2D字符串数组的最后一个元素覆盖先前的数组元素 [英] Last element of 2D array of strings overwriting previous array elements

问题描述

void read_char_from_file(FILE *fp, int formatted)
{
    char sChar[16], eol;
    int col=0, row=0,ind=0,count=0;
    char *str2D[4][4]={{ "NULL" }};
    while ( (fscanf(fp, "%s%c", sChar,&eol)) != EOF)
    {
        if(eol!='\n')
        {
            str2D[row][col]=sChar;
            printf("Row: %d, Col: %d = %s\n",row, col,str2D[row][col]);
            col++;
        }
        else {
            str2D[row][col]=sChar;
            printf("Row: %d, Col: %d = %s\n",row, col,str2D[row][col]);
            row++;
            col=0;
        }
        str2D[row][col]=0;
    }

    int i,j;
    for (i=0; i < 3; i++)
    {
        for(j=0; j <3; j++)
        printf("%s",str2D[i][j]);
    }

}

以上代码显示了str2D的最后一个元素，覆盖了所有先前的数组内容.我需要将sChar复制到二维字符串数组中.

Above code is displaying last element of str2D overwriting all of the previous array contents. I need to copy sChar to 2-D array of strings.

推荐答案

简短答案:

您不将内容复制到由str2D[row][col]指向的存储位置，只是将指针指向sChar.因此，无论sChar的最新内容是什么，都将显示出来.

You are not copying the content into the memory location pointed to by str2D[row][col], you're just making the pointer point to sChar. Thus, whatever is the latest content of the sChar, is being shown.

长答案:

您还有其他问题.首先，

You have other issues. First,

• {{ "NULL" }};与{{ NULL }};
不同
• 您正在创建一个二维指针数组.在将任何内容存储在指针指向的那些存储位置之前，需要确保它们指向一些有效的位置.

• {{ "NULL" }}; is not the same as {{ NULL }};
• You are creating a 2-D array of pointers. Before you can store anything in those memory locations pointed to by the pointers, you need to make sure that they point to some valid locations.

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