Mid-term exam. Scores. Fall term 2012 KAIST EE209 Programming Structures for EE. Thursday Oct 25, Student's name: Student ID:

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1 Fll term 2012 KAIST EE209 Progrmming Structures for EE Mid-term exm Thursdy Oct 25, 2012 Student's nme: Student ID: The exm is closed book nd notes. Red the questions crefully nd focus your nswers on wht hs been sked. You re llowed to sk the instructor/tas for help only in understnding the questions, in cse you find them not completely cler. Be concise nd precise in your nswers nd stte clerly ny ssumption you my hve mde. All your nswers must be included in the ttched sheets. You hve 120 minutes to complete your exm. Be wise in mnging your time. Good luck! Scores Question 1 Question 2 Question 3 Question 4 Totl /20 /25 /20 /30 /90 1

2 1. Deterministic finite stte utomton (DFA). (20 points) ) Wht does the following DFA do? (3 pts) letter (print uppercse equivlent) Not-letter 1 2 Not-letter letter It converts the first chrcter of ech word to n uppercse chrcter. b) Drw DFA tht removes ll single-line comments tht follow // in C source codes. (5 pts) / (do nothing) / (do nothing) Not \n (do nothing) Not / \ Not / (print /, print the letter) 4 5 letter Not nor \ \n (print \n or do nothing) 2

3 (c), (d) It is known tht every regulr expression cn be described by DFA. Assuming we hve chrcters, nd b s input, ^*$ cn be represented s following. Plese specify the regulr expression tht exctly mtches ech of the following DFAs. If stte does not specify prticulr input chrcter, we ssume the DFA utomticlly fils on tht input chrcter (ex. The bove DFA fils when it receives b ). As in the ssignment # 2, * mtches zero or more occurrence of the previous chrcter while + mtches one or more occurrences of the previous chrcter.. mtches either or b. ^ is the strt of the chrcter string while $ is the end of chrcter string. c) (5pts) b b ^b+$ d) (7pts) b or b b b b ^b.+b$ 3

4 2. Explin wht the following does. If progrm crshes, report the reson why. Assume we re on 32-bit mchine. (25pts, () to (g): 2pts ech, (h): 4 pts (i): 7pts ) ) int *p = NULL; p = 1; p hs the vlue 1 (ddress 0x1). b) int *p = NULL; *p = 1; it crshes (with segmenttion fult signl). This is becuse *p = 1 would ttempt to write the vlue t n invlid memory loction (0x0) c) double *k[25][25]; printf( sizeof(k) = %d\n, sizeof(k)); printf( sizeof(k[10][10]) = %d\n, sizeof(k[10][10])); sizeof(k) = 2500 (25*25*4 (sizeof(double *)) sizeof(k[10]k[10]) = 4 (ech element is of the double* type) d) int k = 1; int *p = &k; *p = 0; k = (3 < 2)? (k == 1) : (k == 0); printf( k = %d\n, k); k = (*p = 0)? (k == 1) : (k == 0); printf( k = %d\n, k); k = 1 (3<2 is flse, so (k == 0) is stored t k, which is 1) k = 1 (*p = 0 stores 0 to k, nd it evlutes to flse (0), nd k== 0 is executed which is 1) 4

5 e) int [7] = 0, 1, 2, 3, 4, 5, 6; int *p = &[3]; p += 2; *p += 2; printf( x=%d\n, *p++); printf( size=%d\n,sizeof(p)/sizeof([0])); x = 7 (p points to [5]) size = 1 (sizeof(p) = 4, sizeof([0]) = 4) f) struct L int vl; struct L* link; x, *y; x.vl = 20; y = &x; y->vl = 30; y->link = &x; printf( x.vl = %d\n, x.vl); printf( some vl = %d\n, x.vl + y->link->vl + y->link->link->vl); x.vl = 30 some vl = 90 (x.vl(30), y->link->vl(30),y->link->link->vl(30)) g) chr *nme = kyoungsoo ; nme[sizeof(nme)-2] = Y ; printf( nme=%s\n, nme); it crshes since kyoungsoo is stored t the red-only memory section nd chnging the vlue would violte memory ccess (segmenttion fult). Note tht chr nme[] = kyoungsoo ; is different from chr *nme = kyoungsoo ; where the former is simply initiliztion of the rry, nme[] which sits t the redwrite memory section. The ltter initilizes the pointer, nme, with the ddress of the first byte of kyoungsoo which sits t the red-only memory section. All string literls re stored t the red-only memory section. 5

6 (h), (i) chr *ito(int x) chr buf[5]; sprintf(buf, %d, x); /* see the comment below */ return buf; sprintf() is the sme s printf() except tht the output goes to the first rgument insted of stdout. E.g. sprint(buf, %d, 123); would be the sme s strcpy(buf, 123 ); /* buf[0] = 1, buf[1]= 2, buf[2] = 3, buf[3] = 0; */ (h) Point out two serious bugs with the code. (4 pts) 1. If x is lrge vlue (with more thn four digits), buf would overflow, nd the nswer would be incorrect. 2. The returned vlue is pointer to buf which sits t stck nd disppers fter the function cll. Tht is, the return vlue would point to the loction which would hve grbge vlue fter the function cll. (i) Fix the bugs. (7 pts) chr *ito(int x) chr buf[16]; /* enough to hold 2^32 */ sprintf(buf, %d, x); return strdup(buf); Note: we will give you full points if you ddress the bove two problems (e.g., you cn cll mlloc() nd strcpy() insted of strdup()) 6

