Question

我有一个程序，我读取数据，并将它们存储在1d和2d数组中，然后将它们传递给函数进行分析。我在声明/使用2d数组并将它们传递给函数时遇到问题。

问题： 1）如果我使用malloc来声明2d数组，那么当我尝试在其中存储数据时，我会在第一行的第二列上得到分段错误。但如果我将其声明为array [] []

，它将正常工作

2）我无法将其传递给功能，使用它。在论坛上有不同的选项，尝试过它们但最终都会出现错误，说明声明的参数类型和我传递的变量没有相同的数据类型。

以下完整计划：

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

// Function declaration
void clearNewLines(void);
int match(int numStates, int numAlphabets, char *nameOfStates, char *nameOfAlphabets, char finalState,char *transitionTable, int stringLength,char *string);

int main(int argc, char *argv[]){

    // Number of states and number of alphabets of DFA
    int numStates;
    int numAlphabets;

    // Language string
    char *string;
    int stringLength;

    // Final state
    char finalState;

    // Read numStates 
    printf("Enter the number of STATES:");
    scanf("%d",&numStates);

    // Flush STDIN
    clearNewLines();

    // Array for name of alphabets, and name of states
    char *nameOfStates = malloc(numStates*sizeof(char));

    // Read the nameOfStates 
    int i;
    for(i=0;i<numStates;i++){

        if(i==0)
            printf("Enter name of states as numbers(1-9)\n");   

        printf("Name of STATES:");
        scanf("%c",&nameOfStates[i]);
        clearNewLines();

        //fgets(nameOfStates[i],2,stdin);
        //fgets(nameOfStates[i],2*sizeof(char),stdin);  

    }// End of for-loop to read nameOfStates

    // Read numAlphabets
    printf("Enter the number of ALPHABETS: ");
    scanf("%d", &numAlphabets);

    // Flush STDIN
    clearNewLines();

    // Array for name of alphabets, and name of states
    //char nameOfAlphabets[numAlphabets];   
    char *nameOfAlphabets = malloc(numAlphabets * sizeof(char));
    // Saving transition table
    //char **transitionTable = malloc(sizeof(char*) * numStates * numAlphabets);
    char transitionTable[numStates][numAlphabets];

    // Read name of alphabets
    int j;
    for(j=0;j<numAlphabets;j++){

        // Read the alphabets
        printf("Name of ALPHABETS:");
        scanf("%c",&nameOfAlphabets[j]);

        // Flush STDIN 
        clearNewLines(); 

    }// End for-loop to read alphabets

    // Get the transitionTable[states][alphabets] 
    int row;
    for(row=0;row<numStates;row++){

        int col;
        for(col=0;col<numAlphabets;col++){

            printf("Enter Transition From q%c to %c: ",nameOfStates[row],nameOfAlphabets[col]);
            scanf("%c",&transitionTable[row][col]);
            clearNewLines();
        }

    }// End of (outer) for-loop to store data in transition table

    // Get final state
    printf("Enter final state: ");
    scanf("%c",&finalState);
    clearNewLines(); 

    // Get language string
    printf("Enter the length of string: ");
    scanf("%d",&stringLength);
    clearNewLines();
    string = malloc(stringLength*sizeof(char));
    printf("Enter string: ");
    scanf("%s",string);
    clearNewLines();

    int result =  match(numStates, numAlphabets, nameOfStates, nameOfAlphabets, finalState, (char*)transitionTable, stringLength, string);


    return result;


}// End of main function


/*
*
*   match - check if a string matches a language
*/

int match(int numStates, int numAlphabets, char *nameOfStates, char *nameOfAlphabets, char finalState,char *transitionTable, int stringLength,char *string){


    char state;     // State of the machine
    char stringChar;// Character of string being processed
    int result;     // Result of the character processing

    result = 0;
    int i = 0;
    // initial state
    state = nameOfStates[0];
    stringChar = string[i];

    // Walk through the string, while doing the transition
    while(i < stringLength){

        int row;
        for(row=0;row<numStates;row++){

            if(state == nameOfStates[i]){
                break;
            }           
        }// End of for-loop to find the state

        int col;
        for(col=0;col<numAlphabets;numAlphabets++){

            if(stringChar == nameOfAlphabets[col]){
                break;
            }
        }// End of for-loop to find the alphabet

        state = transitionTable[row][col];
        // Next character   
        i++;
        stringChar = string[i];

    }// End of while-loop to go thorough the string characters of the language    

    // If in final state, then accepted, if not then rejected
    if(state == finalState){

        result = 1;
    }else{

        result = 0;
    }

    return result;

}// End of match function

/*
*
*   clearNewLines - clear any newline character present at the STDIN
*/
void clearNewLines(void)
{
    int c;
    do
    {
        c = getchar();
    } while (c != '\n' && c != EOF);
}

编辑：我根据建议（函数（array [first] [second]）更改了程序。现在它确实传递了函数，但数组是空的。

Answer 1

一个简单的解决方案是将2D数组声明为一维数组。让我们假设您希望将10x20整数数组作为参数传递给函数foo：

void foo(int* a2Darray){
    ...
}

int* my2Darray = malloc(10*20*sizeof(*my2Darray)); //allocate your array as an 1D array
foo(my2Darray); //pass it as an argument

然后让我们说你要访问2D数组的元素[x，y]。这是通过访问1D阵列的元素x + 10 * y来执行的。

如果数组的大小不是常量，那么您也可以将大小作为参数传递，方法是将函数更改为：

void foo(int* a2Darray, int maxX, int maxY)

Answer 2

更改

int match(int numStates, int numAlphabets, char *nameOfStates, char *nameOfAlphabets, char finalState,char *transitionTable, int stringLength,char *string){

到

int match(int numStates, int numAlphabets, char *nameOfStates, char *nameOfAlphabets, char finalState,char transitionTable[numStates][numAlphabets], int stringLength,char *string){

致电

int result =  match(numStates, numAlphabets, nameOfStates, nameOfAlphabets, finalState, transitionTable, stringLength, string);

Answer 3

将此数组传递给函数：

char transitionTable[numStates][numAlphabets];

该功能需要参数：

, size_t numAlphabets, char transitionTable[][numAlphabets],

将二维数组传递给c中的函数

3 个答案: