I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
I have created such an array as followed:
==================================
for(i=0;i<=((size/2)1);i++)
{
if(size%2 != 0)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
}
else
{
t = a[i];
a[i] = a[(size/2)+i];
a[(size/2)+i] = t;
}
}
==================================
NOW, the problem is I want to create a function that does the same
thing, but I want to use a Nested For loop to do it as such without
using IF statements to handle what to do when size is odd or not.
I came up with the following code:
==================================
void reverseHalves(int a[], int size)
{
int i, t, j=0;
for(i=0;i<=((size/2)1);i++)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
for(j=(size(size/2));j<=(size/2)1;j++)
{
t = a[j];
a[j] = a[(size/2)+j1];
a[(size/2)+j1] = t;
}
}
==================================
The problem is the second nestedforloop version I have come up with
doesn't work. It only works for odd numbers. When I enter an even
number, I always end up getting a 24214214 or something like that for
one of the numbers.
It alllllmost works. I have been at this for about three hours and
almost at the end of my rope. Can someone PLEASE FOR THE LOVE OF GOD
GIVE ME SOME HELP BEFORE I BLOW MY BRAINS OUT?!?!!
Thanks for listen. 8 2172
<ch**********@yahoo.comwrote in message
news:11**********************@22g2000hsm.googlegro ups.com...
>I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
I have created such an array as followed:
==================================
for(i=0;i<=((size/2)1);i++)
{
if(size%2 != 0)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
}
else
{
t = a[i];
a[i] = a[(size/2)+i];
a[(size/2)+i] = t;
}
}
==================================
NOW, the problem is I want to create a function that does the same
thing, but I want to use a Nested For loop to do it as such without
using IF statements to handle what to do when size is odd or not.
I came up with the following code:
==================================
void reverseHalves(int a[], int size)
{
int i, t, j=0;
for(i=0;i<=((size/2)1);i++)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
for(j=(size(size/2));j<=(size/2)1;j++)
{
t = a[j];
a[j] = a[(size/2)+j1];
a[(size/2)+j1] = t;
}
}
==================================
The problem is the second nestedforloop version I have come up with
doesn't work. It only works for odd numbers. When I enter an even
number, I always end up getting a 24214214 or something like that for
one of the numbers.
It alllllmost works. I have been at this for about three hours and
almost at the end of my rope. Can someone PLEASE FOR THE LOVE OF GOD
GIVE ME SOME HELP BEFORE I BLOW MY BRAINS OUT?!?!!
This looks like homework, but you have shown a lot of effort. Look at this
code, although you'll have to fix one bug when the arrays are odd sized:
#include <iostream>
void reverseHalves(int a[], int size)
{
int half = size / 2;
bool odd = size % 2 == 1;
int j = half;
if ( odd )
j++;
for ( int i = 0; i < half; ++i )
{
std::swap( a[i], a[j++] );
}
}
int main()
{
int Array[] = { 1, 2, 3, 4, 5, 6 };
reverseHalves( Array, sizeof( Array ) / sizeof( Array[0] ) );
for ( int i = 0; i < sizeof( Array ) / sizeof( Array[0] ); ++i )
std::cout << Array[i] << " ";
std::cout << "\n";
int Array2[] = { 1, 2, 3, 4, 5, 6, 7 };
reverseHalves( Array2, sizeof( Array2 ) / sizeof( Array2[0] ) );
for ( int i = 0; i < sizeof( Array2 ) / sizeof( Array2[0] ); ++i )
std::cout << Array2[i] << " ";
std::cout << "\n";
}
Output:
4 5 6 1 2 3
5 6 7 4 1 2 3
The output is close to what you want except for odd, where the question
becomes... do you want
4567123
or
5671234
Fix the code to express what you want.
On Sep 5, 8:19 pm, "Jim Langston" <tazmas...@rocketmail.comwrote:
<cheetahcl...@yahoo.comwrote in message
news:11**********************@22g2000hsm.googlegro ups.com...
I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
I have created such an array as followed:
==================================
for(i=0;i<=((size/2)1);i++)
{
if(size%2 != 0)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
}
else
{
t = a[i];
a[i] = a[(size/2)+i];
a[(size/2)+i] = t;
}
}
==================================
NOW, the problem is I want to create a function that does the same
thing, but I want to use a Nested For loop to do it as such without
using IF statements to handle what to do when size is odd or not.
I came up with the following code:
==================================
void reverseHalves(int a[], int size)
{
int i, t, j=0;
for(i=0;i<=((size/2)1);i++)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
for(j=(size(size/2));j<=(size/2)1;j++)
{
t = a[j];
a[j] = a[(size/2)+j1];
a[(size/2)+j1] = t;
}
}
==================================
The problem is the second nestedforloop version I have come up with
doesn't work. It only works for odd numbers. When I enter an even
number, I always end up getting a 24214214 or something like that for
one of the numbers.
It alllllmost works. I have been at this for about three hours and
almost at the end of my rope. Can someone PLEASE FOR THE LOVE OF GOD
GIVE ME SOME HELP BEFORE I BLOW MY BRAINS OUT?!?!!