7 3. Plying with C strings. (20 pts) (If you need more spce, plese use the bck of the sheet.) () int hssubstr(const chr *s, const chr *p) returns 1 if s hs p s substring or 0 if not. For exmple, the function would return 1 if s= elephnt nd p= leph since p is substring of s. Plese fill out the function below to implement this. (10 pts) (Note: you should not use ny C runtime librry functions like strchr() or strstr(). Be creful bout error processing (e.g., s nd p could point to NULL). return 1 if *p is \0.) int hssubstr(const chr *s, const chr *p) if (s == NULL p == NULL) if (*p == \0 ) return 1; for (; *s; s++) if (*p == *s) const chr *p = p+1; const chr *s = s+1; while (1) if (*p == \0 ) return 1; if (*s == \0 ) brek; if (*p++!= *s++) brek; 7

8 (b) int is_ngrm(const chr *s, const chr *p) returns 1 if s nd p re ngrms or returns 0 otherwise. s nd p re ngrms if they hve the sme length nd s cn be converted to p by rerrnging the letters of s. For exmple, opts, post, stop, pots, tops re ngrms. (10 pts) (hint: count the number of the sme letters in s nd compre it with tht of the sme letters in p. You cn use n integer rry to remember the number of occurrence of ech letter in s.) int is_ngrm(const chr *s, const chr *p) int cnt[256] = 0; int len = 0; for (; *s; s++, len++) cnt[*s]++; for (; *p; p++, len--) if ((len == 0) (--cnt[*p]) < 0) return (len == 0); 8

9 4. (key, vlue) tble mngement (30pts) We implement simple Tble dt structure tht mintins list of (key, vlue) pirs, where both key nd vlue point to non-null strings. In tble.h, we hve #ifndef TABLE_INCLUDED #define TABLE_INCLUDED typedef struct Tble_t *Tble_T; extern Tble_T Tble_new(void); extern void Tble_free(Tble_T t); extern int Tble_empty(Tble_T t); extern int Tble_dd(Tble_T t, const chr *key, const chr *vl); extern chr* Tble_serch(Tble_T t, const chr *key); extern void Tble_del(Tble_T t, const chr *key); #endif In tble.c, #include <stdlib.h> #include <ssert.h> #include "tble.h" struct item chr *key; chr *vlue; struct item *link; ; struct Tble_t struct item *hed; ; Tble_T Tble_new(void) Tble_T tble = clloc(1, sizeof(tble_t)); ssert(tble!= NULL); return tble; void Tble_free(Tble_T t) struct item *p, *next; ssert(t!= NULL); for (p = t->hed; p; p = next) next = p->next; free(p->key); free(p->vlue); free(p); free(t); 9

10 () Why is struct Tble_t defined in tble.c insted of tble.h? (2pts) It hides the fields of struct Tble_t from the user of the module. Tht is, the user should not be ble to directly ccess the fields of struct Tble_t. This improves the bstrction of the type, Tble_T. (b) Write the code for int Tble_empty(Tble_T t) tht returns 1 if the tble is empty, 0 if not. (3pts) int Tble_empty(Tble_T t) return (t->hed == NULL); (c) Write code int Tble_dd(Tble_T t, const chr *key, const chr *vl)tht stores n item with (key, vlue) t the strt of the list in Tble t. Allocte the memory for key nd vlue nd copy the content from the input so tht the tble opertions won t be disrupted by client s opertions. Return 0 in cse of n error: (1) if either key or vl is NULL, (2) if the tble lredy hs n item with the sme key, or (3) if it bumps into ny other errors. Return 1 if the opertion is successful. You cn cll ny functions declred in tble.h. (10 pts) int Tble_dd(Tble_T t, const chr* key, const chr* vl) struct item *p; if (key == NULL vl == NULL) /* hndle error cses */ if (Tble_serch(key)) /* if key exists, return 0 */ /* llocte the memory for the item */ p = mlloc(sizeof(struct item)); if (p == NULL) /* copy the key: own the key string */ p->key = strdup(key); if (p->key == NULL) free(p); /* copy the vlue: own the vlue string */ p->vl = strdup(vl); if (p->vl == NULL) free(p->key); free(p); 10

11 /* dd to the strt of the list */ p->next = t->hed; t->hed = p; return 1; (d) Write the code for chr* Tble_serch(Tble_T t, const chr *key) tht returns the vlue of the item whose key mtches key. Return NULL if such n item is not found. (7 pts) chr* Tble_serch(Tble_T t, const chr *key) struct item *p; if (key == NULL) return NULL; for (p = t->hed; p; p = p->next) if (strcmp(p->key, key) == 0) return p->vlue; return NULL; (e) Write the code for void Tble_del(Tble_T t, const chr *key) tht removes n item whose key mtches key. It does nothing when such n item does not exist. (8pts) void Tble_del(Tble_T t, const chr *key) struct item *p, *pre = NULL; if (key == NULL) return; for (p = t->hed; p; pre = p, p = p->next) if (strcmp(p->key, key) == 0) if (pre) pre->next = p->next; else t->hed = p->next; free(p->key); free(p->vlue); free(p); return; 11