PROGRAMOWANIE I J
ZYK C++ / ANALIZA PROGRAMÓW NR 2
Temat: Zmienne typu wskaz
nikowego i referencyjnego - arytmetyka wskaz
ników
Zadanie
Wykonać analizę poniższych programów. Nie używając komputera obliczyć wartości zmiennych określonych
jako OUTPUT przy danych wejściowych INPUT. Sprawdzić wyniki uruchamiając odpowiedni kod programu
na komputerze.
Cześć 1: Struktury.
1.
// INPUT : wspolrzedne srodka okregu o2, (x,y) = (2.5, 3.0)
// promien okregu o2, r = 6
// OUTPUT : odl, L1, S2
const double pi = 3.14;
struct okrag
{
double x,y;
int r;
} o1 = {1.5, 0.5, 5};
okrag o2;
cout << "Podaj wspolrzedne srodka oraz promien drugiego okregu ..." << endl;
cout << "x = "; cin >> o2.x;
cout << "y = "; cin >> o2.y;
cout << "r = "; cin >> o2.r;
double odl = sqrt( (o1.x-o2.x)*(o1.x-o2.x) + pow(o1.y-o2.y,2) );
double L1,S2;
L1 = 2.0*pi*o1.r;
S2 = pi*o2.r*o2.r;
2.
// INPUT : -
// OUTPUT : suma.re, suma.im, iloczyn.re, iloczyn.im
struct liczba_zespolona
{
double re;
double im;
} suma, iloczyn;
liczba_zespolona z,w = {2.0,-3.0};
z.re = -5.0;
z.im = 4.0;
// Obliczamy sume liczb zespolonych : z + w
suma.re = z.re + w.re;
suma.im = z.im + w.im;
// Obliczamy iloczyn liczb zespolonych : z * w
iloczyn.re = z.re*w.re - z.im*w.im;
iloczyn.im = z.im*w.re + z.re*w.im;
Cześć 2: Wskaz
niki.
1.
// INPUT : adres zmiennej A = 1000, adres zmiennej B = 2000,
// adres zmiennej c = 3000
// OUTPUT : podac wartosci, ktore wypisze program
int A,B,C=25;
int * pA = &A, * pB = 0, * pC = 0;
A = 5; B = 12;
cout << "A = " << A << endl;
cout << "B = " << B << endl;
cout << "C = " << C << endl << endl;
cout << "pA = " << pA << endl;
cout << "pB = " << pB << endl;
cout << "pC = " << pC << endl << endl;
B = A + 10;
pB = &B;
cout << "B = " << B << endl;
cout << "pB = " << pB << endl << endl;
C = *pA + *pB;
pC = &C;
cout << "C = " << C << endl;
cout << "pC = " << pC << endl << endl;
2.
// INPUT : adres zmiennej a = 1200, adres zmiennej b = 3000,
// adres zmiennej c = 2000
// OUTPUT : podac wartosci, ktore wypisze program
int a,b,c;
int * ptr1 = &a, * ptr2 = &b;
a = 3; b = 10;
c = (a++) - (--b);
cout << "a = " << a << endl;
cout << "b = " << b << endl;
cout << "c = " << c << endl << endl;
cout << "*ptr1 = " << *ptr1 << endl;
cout << "*ptr2 = " << *ptr2 << endl << endl;
*ptr1 = c + 20;
ptr2 = &c;
cout << "ptr1 = " << ptr1 << endl;
cout << "*ptr1 = " << *ptr1 << endl << endl;
cout << "c = " << c << endl;
cout << "*ptr2 = " << *ptr2 << endl;
Cześć 3: Referencje.
1.
// INPUT : -
// OUTPUT : podac wartosci, ktore wypisze program
int x = 4, y = 7;
int &ref_x = x;
int &ref_y = y;
cout << "x = " << x << endl;
cout << "y = " << y << endl;
cout << "ref_x = " << ref_x << endl;
cout << "ref_y = " << ref_y << endl << endl;
ref_x = (++y) - 10;
cout << "x = " << x << endl;
cout << "ref_y = " << ref_y << endl << endl;
2.
