Question

无论出于何种原因，for中的getLetters循环都会导致崩溃。

我无法挑出其中的特定来源，也许有人可以提供帮助？

我想我可能错误地完成了我的指针。我该如何解决这个问题？

#include <stdio.h>
#include <string.h>
#include <time.h>


int main() {
    FILE *ifp;
    ifp = fopen("dictionary.txt", "r");
    int* lValues[26];
    int* lDist[26];
    int* lUsed[26];
    int*fRun = 0;
    int dictLen;
    int i;
    int bestValue;
    fscanf(ifp, "%d", &dictLen);
    char bestWord[7];
    char dictionary[dictLen][7];
    char* letters[7];
    char* userWord[7];

    //reads the dictionary into the array
    for (i = 0; i < dictLen; i++) {
        fscanf(ifp, "%s", &dictionary[i]);
    }

    distribution();
    values();

    while (i != 2) {

        getLetters();

        printf("Welcome to the Scrabble Practice Program!\n");
        printf("\nHere are your letters: %s \n", letters);
        printf("\nWhat would you like to do?\n");
        printf("\t1-Enter Word\n");
        printf("\t2-Quit\n\n");
        printf("User Selection:");
        scanf("%d", &i);
    }

    //fclose(ifp); not sure why, but this crashes the program

    return 0;
}

//Gets the user's set of letters
void getLetters() {
    char* letters[7];
    int* lDist[26];
    int* lUsed[26];
    int lCur;
    int i;

    srand(time(0));

    for (i = 0; i < 7; i++) {

        *letters[i] = 65 + rand()%26;
        lCur = (int)*letters[i] - 65;

        while (*lUsed[lCur] >= *lDist[lCur]){

            *letters[i] = 65 + rand()%26;
            lCur = (int)*letters[i] - 65;
        }

        *lUsed[lCur]++;
    }
}

//Sets the number of each letter available for distribution
void distribution() {
    int* lDist;

    //Number for A
    lDist[0] = 9;

    //Number for B
    lDist[1] = 2;

    //Number for C
    lDist[2] = 2;

    //Number for D
    lDist[3] = 4;

    //Number for E
    lDist[4] = 12;

    //Number for F
    lDist[5] = 2;

    //Number for G
    lDist[6] = 3;

    //Number for H
    lDist[7] = 2;

    //Number for I
    lDist[8] = 9;

    //Number for J
    lDist[9] = 1;

    //Number for K
    lDist[10] = 1;

    //Number for L
    lDist[11] = 4;

    //Number for M
    lDist[12] = 2;

    //Number for N
    lDist[13] = 6;

    //Number for O
    lDist[14] = 8;

    //Number for P
    lDist[15] = 2;

    //Number for Q
    lDist[16] = 1;

    //Number for R
    lDist[17] = 6;

    //Number for S
    lDist[18] = 4;

    //Number for T
    lDist[19] = 6;

    //Number for U
    lDist[20] = 4;

    //Number for V
    lDist[21] = 2;

    //Number for W
    lDist[22] = 2;

    //Number for X
    lDist[23] = 1;

    //Number for Y
    lDist[24] = 2;

    //Number for Z
    lDist[25] = 1;
}


//Sets the value of all of the letters
void values() {

    int* lValues;

    //Value of A
    lValues[0] = 1;

    //Value of B
    lValues[1] = 3;

    //Value of C
    lValues[2] = 3;

    //Value of D
    lValues[3] = 2;

    //Value of E
    lValues[4] = 1;

    //Value of F
    lValues[5] = 4;

    //Value of G
    lValues[6] = 2;

    //Value of H
    lValues[7] = 4;

    //Value of I
    lValues[8] = 1;

    //Value of J
    lValues[9] = 8;

    //Value of K
    lValues[10] = 5;

    //Value of L
    lValues[11] = 1;

    //Value of M
    lValues[12] = 3;

    //Value of N
    lValues[13] = 1;

    //Value of O
    lValues[14] = 1;

