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CS1010E Programming Methodology Tutorial 8 Pointers and Strings

CS1010E Programming Methodology Tutorial 8 Pointers and Strings. C14,A15,D11,C08,C11,A02. Question 1. list1. list2. Question 1 (a). list1. list2. char * ptr ; ptr = list1 [ 2 ]; printf ( "%c" , ptr );. ‘U’. 0x000000055.

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CS1010E Programming Methodology Tutorial 8 Pointers and Strings

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  1. CS1010E Programming MethodologyTutorial 8Pointers and Strings C14,A15,D11,C08,C11,A02

  2. Question 1 list1 list2

  3. Question 1 (a) list1 list2 char*ptr; ptr= list1[2]; printf("%c",ptr); ‘U’ 0x000000055 ptr takes in an address, it will treat ‘u’ as address of ‘0x00000055’. This may be some protected location in your system!! ‘U’ ptr

  4. Question 1 (b) list1 list2 void swap1(char*ptr1,char*ptr2){ char*dummy; dummy = ptr1; ptr1 = ptr2; ptr2 = dummy; } swap1(&list1[2], &list1[4]);

  5. Question 1 (b) list1 list2 void swap1(char*ptr1,char*ptr2){ char*dummy; dummy = ptr1; ptr1 = ptr2; ptr2 = dummy; } swap1(&list1[2], &list1[4]); Scope of swap ptr1 ptr2

  6. Question 1 (b) list1 list2 void swap1(char*ptr1,char*ptr2){ char*dummy; dummy = ptr1; ptr1 = ptr2; ptr2 = dummy; } swap1(&list1[2], &list1[4]); Scope of swap ptr1 ptr2 dummy

  7. Question 1 (b) list1 list2 void swap1(char*ptr1,char*ptr2){ char*dummy; dummy = ptr1; ptr1 = ptr2; ptr2 = dummy; } swap1(&list1[2], &list1[4]); Scope of swap ptr1 ptr2 dummy

  8. Question 1 (b) list1 list2 void swap1(char*ptr1,char*ptr2){ char*dummy; dummy = ptr1; ptr1 = ptr2; ptr2 = dummy; } swap1(&list1[2], &list1[4]); Scope of swap ptr1 ptr2 dummy

  9. Question 1 (b) list1 list2 void swap1(char*ptr1,char*ptr2){ char*dummy; dummy = ptr1; ptr1 = ptr2; ptr2 = dummy; } swap1(&list1[2], &list1[4]); Scope of swap printf("%c %c\n", list1[2], list1[4]); u e ptr1 ptr2 dummy

  10. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4])

  11. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4]) Scope of swap2 ptr1 ptr2

  12. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4]) Scope of swap2 dummy ptr1 ptr2

  13. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4]) Scope of swap2 ‘U’ dummy ptr1 ptr2

  14. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4]) Scope of swap2 ‘U’ dummy ptr1 ptr2

  15. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4]) Scope of swap2 ‘U’ dummy ptr1 ptr2

  16. Question 1 (c) list1 list2 void swap2(char*ptr1,char*ptr2){ chardummy; dummy =*ptr1; *ptr1 =*ptr2; *ptr2 = dummy; } Swap2(&list1[2], &list1[4]); Scope of swap2 ‘U’ dummy printf("%c %c\n", list1[2], list1[4]); E U ptr1 ptr2

  17. Question 1 (d) • printf("%s", list1); • student • printf("%s", &list1[0]); • student • printf("%s", list2[3]); • project • printf("%s", &list2[3][0]); • project list1 list2 “%s” will print everything until the first ‘\0’.

  18. Question 1 (e) list1 list2 char*ptr; ptr= list2[1]; ptr++; printf("%s",ptr); ptr

  19. Question 1 (e) list1 list2 char*ptr; ptr= list2[1]; ptr++; printf("%s",ptr); ptr

  20. Question 1 (e) list1 list2 char*ptr; ptr= list2[1]; ptr++; printf("%s",ptr); ptr

  21. Question 1 (e) list1 list2 char*ptr; ptr= list2[1]; ptr++; printf("%s",ptr); ptr UTORIAL

  22. Question 1 (f) list1 list2 char*ptr; ptr=&list1[2]; *ptr++; printf("%c %c", list1[2],ptr); ptr

  23. Question 1 (f) list1 list2 char*ptr; ptr=&list1[2]; *ptr++; printf("%c %c", list1[2],ptr); ptr

  24. Question 1 (f) list1 list2 char*ptr; ptr=&list1[2]; *ptr++; printf("%c %c", list1[2],ptr); ptr “++” has high priority than “*”!

  25. Question 1 (f) list1 list2 char*ptr; ptr=&list1[2]; *ptr++; printf("%c %c", list1[2],ptr); ptr “++” has high priority than “*”! U D

  26. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually Not in duplicate group Count = 0

  27. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually Not in duplicate group Count = 0

  28. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually In duplicate group Count = 0

  29. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually In duplicate group, about to leave Count = 1

  30. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually Not in duplicate group Count = 1

  31. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually In duplicate group Count = 1

  32. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually In duplicate group, about to leave Count = 2

  33. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually Not in duplicate group Count = 2

  34. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually In duplicate group Count = 2

  35. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually In duplicate group, about to leave Count = 3

  36. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually End of string, return Count = 3

  37. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Before writing program, lets see how we do this manually We increase our counter when: leaving a duplicate group! How do we detect when this? str[i] == str[i-1] && str[i] != str[i+1]

  38. Question 2 • Counting duplicating groups • intrepeat(char str[]); • Algorithm • Count = 0; • For every str[i] in string str[0… len], • if(str[i] == str[i-1] && str[i] != str[i+1]) count++; • Return count; How do we detect when this? str[i] == str[i-1] && str[i] != str[i+1]

  39. Question 2 int repeat(charstr[]){ intcount =0,i=1; for(;i<strlen(str);i++){if(!isspace(str[i]) &&str[i]==str[i-1] &&str[i]!=str[i+1]){ count++; } } returncount; } Code for counting duplicate groups

  40. Question 3 • Delete a substring from main string • Analysis: • To delete a substring, we need to find a string • To find a string, we need to check every character of string for match • After find a string, delete it from string • Suppose we have three functions: • findstr(mainstr, substr); • Find first occurrence of substr in mainstr • delstr(mainstr, index, sublen); • delete from main string at position index

  41. Question 3 char*delete(char*mainstr,char*substr){ intindex,sublen=slength(substr);//find the position of sub string in the main string index =findstr(mainstr,substr); if(index==-1){//no sub string found printf("No sub-string found! Task not successful!\n"); return0; }else{//delete the substring delstr(mainstr, index,sublen); returnmainstr; } } Skeleton of delete function

  42. Question 3 • Compute string length • will be used during deletion • Algorithm: • scanning from string head • count until you find ‘\0’; intslength(char*str){ inti=0; while(str[i]!='\0') i++; returni; }

  43. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Don’t match

  44. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Don’t match

  45. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Don’t match

  46. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Don’t match

  47. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Don’t match

  48. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Don’t match

  49. Question 3 • Find the first occurrence of substring • Algorithm: • Scan from head of string • Check whether there is a substring match substr at each position • Return the matching position Match & return

  50. Question 3 intfindstr(char*mainstr,char*substr){ inti=j=equal=0; intstart=-1, mainlen=slength(mainstr), sublen=slength(substr); intmaxstartindex=mainlen-sublen; //find the first character of substring in the main string for(i=0;i<=maxstartindex&&!equal;i++) if(substr[0]=='*'||mainstr[i]==substr[0]){ equal=1; start=i; //compare the following characters for(j=0; j<sublen; j++) if(substr[j]!='*'&&mainstr[i+j]!=substr[j]){ equal=0; start=-1; } } returnstart; } Code for findstr

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