Swd1316 Tarmoqlanish va uzilishlarni tashkil etish operatorlari. Tarmoqlanuvchi operatorlar(if va switch) va ularni ishlash ketma ketliklari

• MUHAMMAD AL-XORAZMIY NOMIDAGI TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI

• DASTURLASH 1

• SWD1316

• Tarmoqlanish va uzilishlarni tashkil etish operatorlari. Tarmoqlanuvchi operatorlar(if va switch) va ularni ishlash ketma ketliklari.

• 03
• MAVZU

• ISHNIYAZOV ODIL
• OLIMOVICH

• Informatika asoslari kafedrasi katta o’qituvchisi

Mа’ruza rejasi

• Murakkab shart.

• 4

• Tarmoqlanuvchi algoritm

• 1

• Shart opеratori. To`liq tarmoqlanish. To`liq bo’lmagan tarmoqlanish

• 2

• Tanlash opеratori (switch).

• 3

• Xulosa

• 5

• Tarmoqlanuvchi algoritm

• Agar algoritm bajarilish ketma – ketligi bir nechta shartlarga bog’liq bo’lsa u tarmoqlanuvchi deb ataladi.

• Topshiriq. Ikkita butun son kiriting va ulardan kattasini ekranga chiqaring.
• Yechim g’oyasi: Agar birinchi son ikinchi sondan katta bo’lsa, ekranga birinchi son chiqishi kerak, aks holda ekranga ikkinchi son chiqishi kerak.

• 1-variant. Blok-sxema

• Boshlanish

• max = a;

• a, b

• max

• a > b

• max = b;

• Tamom

• ha

• yo’q

• “shart” blogi

• 2-variant. Blok-sxema

• Tarmoqlanishning
• to’liqsiz formasi

• Boshlash

• max = a;

• a, b

• max

• max = b;

• Tamom

• ha

• yo’q

• b > a

Shart operatori

• if ( Shart )
• {
• // agar shart to’g’ri bo’lsa bajariladi
• }
• else
• {
• // agar shart noto’g’ri bo’lsa bajariladi
• }

• O’ziga xosligi:
• Ikkinchi qism (else …) yo’q bo’lishi mumin (To’liqmas forma)
• Agar blokda bitta amal bo’lsa { } belgini olib tashlash mumkin.

Xatolarni toping !

• if a > b {
• a = b;
• }
• else
• b = a;

• if ( a > b ) {
• a = b;
• else
• b = a;

• if ( a > b )
• else
• b = a;

• if ( a > b )
• a = b; c = 2*a;
• else
• b = a;

• }

• (

• )

• }

• {

• a = b;

`

• 1 variant. Dastur (formatli kiritish-chiqarish)

• int main()
• { int a, b, max;
• printf(“Ikkita butun sonni kiriting\n");
• scanf("%d%d", &a, &b );
• if (a > b) {
• }
• else {
• }
• printf(“Katta son %d", max);
• return 0; }

• max = a;

• max = b;

• Shart
• operatorning to’liq ko’rinishi

`

• int main()
• { int a, b, max;
• cout<<"a ni kiriting="; cin>>a;
• cout<<"b ni kiriting="; cin>>b;
• if (a > b) {
• }
• else {
• }
• cout<<"Natija="<
• return 0; }

• max = a;

• max = b;

• Shart
• operatorning to’liq ko’rinishi

• 1 variant. Dastur (oqimli kiritish chiqarish)

• 2 variant. Dastur (formatli kiritish chiqarish)

• int main()
• { int a, b, max;
• printf("a va b sonini kiriting\n");
• scanf("%d%d", &a, &b );
• max = a;
• if (b > a)
• printf(“Katta son %d”, max);
• }

• max = b;

• Shart
• operatoring to’liq ko’rinishi

• 2 variant. Dastur (oqimli kiritish chiqarish)