    //Value of P
    lValues[15] = 3;

    //Value of Q
    lValues[16] = 10;

    //Value of R
    lValues[17] = 1;

    //Value of S
    lValues[18] = 1;

    //Value of T
    lValues[19] = 1;

    //Value of U
    lValues[20] = 1;

    //Value of V
    lValues[21] = 4;

    //Value of W
    lValues[22] = 4;

    //Value of X
    lValues[23] = 8;

    //Value of Y
    lValues[24] = 4;

    //Value of Z
    lValues[25] = 10;
}

Answer 1

*letters[i] = 65 + rand()%26;

字母[i]指的是无意义的（未初始化的），您最终可以使用Access violation fault。

进一步

char* letters[7];
int* lDist[26];
int* lUsed[26];

所有这些都是arrays of pointer而不是基本类型的数组（即int，char等）。我想你可能想要

char letters[7];//Array of seven characters
int lDist[26];//Array of 26 ints
int lUsed[26];//Array of 26 ints

并开始将*letters[i]简称为letters[i]

Answer 2

letters，lDist和lUsed是指针数组（未初始化）。为什么不使用数组呢？

#include <stdlib.h>
#include <time.h>

void getLetters(void)
{
    char letters[7];
    int lDist[26];
    int lUsed[26];
    int lCur;
    int i;

    srand(time(NULL));

    for (i = 0; i < 7; i++) {
        letters[i] = 65 + rand() % 26;
        lCur = (int) letters[i] - 65;

        while (lUsed[lCur] >= lDist[lCur]) {
            letters[i] = 65 + rand() % 26;
            lCur = (int) letters[i] - 65;
        }

        lUsed[lCur]++;
    }
}

Answer 3

getLetters函数中的每个单一数组访问都可能导致问题：您声明了三个从未初始化的指针数组，因此每次取消引用任何数组的元素时都会取消引用垃圾内存地址，有时试图写入该内存地址。

从您的代码中不清楚为什么需要指针数组。

为什么不简单：

char letters[7];
int lDist[26];
int lUsed[26];

摆脱解除引用？

但即使这样，该功能也存在问题。您需要初始化数组值，至少lDist和lUsed。您可以与lDist的元素进行比较，但不会更改这些值。

Answer 4

上述代码似乎没有按照您的意图执行。声明变量

时

char* letters[7];

...你宣布一个7 char * s的数组，而不是7 char s。因此，当你说

时

*letters[i] = 65 + rand()%26;

...您首先索引数组并检索索引i处的指针。然后取消引用该指针并尝试在那里存储一个字符。问题是letters永远不会被初始化，并且最有可能充满垃圾：对于任何值i，letters[i]都会返回一个随机值。因此，当您尝试将其解除引用作为地址时，您将索引到其他人的内存中，并且写入它将导致分段错误。

您似乎想要声明letters（以及lDist和lUsed），如下所示：

char letters[7];
int lDist[26];
int lUsed[26];

这意味着一个长度为7的char数组。然后，您将在不解除引用的情况下解决它们：

letters[i] = 65 + rand()%26;

你也可以做以上......

*(letters + i) = 65 + rand()%26;

...但你通常不是要将两者混合在一起。

Answer 5

不需要使用指针，使用普通数组就行了，就像这样（我还为lUsed和lDist添加了初始化）：

void getLetters() {
    char letters[7];
    int lDist[26];
    int lUsed[26];
    int lCur;
    int i;


    srand(time(0));

    // initialize the arrays
    for (i=0;i<26;i++) {
        lUsed[i] = lDist[i] = 0;
    }

    for (i = 0; i < 7; i++) {

        letters[i] = 65 + rand()%26;
        lCur = (int)letters[i] - 65;

        while (lUsed[lCur] >= lDist[lCur]){

            letters[i] = 65 + rand()%26;
            lCur = (int)letters[i] - 65;

        }

        lUsed[lCur]++;

    }
}

For循环崩溃我的程序？

5 个答案: