c - 将二维数组传递给 c 中的函数

Question

我有一个程序，我在其中读取数据，并将它们存储在一维和二维数组中，然后将它们传递给一个函数来分析它。我在声明/使用二维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 来声明二维数组，那么当我试图在其中存储数据时，我会在第一行的第二列出现段错误。但如果我将它声明为 array[][]

它会工作正常

2) 我无法将它传递给函数，无法使用它。论坛上有不同的选项，我试过了，但都以错误结束，指出声明的参数类型和我传递的变量没有相同的数据类型。

完整程序如下:

#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:我根据建议更改了程序( function(array[first][second] )。现在它确实传递了函数，但数组为空。

Answer 1

一个简单的解决方案是将二维数组声明为一维数组。 假设您想将一个 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

然后假设您要访问二维数组的元素 [x,y]。这是通过访问一维数组的元素 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，方法是将您的函数更改为:

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