편집 기록

편집 기록
  • 프로필daewon님의 편집
    날짜2021.11.23

    언어의 예외 처리 문제를 해결하고 싶습니다.


    c언어의 예외 처리 문제를 해결하고 싶습니다.

    아래는 코드입니다.

    addr.c

    #define _CRT_SECURE_NO_WARNINGS
    #include "backend-hash.h"
    #include <stdio.h>
    #include <stdlib.h>   // Necessary for exit()
    #include <string.h>
    
    
    #define WELCOME_MESSAGE "Welcome to Address Book!\n"
    #define COMMAND_PROMPT "ADDR>"
    #define ADD_PROMPT "ADD NAME>>"
    #define SEARCH_PROMPT "SEARCH NAME>>"
    #define NUMBER_PROMPT "ENTER NUMBER>>>"
    #define DELETE_PROMPT "DELETE NAME>>"
    #define MAX 100
    int prompt_command(char*, char*);
    int prompt_name(char*, char*);
    int prompt_number(char*, char*);
    void add_words();
    const int SIZE = 5;
    const int ONE = 1;
    int main()
    {
    
        char command[1];
        char name[5];
    
        init();
    
        printf(WELCOME_MESSAGE);
    
        while (1) {
        restart:
            if (prompt_command(COMMAND_PROMPT, command) != 0)
                goto restart;
            switch (command[0]) {
            case 'A':
            case 'a':
                add_words(); //a를 누르면 words.dat가 실행이 된다.
                add();
                //이윽고 입력받은 데이터들을 배치하는 함수가 실행이 된다.
                break;
            case 'S':
            case 's':
                if (prompt_name(SEARCH_PROMPT, name) != 0)
                    goto restart;
    
                break;
            case 'D':
            case 'd':
                if (prompt_name(DELETE_PROMPT, name) != 0)
                    goto restart;
    
                break;
            case 'Q':
            case 'q':
                printf("Quitting ... \n");
                return 0;
                //  Prints the number of keys in each chain.
            case 'P':
            case 'p':
                print_dist();
                break;
            default:
                printf("Please enter a valid command.\n");
                goto restart;
            }
        }
    }
    
    void add_words() //words.dat을 입력받는 함수이다.
    {
    
        char* p_word_list[5762] = { 0, }; //단어를 읽어올 공간
        char* p[5757][5]; //진짜 최종 저장 공간
    
        int i = 0;
        FILE* fp = NULL;
    
        int f = 0;
        int k = 0;
        fp = fopen("words.dat", "r");
        if (fp == NULL) //파일이 비어 있으면 긴급 탈출.
        {
            fprintf(stderr, "File Open Error!\n");
            return;
        }
        //메모리 확보
    
    
    
        for (i = 0; i < 5762; ++i) //5줄의 문자를 읽어옴.
        {
            p_word_list[i] = (char*)malloc(sizeof(char) * 6); //5글자+널문자
        }
        for (i = 0; i < 5762; ++i)
        {
            for (f = 0; f < 5; ++f)
            {
    
    
                    fscanf(fp, "%s", &p_word_list[i][f]);
    
            }
            //malloc 함수는 실패시 null을 반환합니다.
    
            if (p_word_list[i][0] != '*'&& p_word_list[i][0]!='\0')
            {
                for (f = 0; f < 5; ++f)
                {
                    wcscpy(p[k][f], p_word_list[i][f]);
    
                }
                k++;
            }
        }
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
        printf("The name was successfully added!\n");
        return;
    }
    
    
    int prompt_command(char* s, char* p)
    {
    
        char* q;
        char c;
    
        while (1) {
            printf("%s ", s);
            fflush(stdout);
            q = p;
            while (1) {
                c = getchar();
                if (c == EOF)
                    exit(-1);      // Should let exit() take care of the sudden EOF.
                if (c == '\n')
                    break;
                if (q < p + 1)
                    *q = c;
                q++;
            }
            if (q == p + 1)
                return 0;   // got a good command //
            if (q == p)
                return 1;   // just a <return>  //
            printf("Please type a command.\n");
        }
    }
    
    
    int prompt_name(char* s, char* p)
    {
    
        char* q;
        char c;
    
        while (1) {
            printf("%s ", s);
            fflush(stdout);
            q = p;
            while (1) {
                c = getchar();
                if (c == EOF)
                    exit(-1);      // Should let exit() take care of the sudden EOF.
                if (c == '\n')
                    break;
                if (q < p + 3)
                    *q = c;
                q++;
            }
            if (q == p + 3)
                return 0;   // got a good name //
            if (q == p)
                return 1;   // just a <return>  //
            printf("Please type a three-letter name.\n");
        }
    }
    
    
    int prompt_number(char* s, char* p)
    {
        char* q;
        char c;
    
        while (1) {
            printf("%s ", s);
            fflush(stdout);
            q = p;
            while (1) {
                c = getchar();
                if (c == EOF)
                    exit(-1);            // Should let exit() take care of the sudden EOF.
                if (c == '\n')
                    break;
                if (q < p + 4)
                    *q = c;
                q++;
            }
            if (q == p + 4)
                return 0;   // got a good number //
            if (q == p)
                return 1;   // just a <return>  //
            printf("Please type a four-digit number.\n");
        }
        return 0;
    }
    

    backend-hash.c

    /*
     Data Structure programming assignment.  The same address book program with hashing with chaining.
    
    Will take a hash prime p, not too big, probably 17, for an easy test.
    
    Give a function name_to_num().
    
    Define the hash function hash(name)= name_to_num(name) mod p.
    
    
    The list of first three hundred primes:
    
    {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, \
    67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, \
    139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, \
    223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, \
    293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, \
    383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, \
    463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, \
    569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, \
    647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, \
    743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, \
    839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, \
    941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, \
    1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, \
    1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, \
    1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, \
    1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, \
    1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, \
    1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, \
    1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, \
    1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, \
    1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, \
    1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, \
    1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, \
    1931, 1933, 1949, 1951, 1973, 1979, 1987}
    
    */
    
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include "backend-hash.h"
    
    #define HASH_PRIME 7000
    #define POOL_SIZE 100
    int hh = 0;
    int name_to_num(char name[3])
    {
        int x;
    
        x = name[0];
        x = x << 8;
        x = x + name[1];
        x = x << 8;
        return x + name[2];
    }
    
    int hash(char key[5]) //이번 과제에서는 hash함수도 바뀐다.
    {
        int i;
        long long x;
        x = 0;
        for (i = 0; i < 4; i++) {
            x = x + key[i];
            x = x << 8;
        }
        x = x + key[4];
        return x % HASH_PRIME;
    }
    
    
    // record structure
    struct record {
        char name[5];
        char number[4];
        struct record* next;
    };
    
    // pool of memory
    struct record pool[POOL_SIZE];
    struct record* top = pool;  // a pointer variable for stack top.
    
    // utility function declarations
    
    void print_name(struct record*);
    void print_number(struct record*);
    void print_data(char*, int);
    
    // Hash table: an array of hash values.  
    struct record* hash_table[HASH_PRIME];
    
    // Initially, each pointer of the hash table is valued NULL.
    void init_hash_table()
    {
        int i;
    
        for (i = 0; i < HASH_PRIME; i++)
            hash_table[i] = NULL;
    }
    
    // Printing all the addresses does not make a good sense in hashing...
    
    
    // Still use the same memory management.
    void init_pool() // Initialize the pool. 
    {
        int i;
        struct record* r = pool;
        struct record* s;
    
        pool[POOL_SIZE - 1].next = NULL;
    
        for (i = 1; i < POOL_SIZE; i++) {
            s = r++;
            s->next = r;
        }
    }
    
    
    void init()
    {
        init_pool();
        init_hash_table();
    }
    
    
    // Get a node from the pool. Returns NULL if pool is empty. (A BETTER DESIGN)
    // When calling new(), make sure to check if it is NULL. 
    struct record* new_node()
    {
        struct record* r;
    
        if (top == NULL)
            return NULL;
    
        r = top;
        top = r->next;
        return r;
    }
    
    // Push a node to the pool.
    void free_node(struct record* r)
    {
        r->next = top;
        top = r;
    }
    
    /**************************
    IMPORTANT NOTE on add()
     1. Allow duplicate names.
     2. Put in front of the linked list.
    **************************/
    void add() //이번 숙제에서의 add는 ()안에 아무 것도 없는 void 형식이다.
    {
    
        int i = 0;
    
        for (i = 0; i < 5757; i++) {
            hh = hash(*p[i]); //는 0부터 16까지의 값을 받게 된다. 
    