This looks like homework, but you have shown a lot of effort. Look at this
code, although you'll have to fix one bug when the arrays are odd sized:
#include <iostream>
void reverseHalves(int a[], int size)
{
int half = size / 2;
bool odd = size % 2 == 1;
int j = half;
if ( odd )
j++;
for ( int i = 0; i < half; ++i )
{
std::swap( a[i], a[j++] );
}
}
int main()
{
int Array[] = { 1, 2, 3, 4, 5, 6 };
reverseHalves( Array, sizeof( Array ) / sizeof( Array[0] ) );
for ( int i = 0; i < sizeof( Array ) / sizeof( Array[0] ); ++i )
std::cout << Array[i] << " ";
std::cout << "\n";
int Array2[] = { 1, 2, 3, 4, 5, 6, 7 };
reverseHalves( Array2, sizeof( Array2 ) / sizeof( Array2[0] ) );
for ( int i = 0; i < sizeof( Array2 ) / sizeof( Array2[0] ); ++i )
std::cout << Array2[i] << " ";
std::cout << "\n";
}
Output:
4 5 6 1 2 3
5 6 7 4 1 2 3
The output is close to what you want except for odd, where the question
becomes... do you want
4567123
or
5671234
Fix the code to express what you want.
Thanks for the help. I'm starting to put together a picture of
something here. I was just wondering, how does the swap function work?
If I wanted to exclude it from the ForLoop and use something like:
t=a[i];
a[i] = a[j];
a[j] = t;
j++;
where t is a temporary variable, would that be the same?
In article <11**********************@22g2000hsm.googlegroups. com>, ch**********@yahoo.com says...
I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
Already posted on c.l.c++.m, but since you seem a bit stressed and this
will get to you a bit sooner:
#include <algorithm>
void swap_halves(int *a, size_t size) {
int pivot = size/2;
std::swap_ranges(a, a+pivot, a+pivot+size%2);
}

Later,
Jerry.
The universe is a figment of its own imagination.
On 9 6 , 7 50 , cheetahcl...@yahoo.com wrote:
I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
I have created such an array as followed:
==================================
for(i=0;i<=((size/2)1);i++)
{
if(size%2 != 0)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
}
else
{
t = a[i];
a[i] = a[(size/2)+i];
a[(size/2)+i] = t;
}
}
==================================
NOW, the problem is I want to create a function that does the same
thing, but I want to use a Nested For loop to do it as such without
using IF statements to handle what to do when size is odd or not.
I came up with the following code:
==================================
void reverseHalves(int a[], int size)
{
int i, t, j=0;
for(i=0;i<=((size/2)1);i++)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
for(j=(size(size/2));j<=(size/2)1;j++)
{
t = a[j];
a[j] = a[(size/2)+j1];
a[(size/2)+j1] = t;
}
}
==================================
The problem is the second nestedforloop version I have come up with
doesn't work. It only works for odd numbers. When I enter an even
number, I always end up getting a 24214214 or something like that for
one of the numbers.
It alllllmost works. I have been at this for about three hours and
almost at the end of my rope. Can someone PLEASE FOR THE LOVE OF GOD
GIVE ME SOME HELP BEFORE I BLOW MY BRAINS OUT?!?!!
Thanks for listen.
I have a silly idea, what about to use a stack to make it? I think you
understand what I mean.
<ch**********@yahoo.comwrote in message
news:11**********************@w3g2000hsg.googlegro ups.com...
On Sep 5, 8:19 pm, "Jim Langston" <tazmas...@rocketmail.comwrote:
><cheetahcl...@yahoo.comwrote in message
news:11**********************@22g2000hsm.googlegr oups.com...
>I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
I have created such an array as followed:
==================================
for(i=0;i<=((size/2)1);i++)
{
if(size%2 != 0)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
}
else
{
t = a[i];
a[i] = a[(size/2)+i];
a[(size/2)+i] = t;
}
}
==================================
NOW, the problem is I want to create a function that does the same
thing, but I want to use a Nested For loop to do it as such without
using IF statements to handle what to do when size is odd or not.
I came up with the following code:
==================================
void reverseHalves(int a[], int size)
{
int i, t, j=0;
for(i=0;i<=((size/2)1);i++)
{
t = a[i];
a[i] = a[(size/2)+i+1];
a[(size/2)+i+1] = t;
for(j=(size(size/2));j<=(size/2)1;j++)
{
t = a[j];
a[j] = a[(size/2)+j1];
a[(size/2)+j1] = t;
}
}
==================================
The problem is the second nestedforloop version I have come up with
doesn't work. It only works for odd numbers. When I enter an even
number, I always end up getting a 24214214 or something like that for
one of the numbers.
It alllllmost works. I have been at this for about three hours and
almost at the end of my rope. Can someone PLEASE FOR THE LOVE OF GOD
GIVE ME SOME HELP BEFORE I BLOW MY BRAINS OUT?!?!!