• int main()
• {
• int a, b, max;
• cout<<"a va b sonini ni kiriting=";
• cin>>a>>b;
• max = a;
• if (b > a)
• cout<<“Katta son ="<
• }

• max = b;

• Shart
• operatoring to’liq ko’rinishi

• Topshiriq

• “1": Uchta sonni kiriting va ular ichidan eng kattasini toping.
• Masalan:
• Uchta sonni kiriting:
• 5 20 9
• Eng katta son 20
• “2": Beshta sonni kiriting va ular ichidan eng kattasini toping.
• Masalan:
• Beshta sonni kiriting :
• 5 27 20 26 24
• Eng katta son 27

Tanlash operatori

• Masala: Oy tartib raqamini kiriting va shu oy kunlarini chiqaring.
• Yechim: Oy bo’yicha kunlar:
• 28 kun – 2 (fevral)
• 30 kun – 4 (aprel), 6 (iyun), 9 (sentabr), 11 (noyabr)
• 31 kun – 1 (yanvar), 3 (mart), 5 (may), 7 (iyul), 8 (avgust), 10 (oktabr), 12 (dekabr)
• Хосса: Oy tartib raqamidan ikkita emas, balki bir nechta farqli bog’lanishlarni tanlash kerak.

• Algoritm. Blok-sxema

• Boshi

• Tanlash operatori

• M ni kirit

• ha

• yo’q

• M = 1

• D = 31;

• yo’q

• M = 2

• D = 28;

• ha

• D ni chop
• etish

• Xato

• Tamom

• A

• A

• B

• B

• M = 12

• D = 31;

• yo’q

• ha

• Birorta ham variant to’g’ri kelmadi

• Masala yechimi. Dastur

• int main()
• { int M, D;
• cout<<“ Oy tartib raqamini kiriting:\n";
• cin>>M;
• if (D > 0)
• cout<<"Bu oyda " <
• else cout<<"oyning raqami noma’lum";
• }

• Masala yechimi. Dastur

• break;

• Switch dan chiqish

• switch ( M ) {
• case 2: D = 28;
• case 4: case 6: case 9:
• case 11:D = 30;
• case 1: case 3: case 5: case 7:
• case 8: case 10: case 12: D = 31;
• default: D = -1;
• }

• break;

• break;

• Masala yechimi. Dastur

• Birorta ham variant to’g’ri kelmadi

Tanlash operatori

• int main()
• {
• char c;
• cout<<"Hayvonlar nomini 1-harfini kiriting:\n";
• c = getch();//conio.h kutubxonasi kerak bo’ladi
• switch ( c ) {
• case 'a': cout<<"Ayiq"; break;
• case 'b': cout<<"Bo’ri"; break;
• case 'k': cout<<"Kiyik"; break;
• default: cout<<"Bilmayman!";
• }
• }

• hamma joydan break ni olib tashlasa nima bo’ladi?

• ?

Tanlash operatori

• Xossalar:
• switch dan keyin o’zgaruvchining nomi bo’lishi mumkin yoki butun tipli arifmetik ifoda bo’ishi mumkin (int, long)

• switch ( i+3 ) {
• case 1: a = b; break;
• case 2: a = c;
• }

Tanlash operatori

• Simvolli (belgili) (char) tip bo’lishi mumkin ikkita bir xil qiymatlarni qo’yish mumkin emas:

• switch ( x ) {
• case 1: a = b; break;
• case 1: a = c;
• }

Murakkab shart

• Murakab shart – bu mantiqiy amallar bilan bog’langan bir nechta oddiy shartlardan (munisabatlardan) tashkil topadi.
• ! - inkor (not, inkor qilish, inversiya)
• && - va (and, mantiqiy ko’paytma kon'yunksiya)
• || - yoki (or, mantiqiy qo’shish, diz'yunksiya)
• Masala. Tashkilot 25 dan 40 yoshgacha bo’lgan dasturchilarni ishga oladi. Nomzodning yoshini kiriting va tashkilotga to’g’ri keladimi (“to’g’ri keladi” yoki “to’g’ri kelmaydi” deb ekrakga chiqaring).