// INPUT : -
// OUTPUT : podac wartosci, ktore wypisze program
int x = 4, y, z = 12;
int &ref_x = x;
int &ref_z = z;
cout << "x = " << x << endl;
cout << "z = " << z << endl;
cout << "ref_x = " << ref_x << endl;
cout << "ref_z = " << ref_z << endl << endl;
int &ref = y;
ref = (ref_x < z) ? 3:30;
cout << "y = " << y << endl;
cout << "ref = " << ref << endl << endl;
ref_x = (--z) - (ref++);
cout << "x = " << x << endl;
cout << "y = " << y << endl;
cout << "z = " << z << endl;
cout << "ref_x = " << ref_x << endl;
cout << "ref = " << ref << endl;
cout << "ref_z = " << ref_z << endl << endl;
Cześć 4: Arytmetyka wskaz
ników.
1.
// INPUT : adres pierwszego elementu tablicy jest rowny 1000
// OUTPUT : podac wartosci, ktore wypisze program
int A[6] = {2, 4, 6, 8, 10, 12};
int * pA0, * pA1;
int diff;
pA0 = &A[0];
pA1 = A;
cout << "A[0] = " << A[0] << endl;
cout << "pA0 = " << pA0 << endl;
cout << "pA1 = " << pA1 << endl << endl;
pA0 = pA0++;
pA1 = pA1 + 4;
cout << "pA0 = " << pA0 << endl;
cout << "pA1 = " << pA1 << endl << endl;
diff = pA1 - pA0;
cout << "pA1 - pA0 = " << diff << endl;
2.
// INPUT : adres pierwszego elementu tablicy jest rowny 2000
// OUTPUT : podac wartosci, ktore wypisze program
double A[6] = {2, 4, 6, 8, 10, 12};
double * ptrA;
ptrA = &A[0];
// Tablica w wersji pierwszej
for (int i=0; i<=5; i++)
cout << "A[" << i << "] = " << A[i] << endl;
cout << endl;
// I - Operacje na tablicy A
for (int i=0; i<=5; i++)
*(A + i) = A[i] + 5;
for (int i=0; i<=5; i++)
cout << "A[" << i << "] = " << A[i] << endl;
cout << endl;
// II - Operacje na tablicy A
for (int i=0; i<=5; i++)
ptrA[i] = A[i] + 10;
for (int i=0; i<=5; i++)
cout << "A[" << i << "] = " << A[i] << endl;
cout << endl;
// III - Operacje na tablicy A
for (int i=0; i<=5; i++)
*(ptrA + i) = A[i] - 1;
for (int i=0; i<=5; i++)
cout << "A[" << i << "] = " << A[i] << endl;
cout << endl;
3.
// INPUT : adres pierwszego elementu tablicy jest rowny 1500
// OUTPUT : podac wartosci, ktore wypisze program
double A[6] = {2, 4, 6, 8, 10, 12};
double * ptr1, * ptr2, * ptr3;
int diff1,diff2;
ptr1 = A;
ptr2 = &A[2];
ptr3 = &A[5];
cout << "ptr1 = " << ptr1 << endl;
cout << "ptr2 = " << ptr2 << endl;
cout << "ptr3 = " << ptr3 << endl << endl;
cout << "*ptr1 = " << *ptr1 << endl;
cout << "*ptr2 = " << *ptr2 << endl;
cout << "*ptr3 = " << *ptr3 << endl << endl;
for (int i=3; i<=5; i++)
*(ptr1 + i) = A[i] + 10;
for (int i=3; i<=5; i++)
cout << "*(ptr1 + " << i << ") = " << *(ptr1 + i) << endl;
cout << endl;
for (int i=0; i<=5; i++)
cout << "A[" << i << "] = " << A[i] << endl;
cout << endl;
ptr3 = ptr3 - 1;
diff1 = ptr3 - ptr1;
cout << "ptr3 = " << ptr3 << endl;
cout << "*ptr3 = " << *ptr3 << endl;
cout << "diff1 = ptr3 - ptr1 = " << diff1 << endl << endl;
ptr2 = ptr2 + 3;
diff2 = ptr2 - ptr1;
cout << "ptr2 = " << ptr2 << endl;
cout << "*ptr2 = " << *ptr2 << endl;
cout << "diff2 = ptr2 - ptr1 = " << diff2 << endl << endl;