if -Uaje

cout ob.at(i) cout \n\n ;

Используется функция the clearO. cout Clear ob.\n ; ob.clearO ;

for each(ob.begin.(), ob.endO, display); no effect! cout Size of ob after clear: ob.sizeO

\nBounds: ob. getiowerbound О

to ob.getupperboundO \n\n ;

Создает копию объекта, cout Malce a copy of ob2.\n ; RangeArray<int> оЬ5(оЬ2); for each(ob5.beginO, obS.endO, display); cout \n\n ;

Конструирует новый объект из заданного диапазона.

cout Construct object from a range.\n ;

RangeArray<int> оЬб(&ob2[-2], ob2.end());

cout Size of оЬб: ob6.size() endl; for each(ob6.beginO, ob6.end(), display);

cout endl; retium 0;

Далее приведен вывод программы.

Size of ob is: 11

Initial contents of ob:

00000000000

New values for ob:

-5 -4 -3 -2 -1 0 1 2 3 4 5

Sum of values with negative subscripts is: -15

Copy ob to ob2 using copy О algorithm. Contents of ob2: -5-4-3-2-1012345

Replace values less than zero with zero.

Pt the result into оЬЗ. Ccjotents of оЬЗ: (,0000012345

gp ob and оЬЗ. неге is оЬЗ:

,5-4-3-2-1012345

gp again to restore.

Here is оЬЗ after second swap:

00000012345

Element at ob[01 is 0 insert values into ob.

-5 -4 -99 -3 -2 -1 0 99 1 2 3 4 5 -9999 Element at ob[0] is 0

Insert -7 three times to front of ob.

-7 -7 -7 -5 -4 -99 -3 -2 -1 0 99 1 2 3 4 5 -9999

Element at ob[0] is 0

Push back the value 40 onto ob.

-7 -7 -7 -5 -4 -99 -3 -2 -1 0 99 1 2 3 4 5 -9999 40

fop back two values from ob.

-7 -7 -7 -5 -4 -99 -3 -2 -1 0 99 1 2 3 4 5

Push front the value 19 onto ob.

19 -7 -7 -7 -5 -4 -99 -3 -2 -1 0 99 1 2 3 4 5

Pop front two values from ob.

-7 -7 -5 -4 -99 -3 -2 -1 0 99 1 2 3 4 5

Ob.frontO: -7 Ob.backO: 5

Erase element at 0.

-7 -5 -4 -99 -3 -2 -1 99 1 2 3 4 5 dement at ob[0] is 99

ase many elements in ob.

Insert оЬ4 into ob.

-7 -7 -5 -4 -99 -3 100 101 102 3 4 5 Element at ob[01 is 100

Here is ob shovm with its indices:

[-6]: -7

[-51: -7

[-4]: -5

[-3]: -4

[-21: -99

[-11: -3

[01: 100

[11: 101

[21: 102

[31: 3

[41: 4

[51: 5

Use the atO function.

-66 -55 -44 -33 -22 -11 0 11 22 33 44 55 Clear ob.

Size of ob after clear: 0 Bounds: 0 to 0

Make a copy of ob2.

-5 -4 -3 -2 -1 0 1 2 3 4 5

Construct object from a range. Size of оЬб: 8 -2-1012345

В программе листинга 8.3 демонстрируется применение операций отношения.

7 7 -5 -4 -99 -3345 Element at ob[0] is 3