error when I simply make the function non-template and use
vector<int> instead.
So What could be the cause?
Thanks,
Lawal. O
#include <iostream>
#include <string>
#include<cstdlib>
#include<vector>
using namespace std;
template <typename T>
bool isNumberInVector( vector<T>&v, T numToCheck)
{
bool isPresent=false;
vector<T>::iterator p;
for(p=v.begin( );p!=v.end( );p++)
if((*p)==numToCheck ) isPresent=true;
return isPresent;
}
int main(){
vector<int> vect;
for(int i=0;i<20;i++ )
if(i%2==0)
vect.push_back( i);
int number=4;
if(isNumberInVector (vect,number) )cout<<"Yes It is";
else cout<<"No it isn't";
}
[Non-text portions of this message have been removed]
