Question

Part #2 Trace and show your steps and the output of the following code segment const int TEN = 10; int a(TEN), n = 354, num, c = 0; num = n; a[C++] = num % TEN; num = num/TEN; } while (num); cout << "The number" <<n<<" is now: for (int k = -1;k >= 0; k--) cout << a[k]; cout << endl; Trace and show the content of the arrays A and B int A[7] = {66, 55, 77, 55, 88, 77, 66}; A int B[7]; int num = 55; int k = 0; for (int j = 0; j < 6; j++) if (AU) > num) B[k++] = AD]; AC) = 0;

Answer 1

1st question answer

chang of Nalu of Voiables Line (Means Line Trainy 10 nt TEN= TEN, aTEN n num c Const n354,num,c=0 araTENJ aiEN] That is all ha vaa Valus of asignsd to dafault value o's "aCioJ=f0,0,0,0,0,0,0,0,0,0 a is n-354, Vatiabl of n haue he vabe tham 35y Il num is daclard and dalault valu o is Ilc=o the fnteger variable fhalged ho He alus os manually num= n rf actt =nun .TEN alcJ num /7EN artt means Combinahon af C Ctl; =num/TEN arhum/.TEN alo- 354 10 us Remain den oken dewidy with ID C=Ct1 C OT1 Ce num hum/TEN 35/10 wnu num 35 whila (35 It loop Continus until o' elament encounlens

atct) num % TEN 5 Caut es akkJ a1 =num/ TEN aG 35 10 a0-5 k 1-10 20 k=k-1; Czlt k>-0 00oop itus krminals CE 2 nume num/TEN aE 35/10 *ls a2J 3 elanont Dwhil (s loop Coninus becaune of hon-gero alcrt3-num TEN aEJ 3/10 aJ-3 are- hum1-TEN C=Ctt E2+ C 3 num num/rEN fhum 3/10 12 num-D ohile (o) loop lenminala beaune o eimnt enlaunk n is now (M Cout < The number for (int k-c-i kseo; k--) K C- 1 couk afk] k tk-1 k- 14 k so oTrue (oop Contus

in simple steps

TEN, aTEN, n, num to 0000, 35 0 1o fo,000,0,0,35 3 loop started 1o 4,0,0,0,0 3 3y 1 -0,01010,03 35 flerations Complated- first loop lo fs00,03 2 35 0,0,0,0,0 0,O,0,0,0 Second loop lerahon Complekd- 10 0,02 3 f2%0.0.0.035y Lo0,010,0 3 loop terminatd he number 354 is now.

TEN ahEN C n num k O-I0 false leminakd loop for loop shted I6 from step aften now a] d is prihed Cout endend w new Gn qkJ- afs af1 3 The number 35y is now 3 17 lo,0,0,0,0 3 first terahon Completed 12 fron step after Senlenta a valu is pr ie 5 The number 3ry is now 3 Second tenatien Completid S4S3,0,02 3 20 0 frem Sep aPer e senlen ca a fo) value s priked. te, 4 number 3ry is now 3S4 k-k-t, R6-1 k-

program and output

.Cpp #include <iostream> 2 using namespace std; 3 int main() 4 1 const int TEN 10; int a[TEN],n-354, num, c-0; 5 6 7 num-n; 8O dof a[c++]-num%TEN; 9T 10 num=num/TEN; 11 12 Jwhile(num); cout<<"The number "K<n<<"is now"; for (int k=c-1;k>-0;k--) cout<<a[k]; 13 14 15 16 17 cout<<endl; return 0; 18 19 20 The number 354is now354

2nd question answer

int A=6,sT, ss, 88, 73,063 for loop beg & k num AF /Fntsf 4,r,73, ,0,0p.0 ACGI AC] £6 66 55 Ab hum &07- f0,0,0,0,0,0,0 A (03-0. jejet AC3J-D O, ST,,,C,0,0,0 Lt ST B ES]-0 B]-0. A ST>55 false A]>num int num ST 1 2 int k O A(2>num 7 Bk] ACi k-0 B[1] = ヨ子; k=k+ k 2 foss,0,,,7,0,0,0 3 83, 7,66 0/03 ACSJ> num j-j+j4 S555false

Afu snum 88>55 B(ki+]A 2J=8k2l=3 A 0 jtl ST 3 6131,463 0,01 -Toa oTOn Afr]snum 0-UA B(ktt) ACi k 3tl BC3] 77 j-5H6 AGJ-0 fo,T,0,5,7, 23, 79, SS స for (i<6) 6x6 Herminatid Affen e Trauing e valu of false So loop Aand (0T,0, ST,0,0, 6-f 46, 77/23,7700 0F A= -May holds mbage valau alsor

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