    
    
            struct record* NewNode = new_node();
    
            NewNode->next = NULL;
    
    
            struct record* r = hash_table[hh];
    
            if (hash_table[hh] == NULL) //만약 텅 비어있다면
            {
                hash_table[hh] = NewNode;
    
    
            }
            //텅 비어 있지 않다면
            else { //whw3숙제는 각 체인의 순서가 중요하지는 않으므로 그냥 맨 앞에 추가한다.
    
    
                NewNode->next = r;
                hash_table[hh] = NewNode;
    
    
            }
    
        }
    
    
    
    }
    
    /**************************
    IMPORTANT NOTE on search()
    Because we allow duplicate names, search() finds the most recently-added name.
    Prints the number of comparisons.
    **************************/
    void search(char name[5])
    {//이 함수에서는 search도 중요한 기능이 아니기 때문에 기능을 사실상 없앴다.
        printf("Couldn't find the name.\n");
    }
    
    
    
    
    
    void print_name(struct record* r)
    {
        print_data(r->name, 3);
    }
    
    void print_number(struct record* r)
    {
        print_data(r->number, 4);
    }
    
    void print_data(char* s, int n)
    {
        int i;
        for (i = 0; i < n; i++)
            putchar(s[i]);
    }
    
    void print_dist()
    {
        int i;
        for (i = 0; i < HASH_PRIME; i++)
        {
            int Count = 0;
            struct record* r = hash_table[i];
    
            printf("%d :", i);
            while (r != NULL) {
                Count = Count + 1;
    
    
                r = r->next;
                //whw3에서는 숫자만 count하면 된다.
            }
            printf(":%d  ", Count);
    
            printf("\n");
        }
        printf("\n");
    }
    

    backend-hash,h

    void search(char[5]);
    void init();
    void print_dist();
    void add_words();
    char* p[5757][5]; //진짜 최종 저장 공간
    char* p_word_list[5762]; //단어를 읽어올 공간
    void add();
    int hash(char[5]);
    

    이 함수를 실행시켜 보았더니

    nt hash(char key[5]) //이번 과제에서는 hash함수도 바뀐다.
    {
        int i;
        long long x;
        x = 0;
        for (i = 0; i < 4; i++) {
            x = x + key[i];   (예외가 발생한 부분)
            x = x << 8;
        }
        x = x + key[4];
        return x % HASH_PRIME;
    }
    
    예외가 throw됨: 읽기 액세스 위반입니다.
    key이(가) 0x1110112였습니다.
    

    같은 예외가 발생했습니다.

  • 프로필이장한님의 편집
    날짜2021.11.23

    언어의 예외 처리 문제를 해결하고 싶습니다.


    c언어의 예외 처리 문제를 해결하고 싶습니다.

    아래는 코드입니다.

    addr.c

    define _CRT_SECURE_NO_WARNINGS

    include "backend-hash.h"

    include

    include // Necessary for exit()

    include

    define WELCOME_MESSAGE "Welcome to Address Book!\n"

    define COMMAND_PROMPT "ADDR>"

    define ADD_PROMPT "ADD NAME>>"

    define SEARCH_PROMPT "SEARCH NAME>>"

    define NUMBER_PROMPT "ENTER NUMBER>>>"

    define DELETE_PROMPT "DELETE NAME>>"

    define MAX 100

    int prompt_command(char*, char*); int prompt_name(char*, char*); int prompt_number(char*, char*); void add_words(); const int SIZE = 5; const int ONE = 1; int main() {

    char command[1];
    char name[5];
    
    init();
    
    printf(WELCOME_MESSAGE);
    
    while (1) {
    restart:
        if (prompt_command(COMMAND_PROMPT, command) != 0)
            goto restart;
        switch (command[0]) {
        case 'A':
        case 'a':
            add_words(); //a를 누르면 words.dat가 실행이 된다.
            add();
            //이윽고 입력받은 데이터들을 배치하는 함수가 실행이 된다.
            break;
        case 'S':
        case 's':
            if (prompt_name(SEARCH_PROMPT, name) != 0)
                goto restart;
    
            break;
        case 'D':
        case 'd':
            if (prompt_name(DELETE_PROMPT, name) != 0)
                goto restart;
    
            break;
        case 'Q':
        case 'q':
            printf("Quitting ... \n");
            return 0;
            //  Prints the number of keys in each chain.
        case 'P':
        case 'p':
            print_dist();
            break;
        default:
            printf("Please enter a valid command.\n");
            goto restart;
        }
    }
    

