"vector<int>::i terator" problem

[arnuld]

i wrote a programme to create a vector of 5 elements (0 to 4), here is
the code & output:

#include <iostream>
#include <vector>

int main() {
std::vector<int ivec;

// dynamically create a vector
for(std::vector <int>::size_typ e ix = 0; ix <= 4; ++ix)
{
ivec.push_back( ix);
}
// print-out the elements
for(std::vector <int>::iterat or iter = ivec.begin();
iter != ivec.end(); ++iter)
{
std::cout << "Element: " << *iter << "..." << std::endl;
}

// create a middle iterator
std::vector<int >::iterator mid_iter = (ivec.begin() + ivec.size() /
2);
// print it out to check where it points
std::cout << "middle element: " << *mid_iter << std::endl;

return 0;
}

OUTPUT:

Element: 0...
Element: 1...
Element: 2...
Element: 3...
Element: 4...
middle element: 2 // fine :-)
now if i change the "mid_iter" to:

/std::vector<int >::iterator mid_iter = (ivec.size() / 2);/

then i got an ERROR:
----------------------------------------------------------------
unix@debian:~/Desktop$ g++ test.cpp

test.cpp: In function `int main()':
test.cpp:22: error: conversion from `unsigned int' to non-scalar type `
__gnu_cxx::__no rmal_iterator<i nt*, std::vector<int ,
std::allocator< int >'
requested

unix@debian:~/Desktop$
----------------------------------------------------------------
it is showing "non-scalar type". what is that? what is happening
"behind the curtains"?
"arnuld"

[Kai-Uwe Bux]

arnuld wrote:

i wrote a programme to create a vector of 5 elements (0 to 4), here is
the code & output:

#include <iostream>
#include <vector>

int main() {
std::vector<int ivec;

// dynamically create a vector
for(std::vector <int>::size_typ e ix = 0; ix <= 4; ++ix)
{
ivec.push_back( ix);
}
// print-out the elements
for(std::vector <int>::iterat or iter = ivec.begin();
iter != ivec.end(); ++iter)
{
std::cout << "Element: " << *iter << "..." << std::endl;
}

// create a middle iterator
std::vector<int >::iterator mid_iter = (ivec.begin() + ivec.size() /
2);
// print it out to check where it points
std::cout << "middle element: " << *mid_iter << std::endl;

return 0;
}

OUTPUT:

Element: 0...
Element: 1...
Element: 2...
Element: 3...
Element: 4...
middle element: 2 // fine :-)
now if i change the "mid_iter" to:

/std::vector<int >::iterator mid_iter = (ivec.size() / 2);/

then i got an ERROR:
----------------------------------------------------------------
unix@debian:~/Desktop$ g++ test.cpp

test.cpp: In function `int main()':
test.cpp:22: error: conversion from `unsigned int' to non-scalar type `
__gnu_cxx::__no rmal_iterator<i nt*, std::vector<int ,
std::allocator< int >'
requested

unix@debian:~/Desktop$
----------------------------------------------------------------
it is showing "non-scalar type". what is that? what is happening
"behind the curtains"?

Well, look at the statement:

std::vector<int >::iterator mid_iter = (ivec.size() / 2);

The variable mid_iter has type std::vector<int >::iterator. Note that
std::vector<int >::iterator, according to the standard, is an implementation
defined type. What it is or is not 'behind the curtains' is of no concern
to the programmer.

Now, the expression (ivec.size() / 2) has type std::vector<int >::size_type.
The compiler is telling you that you cannot initialize an iterator from a
size_type. The compiler it correct: the interface of iterator does not
contain such a conversion (for good reasons!).

Best

Kai-Uwe Bux

[yoge]

Kai-Uwe Bux wrote:

Well, look at the statement:

std::vector<int >::iterator mid_iter = (ivec.size() / 2);

The variable mid_iter has type std::vector<int >::iterator. Note that
std::vector<int >::iterator, according to the standard, is an implementation
defined type. What it is or is not 'behind the curtains' is of no concern
to the programmer.

ok, no concern :-)

Now, the expression (ivec.size() / 2) has type std::vector<int >::size_type.
The compiler is telling you that you cannot initialize an iterator from a
size_type. The compiler it correct: the interface of iterator does not
contain such a conversion (for good reasons!).

thanks Kai.

"arnuld"
www.arnuld.blogspot.com

[Ron Natalie]

Kai-Uwe Bux wrote:

arnuld wrote:
std::vector<int >::iterator mid_iter = (ivec.size() / 2);

The variable mid_iter has type std::vector<int >::iterator. Note that
std::vector<int >::iterator, according to the standard, is an implementation
defined type. What it is or is not 'behind the curtains' is of no concern
to the programmer.
=

mid_iter = ivec.begin() + (ivec.size() / 2)

would however do what the OP was trying to accomplish.

[Kai-Uwe Bux]

Ron Natalie wrote:

Kai-Uwe Bux wrote:

>arnuld wrote:
std::vector<int >::iterator mid_iter = (ivec.size() / 2);

The variable mid_iter has type std::vector<int >::iterator. Note that
std::vector<in t>::iterator, according to the standard, is an
implementati on defined type. What it is or is not 'behind the curtains'
is of no concern to the programmer.
=

mid_iter = ivec.begin() + (ivec.size() / 2)

would however do what the OP was trying to accomplish.

True, but the OP already knew that [from the OP as quoted in my previous
posting]:

arnuld wrote:

i wrote a programme to create a vector of 5 elements (0 to 4), here is
the code & output:

[snip]

// create a middle iterator
std::vector<int >::iterator mid_iter = (ivec.begin() + ivec.size()/2);

[snip]

now if i change the "mid_iter" to:

/std::vector<int >::iterator mid_iter = (ivec.size() / 2);/

then i got an ERROR:

Best

Kai-Uwe Bux