BDII_tema02

Tema nr. 2 Observa Observație! Scrieți rezolvarea rezolvarea direct în acest document! document!

1. Write Write an anonymous anonymous block block that uses a country country name name as input and prints prints the highest highest and lowest lowest elevations for that country. Use the wfcountries table. "ecute your block three times using United States of #merica$ %rench &epublic$ and 'apan. ()*#& vlowestelevation vlowestelevation number+,- S#U vlowestelevation vlowestelevation wfcountries.lowestelevation wfcountries.lowestelevation /type vhighestelevation vhighestelevation wfcountries.highestelevation wfcountries.highestelevation /type 023 S*)4 lowestelevation$ highestelevation 2345 vlowestelevation$ vhighestelev vhighestelevation ation %&56 wfcountries W7& countryname 8 9'apan: (06S5U4;U4.;U4*23 (06S5U4;U4.;U4*23+9the +9the lowest elevation is : << vlowestelevations- vlowestelevations- (06S5U4;U4.;U4*23 (06S5U4;U4.;U4*23+9the +9the highest elevation is : << vhighestelevations- vhighestelevations- 3( =. valuate the variables in the following code. #nswer the following >uestions about each variable. 2s it named well? Why or why not? 2f it is not named well$ what would be a better name and why? ()*#& countryname countryna me @#&)7#&= +,A- medianage 3U60&+B$=- 023 S*)4 countryname$ medianage 2345 countryname$ medianage %&56 wfcountries W7& countryname 8 CUnited States of #mericaC (06S5U4;U4.;U4*23 (06S5U4;U4.;U4*23+C +C 4he median age in C<uestion =. )hange the declarations so that they use the /4I;

attribute. ()*#& vcountryname wfcountries.countryname/4I; vmedianage wfcountries.medianage/4I; 023 S*)4 countryname$ medianage 2345 vcountryname$ vmedianage %&56 wfcountries W7& countryname 8 CUnited States of #mericaC (06S5U4;U4.;U4*23+C 4he median age in C<uestions. ()*#& " @#&)7#&=+=A- 023 "K8 C1=HC O CJ,BC  (06S5U4;U4.;U4*23+"- 3(

#. What do you think the output will be when you run the above code?  3U60& DDD initial este char dar se face conversia implicita la number  0. 3ow$ run the code. What is the output? ,PQ ). 2n your own words$ describe what happened when you ran the code. (id any implicit conversions take place? )onversie implicita din char in number$ noutate in plLs>l B. Write an anonymous ;*LSM* block that assigns the programmer:s full name to a variable$ and then displays the number of characters in the name. ()*#& " @#&)7#&=+=A- 023 "K8 C;avel 6ariu 2ulianC (06S5U4;U4.;U4*23+*347+"-- 3( P. Write an anonymous ;*LSM* block that uses todayCs date and outputs it in the format of 96onth dd$ yyyy:. Store the date in a (#4 variable called mydate. )reate another variable of the (#4 type called vlastday. #ssign the last day of this month to vlastday. (isplay the value of vlastday. ()*#& 6I(#4 (#4K8SIS(#4 vlastday (#4K8*#S4(#I+6I(#4- 023 (06S5U4;U4.putline +45)7#& +6I(#4$ C6onth (( IIII C-<uestion H to add J, days to today:s date and then calculate and display the number of months between the two dates. ()*#& 6I(#4 (#4K8SIS(#4 6I(#4= (#4K86I(#4 O J, 023   (06S5U4;U4.putline +45)7#&+65347S04W3+6I(#4=$6I(#4--- (06S5U4;U4.putline +6I(#4<
weight K8 weight O 1 newlocn K8 CWestern C << newlocn DD ;osition 1 DD 3( weight K8 weight O 1 message K8 message << C is in stockC DD ;osition = DD 3( #. 4he value of weight at position 1 isK = 0. 4he value of newlocn at position 1 isK Western urope ). 4he value of weight at position = isKBA1 (. 4he value of message at position = isK ;roduct 1AA1= is in stock  . 4he value of newlocn at position = isK eroare$ nu este declarant in blocul respectiv ()*#& weight 3U60&+H- K8 BAA message @#&)7#&=+=,,- K8 C;roduct 1AA1=C 023 ()*#& weight 3U60&+H- K8 1 message @#&)7#&=+=,,- K8 C;roduct 11AA1C newlocn @#&)7#&=+,A- K8 CuropeC 023 weight K8 weight O 1 newlocn K8 CWestern C << newlocn (06S5U4;U4.putline +C#. 4he value of weight at position 1 isK C << weight- (06S5U4;U4.putline +C0. 4he value of newlocn at position 1 isK C << newlocn - DD ;osition 1 DD 3( weight K8 weight O 1 message K8 message << C is in stockC (06S5U4;U4.putline +C). 4he value of weight at position = isK C << weight- (06S5U4;U4.putline +C(. 4he value of message at position = isK C << message- DD ;osition = DD 3( 1A. nter and run the following ;*LSM* block$ which contains a nested block. *ook at the output and answer the >uestions. ()*#& vemployeeid employees.employeeid/4I; vob employees.obid/4I; 023 S*)4 employeeid$ obid 2345 vemployeeid$ vob %&56 employees W7& employeeid 8 1AA ()*#& vemployeeid employees.employeeid/4I; vob employees.obid/4I;

  023 S*)4 employeeid$ obid 2345 vemployeeid$ vob %&56 employees W7& employeeid 8 1AH (06S5U4;U4.;U4*23+vemployeeid<< C is a C<
1=. 6odify the block in >uestion H to omit the e"ception handler$ then reDrun the block. "plain the output. ()*#& vnumber 3U60&+J- 023 vnumber K8 QQQQ );4253 W73 547&S 473 (06S5U4;U4.;U4*23+C#n e"ception has occurredC << vnumber- 3( 1H. nter and run the following code and e"plain the output. ()*#& vnumber 3U60&+J- 023 vnumber K8 1=HJ ()*#& vnumber 3U60&+J- 023 vnumber K8 ,BPR vnumber K8 C# character stringC 3( );4253 W73 547&S 473 (06S5U4;U4.;U4*23+C#n e"ception has occurredC- (06S5U4;U4.;U4*23+C4he number isK C<