    }

    void add_words() //words.dat을 입력받는 함수이다. {

    char* p_word_list[5762] = { 0, }; //단어를 읽어올 공간
    char* p[5757][5]; //진짜 최종 저장 공간
    
    int i = 0;
    FILE* fp = NULL;
    
    int f = 0;
    int k = 0;
    fp = fopen("words.dat", "r");
    if (fp == NULL) //파일이 비어 있으면 긴급 탈출.
    {
        fprintf(stderr, "File Open Error!\n");
        return;
    }
    //메모리 확보
    
    
    
    for (i = 0; i < 5762; ++i) //5줄의 문자를 읽어옴.
    {
        p_word_list[i] = (char*)malloc(sizeof(char) * 6); //5글자+널문자
    }
    for (i = 0; i < 5762; ++i)
    {
        for (f = 0; f < 5; ++f)
        {
    
    
                fscanf(fp, "%s", &p_word_list[i][f]);
    
        }
        //malloc 함수는 실패시 null을 반환합니다.
    
        if (p_word_list[i][0] != '*'&& p_word_list[i][0]!='\0')
        {
            for (f = 0; f < 5; ++f)
            {
                wcscpy(p[k][f], p_word_list[i][f]);
    
            }
            k++;
        }
    }
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    printf("The name was successfully added!\n");
    return;
    

    }

    int prompt_command(char* s, char* p) {

    char* q;
    char c;
    
    while (1) {
        printf("%s ", s);
        fflush(stdout);
        q = p;
        while (1) {
            c = getchar();
            if (c == EOF)
                exit(-1);      // Should let exit() take care of the sudden EOF.
            if (c == '\n')
                break;
            if (q < p + 1)
                *q = c;
            q++;
        }
        if (q == p + 1)
            return 0;   // got a good command //
        if (q == p)
            return 1;   // just a <return>  //
        printf("Please type a command.\n");
    }
    

    }

    int prompt_name(char* s, char* p) {

    char* q;
    char c;
    
    while (1) {
        printf("%s ", s);
        fflush(stdout);
        q = p;
        while (1) {
            c = getchar();
            if (c == EOF)
                exit(-1);      // Should let exit() take care of the sudden EOF.
            if (c == '\n')
                break;
            if (q < p + 3)
                *q = c;
            q++;
        }
        if (q == p + 3)
            return 0;   // got a good name //
        if (q == p)
            return 1;   // just a <return>  //
        printf("Please type a three-letter name.\n");
    }
    

    }

    int prompt_number(char* s, char* p) { char* q; char c;

    while (1) {
        printf("%s ", s);
        fflush(stdout);
        q = p;
        while (1) {
            c = getchar();
            if (c == EOF)
                exit(-1);            // Should let exit() take care of the sudden EOF.
            if (c == '\n')
                break;
            if (q < p + 4)
                *q = c;
            q++;
        }
        if (q == p + 4)
            return 0;   // got a good number //
        if (q == p)
            return 1;   // just a <return>  //
        printf("Please type a four-digit number.\n");
    }
    return 0;
    

    }

    backend-hash.c

    /* Data Structure programming assignment. The same address book program with hashing with chaining.

    Will take a hash prime p, not too big, probably 17, for an easy test.

    Give a function name_to_num().

    Define the hash function hash(name)= name_to_num(name) mod p.

    The list of first three hundred primes:

    {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, \ 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, \ 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, \ 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, \ 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, \ 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, \ 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, \ 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, \ 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, \ 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, \ 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, \ 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, \ 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, \ 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, \ 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, \ 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, \ 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, \ 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, \ 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, \ 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, \ 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, \ 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, \ 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, \ 1931, 1933, 1949, 1951, 1973, 1979, 1987}

    */

    include

    include

    include

    include "backend-hash.h"

    define HASH_PRIME 7000

    define POOL_SIZE 100

    int hh = 0; int name_to_num(char name[3]) { int x;

    x = name[0];
    x = x << 8;
    x = x + name[1];
    x = x << 8;
    return x + name[2];
    

    }

    int hash(char key[5]) //이번 과제에서는 hash함수도 바뀐다. { int i; long long x; x = 0; for (i = 0; i < 4; i++) { x = x + key[i]; x = x << 8; } x = x + key[4]; return x % HASH_PRIME; }

    // record structure struct record { char name[5]; char number[4]; struct record* next; };

    // pool of memory struct record pool[POOL_SIZE]; struct record* top = pool; // a pointer variable for stack top.