Masalani yechish algoritm blok-sxemasi

• Boshlash

• x

• “to’g’ri keladi”

• tamom

• ha

• yo’q

• x>= 25

• x<= 40

• ha

• “to’g’ri kelmaydi”

• yo’q

Masala dasturi. 1-variant (1-usul)

• int main()
• { int x;
• printf("Yoshni kiriting\n");
• scanf("%d", &x);
• if (x >= 25)
• if (x <= 40)
• printf("to'g'ri keladi");
• else printf("to'g'ri kelmaydi“);
• else
• printf("to'g'ri kelmaydi");
• }

Masala dasturi. 1-variant (2-usul)

• int main()
• { int x;
• cout<<"Yoshni kiriting=";
• cin>>x;
• if (x >= 25)
• if (x <= 40)
• cout<<"to'g'ri keladi";
• else cout<<"to'g'ri kelmaydi";
• else
• cout<<"to'g'ri kelmaydi";
• }

• Boshlash

• x

• x >= 25 va
• x <= 40

• Tamom

• “to’g’ri keladi”

• ha

• yo’q

• “to’g’ri kelmaydi”

• Masalani yechish algoritm blok-sxemasi

Masala dasturi. 2-variant

• int main()
• {
• int x;
• printf("Yoshni kiriting\n");
• scanf("%d", &x);
• if ( x >= 25 && x <= 40 )
• printf("to'g'ri keladi");
• else printf("to'g'ri kelmaydi");
• }

• Murakkab shart

Murakkab shart

• Murakkab shartning bajarilish tartibi:
• Qavs ichidagi ifoda
• ! (inkor, inkor qilish)
• <, <=, >, >=
• ==, !=
• && (va)
• || (yoki)

• Мisol:
• 2 1 6 3 5 4
• if ( !(a > b) || c != d && b == a)
• { ... }

Murakkab shart

• a = 2; b = 3; c = 4 bo’lganda natija rost yoki yolg’on ekanligini aniqlaymiz.
• !(a > b)
• a < b && b < c
• !(a >= b) || c == d
• a < c || b < c && b < a
• a > b || !(b < c)

• 1

• 1

• 0

• 1

• 1

• 14:44

Murakkab shart

• x ning qanday qiymatlarida shart rost bo’ladi:
• x < 6 && x < 10
• x < 6 && x > 10
• x > 6 && x < 10
• x > 6 && x > 10
• x < 6 || x < 10
• x < 6 || x > 10
• x > 6 || x < 10
• x > 6 || x > 10

• (-, 6)

• 

• (6, 10)

• (10, )

• (-, 10)

• (-, 6)  (10,)

• (-, )

• (6, )

• x < 6

• x > 10

• x < 10

• x > 6

Masala. Kiritilgan son musbat yoki manfiy?

• #include
• using namespace std;
• int main()
• { int n; cout << "Butun son kiriting:";
• cin >> n;
• if ( n > 0)
• {cout<<"Siz kiritgan son musbat:"<
• if ( n < 0)
• {cout <<"Siz kiritgan son manfiy:"<
• else
• {cout <<"Siz kiritgan son nol: "<
• return 0;
• }

XULOSA

• Shartga oid masalalarni algoritm va dasturini tuzishda. Masala yechimini ishlab chiqish holatiga qarab
• shartni tekshirish operatori (if () else ) yoki
• tanlash operatori (switch() case )
• dan foydalanish maqsadga muvofiq.

• MUHAMMAD AL-XORAZMIY NOMIDAGI TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI

• ISHNIYAZOV ODIL
• OLIMOVICH

• Informatika asoslari kafedrasi katta o’qituvchisi

• E’TIBORINGIZ UCHUN RAXMAT!