This looks like homework, but you have shown a lot of effort. Look at this code, although you'll have to fix one bug when the arrays are odd sized:
#include <iostream>
void reverseHalves(int a[], int size) { int half = size / 2; bool odd = size % 2 == 1;
int j = half; if ( odd ) j++;
for ( int i = 0; i < half; ++i ) { std::swap( a[i], a[j++] ); }
}
int main() { int Array[] = { 1, 2, 3, 4, 5, 6 }; reverseHalves( Array, sizeof( Array ) / sizeof( Array[0] ) ); for ( int i = 0; i < sizeof( Array ) / sizeof( Array[0] ); ++i ) std::cout << Array[i] << " "; std::cout << "\n";
int Array2[] = { 1, 2, 3, 4, 5, 6, 7 }; reverseHalves( Array2, sizeof( Array2 ) / sizeof( Array2[0] ) ); for ( int i = 0; i < sizeof( Array2 ) / sizeof( Array2[0] ); ++i ) std::cout << Array2[i] << " "; std::cout << "\n";
}
Output: 4 5 6 1 2 3 5 6 7 4 1 2 3
The output is close to what you want except for odd, where the question becomes... do you want 4567123 or 5671234
Fix the code to express what you want.
Thanks for the help. I'm starting to put together a picture of
something here. I was just wondering, how does the swap function work?
If I wanted to exclude it from the ForLoop and use something like:
t=a[i];
a[i] = a[j];
a[j] = t;
j++;
where t is a temporary variable, would that be the same?
Yes. Swapping two variables is a very common programming in task, where you
save one variable to a temp, assign the other to it, then save the temp to
the second. It is so common, in fact, that it's in the stl. Realistically,
it's probably supposed to be std::swap and I'm not even sure what header
it's supposed to be in. Whichever header, apparently <iostreamincludes
it.
On Sep 6, 3:03 am, Jerry Coffin <jcof...@taeus.comwrote:
In article <1189036219.364561.317...@22g2000hsm.googlegroups. com>,
cheetahcl...@yahoo.com says...
I am trying to create a function that reverses the halves of an
integer array. For example. if the array contains 1,2,3,4  after it
has gone to the function I constructed, the array will look like
3,4,1,2. In the case the array contains an odd amount of elements e.g.
1,2,3,4,5, it will look like 4,5,3,1,2.
Already posted on c.l.c++.m, but since you seem a bit stressed and this
will get to you a bit sooner:
#include <algorithm>
void swap_halves(int *a, size_t size) {
int pivot = size/2;
std::swap_ranges(a, a+pivot, a+pivot+size%2);
}
Interesting. Yet another standard function which I didn't know
about. Using std::vector (of course), something like:
std::swap_ranges( v.begin(),
v.begin() + v.size() / 2,
v.end  v.size() / 2 ) ;
should do the trick.
My first reaction, not knowing of this function, was to ask why
he didn't use the classical solution:
std::reverse( v.begin(), v.end() ) ;
std::reverse( v.begin(), v.begin() + v.size() / 2 ) ;
std::reverse( v.end()  v.size() / 2, v.end() ) ;
And of course, std::rotate can also be used, although it would
require an extra test:
std::rotate( v.begin(), v.begin() + v.size() / 2, v.end() ) ;
if ( v.size() % 2 != 0 ) {
std::rotate( v.begin(), v.begin() + 1, v.end()  v.size() /
2 ) ;
}

James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.Cyrl'École, France, +33 (0)1 30 23 00 34
In article <11**********************@g4g2000hsf.googlegroups. com>, ja*********@gmail.com says...
On Sep 6, 3:03 am, Jerry Coffin <jcof...@taeus.comwrote:
[ ... ]
std::swap_ranges(a, a+pivot, a+pivot+size%2);
}
Interesting. Yet another standard function which I didn't know
about.
Quite honestly, I didn't either until this came up. What I orginally
thought of (and was looking for) was whether there was a version of
remove that took a pair of iterators to specify what was to be removed.
It doesn't exist, but opening the reference for algorithm, I noticed
swap_ranges, which sounded quite promising...
Using std::vector (of course), something like:
std::swap_ranges( v.begin(),
v.begin() + v.size() / 2,
v.end  v.size() / 2 ) ;
You meant 'v.end()' of course.
should do the trick.
Yup. Roughly the same idea would work here, using a+sizesize/2 instead
of a+size+size%2 [or, equivalently, a+size+(size&1)].
My first reaction, not knowing of this function, was to ask why
he didn't use the classical solution:
[ ... reversals ]
And of course, std::rotate can also be used, although it would
require an extra test:
Right  but at least in this case, the extra test is only executed
once, instead of inside a loop.

Later,
Jerry.
The universe is a figment of its own imagination.
Jim Langston wrote in message...
>
Yes. Swapping two variables is a very common programming in task, where
you
save one variable to a temp, assign the other to it, then save the temp to
the second. It is so common, in fact, that it's in the stl.
Realistically,
it's probably supposed to be std::swap and I'm not even sure what header
it's supposed to be in. Whichever header, apparently <iostreamincludes
it.
#include <algorithm>

Bob R
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