    // utility function declarations

    void print_name(struct record*); void print_number(struct record*); void print_data(char*, int);

    // Hash table: an array of hash values.
    struct record* hash_table[HASH_PRIME];

    // Initially, each pointer of the hash table is valued NULL. void init_hash_table() { int i;

    for (i = 0; i < HASH_PRIME; i++)
        hash_table[i] = NULL;
    

    }

    // Printing all the addresses does not make a good sense in hashing...

    // Still use the same memory management. void init_pool() // Initialize the pool. { int i; struct record* r = pool; struct record* s;

    pool[POOL_SIZE - 1].next = NULL;
    
    for (i = 1; i < POOL_SIZE; i++) {
        s = r++;
        s->next = r;
    }
    

    }

    void init() { init_pool(); init_hash_table(); }

    // Get a node from the pool. Returns NULL if pool is empty. (A BETTER DESIGN) // When calling new(), make sure to check if it is NULL. struct record* new_node() { struct record* r;

    if (top == NULL)
        return NULL;
    
    r = top;
    top = r->next;
    return r;
    

    }

    // Push a node to the pool. void free_node(struct record* r) { r->next = top; top = r; }

    /************************** IMPORTANT NOTE on add()

    1. Allow duplicate names.
    2. Put in front of the linked list. **************************/ void add() //이번 숙제에서의 add는 ()안에 아무 것도 없는 void 형식이다. {

      int i = 0;

      for (i = 0; i < 5757; i++) { hh = hash(*p[i]); //는 0부터 16까지의 값을 받게 된다.

      struct record* NewNode = new_node();
      
      NewNode->next = NULL;
      
      struct record* r = hash_table[hh];
      
      if (hash_table[hh] == NULL) //만약 텅 비어있다면
      {
          hash_table[hh] = NewNode;
      
      }
      //텅 비어 있지 않다면
      else { //whw3숙제는 각 체인의 순서가 중요하지는 않으므로 그냥 맨 앞에 추가한다.
      
          NewNode->next = r;
          hash_table[hh] = NewNode;
      
      }
      

      }

    }

    /************************** IMPORTANT NOTE on search() Because we allow duplicate names, search() finds the most recently-added name. Prints the number of comparisons. **************************/ void search(char name[5]) {//이 함수에서는 search도 중요한 기능이 아니기 때문에 기능을 사실상 없앴다. printf("Couldn't find the name.\n"); }

    void print_name(struct record* r) { print_data(r->name, 3); }

    void print_number(struct record* r) { print_data(r->number, 4); }

    void print_data(char* s, int n) { int i; for (i = 0; i < n; i++) putchar(s[i]); }

    void print_dist() { int i; for (i = 0; i < HASH_PRIME; i++) { int Count = 0; struct record* r = hash_table[i];

        printf("%d :", i);
        while (r != NULL) {
            Count = Count + 1;
    
    
            r = r->next;
            //whw3에서는 숫자만 count하면 된다.
        }
        printf(":%d  ", Count);
    
        printf("\n");
    }
    printf("\n");
    

    }

    backend-hash,h

    void search(char[5]); void init(); void print_dist(); void add_words(); char* p[5757][5]; //진짜 최종 저장 공간 char* p_word_list[5762]; //단어를 읽어올 공간 void add(); int hash(char[5]);

    이 함수를 실행시켜 보았더니

    nt hash(char key[5]) //이번 과제에서는 hash함수도 바뀐다. { int i; long long x; x = 0; for (i = 0; i < 4; i++) { x = x + key[i]; (예외가 발생한 부분) x = x << 8; } x = x + key[4]; return x % HASH_PRIME; }

    예외가 throw됨: 읽기 액세스 위반입니다. key이(가) 0x1110112였습니다.

    같은 예외가 발생했습니다.

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