C 未正确存储数组中的第一个条目

Question

我正在尝试编写一个程序来读取文本文件并将其内容放入数组中。这样，您可以读取任何文件，并且无论字符串长度如何，它都会动态构建一个数组并将其填充为一份文件。我将此作为练习 C 的练习，并希望将其推广到其他类型和结构。

但是，由于某种原因，我的第一个条目不匹配，导致出现意外的行为。我知道使用 C 语言，您基本上需要微观管理所有内存，并且使用代码，我尝试为每个条目分配内存，但这是正确的方法吗？我在脑子里运行了代码，从 0 个条目开始，逻辑上是有意义的，但我不明白为什么第一个条目失败，而其余条目工作。

代码:

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

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

    //Initialize variables and pointers
    //Create an array of chars to use when reading in
    //Create an array of strings to store
    //i : use to keep track of the number of strings in array
    //j : loop variable
    //size: size of string
    char *s = (char *) malloc(sizeof(char));
    int i=0,j=0;
    int size = 0;
    char **a = (char **) malloc(sizeof(char *));

    //Read in string, assign string to an address at a[]
    while( scanf("%79s",s) == 1){

        //Get the size of the input string
        size = (unsigned) strlen(s);

        //Print some notes here
        printf("\nString is \"%-14s\"\tSize is %-3d, i is currently %d\n",s,size,i);
        printf("Using MALLOC with %d bytes\n",size+1);

        //Allocate memory to copy string
        //
        //For some reason, the commented code works
        //a[i] = (char *) (malloc(sizeof(char)*(size+1)) + 'a');
        a[i] = (char *) (malloc(sizeof(char)*(size+1)) );

        //Go and allocate memory for each character in string to store
        for(j=0; j<(size+1); j++)   a[i][j] = (char) (malloc(sizeof(char)));

        //Print some more notes here
        printf("Size: a[%2d] is %3d bytes, *a[%2d] is %3d bytes, Length of a[%2d] is %d\n",i,(int) sizeof(a[i]),i,(int) sizeof(*a[i]),i,(unsigned) strlen(a[i]));

        //Copy over string and set last char to be end
        for(j=0; j<size; j++)       a[i][j] = (char) s[j];
        a[i][size] = '\0';

        //Print it out and increase i
        printf("a[%3d] is now %s\n",i,a[i]);

        i++;
    }
    printf("I is now %d\n\n\n",i);
    a[i] = NULL;

    //print out array
    for(j=0; j<i; j++)      printf("%3d. %-40s\n",j,a[j]);


    return 0;
}

测试文本文件(numbers.txt):

1
22
333
4444
55555
666666
7777777
88888888
9999999
0000000000
11111111111
222222222

命令:

./a.out

结果:

String is "1             "      Size is 1  , i is currently 0
Using MALLOC with 2 bytes
Size: a[ 0] is   8 bytes, *a[ 0] is   1 bytes, Length of a[ 0] is 2
a[  0] is now 1

String is "22            "      Size is 2  , i is currently 1
Using MALLOC with 3 bytes
Size: a[ 1] is   8 bytes, *a[ 1] is   1 bytes, Length of a[ 1] is 3
a[  1] is now 22

String is "333           "      Size is 3  , i is currently 2
Using MALLOC with 4 bytes
Size: a[ 2] is   8 bytes, *a[ 2] is   1 bytes, Length of a[ 2] is 4
a[  2] is now 333

String is "4444          "      Size is 4  , i is currently 3
Using MALLOC with 5 bytes
Size: a[ 3] is   8 bytes, *a[ 3] is   1 bytes, Length of a[ 3] is 5
a[  3] is now 4444

String is "55555         "      Size is 5  , i is currently 4
Using MALLOC with 6 bytes
Size: a[ 4] is   8 bytes, *a[ 4] is   1 bytes, Length of a[ 4] is 6
a[  4] is now 55555

String is "666666        "      Size is 6  , i is currently 5
Using MALLOC with 7 bytes
Size: a[ 5] is   8 bytes, *a[ 5] is   1 bytes, Length of a[ 5] is 7
a[  5] is now 666666

String is "7777777       "      Size is 7  , i is currently 6
Using MALLOC with 8 bytes
Size: a[ 6] is   8 bytes, *a[ 6] is   1 bytes, Length of a[ 6] is 8
a[  6] is now 7777777

String is "88888888      "      Size is 8  , i is currently 7
Using MALLOC with 9 bytes
Size: a[ 7] is   8 bytes, *a[ 7] is   1 bytes, Length of a[ 7] is 9
a[  7] is now 88888888

String is "9999999       "      Size is 7  , i is currently 8
Using MALLOC with 8 bytes
Size: a[ 8] is   8 bytes, *a[ 8] is   1 bytes, Length of a[ 8] is 8
a[  8] is now 9999999

String is "0000000000    "      Size is 10 , i is currently 9
Using MALLOC with 11 bytes
Size: a[ 9] is   8 bytes, *a[ 9] is   1 bytes, Length of a[ 9] is 11
a[  9] is now 0000000000

String is "11111111111   "      Size is 11 , i is currently 10
Using MALLOC with 12 bytes
Size: a[10] is   8 bytes, *a[10] is   1 bytes, Length of a[10] is 12
a[ 10] is now 11111111111

String is "222222222     "      Size is 9  , i is currently 11
Using MALLOC with 10 bytes
Size: a[11] is   8 bytes, *a[11] is   1 bytes, Length of a[11] is 10
a[ 11] is now 222222222
I is now 12


  0. ▒"▒
  1. 22
  2. 333
  3. 4444
  4. 55555
  5. 666666
  6. 7777777
  7. 88888888
  8. 9999999
  9. 0000000000
 10. 11111111111
 11. 222222222

Answer 1

代码中有很多冗余。首先，您只为字符串 s 分配一个字节并将字符串读入其中，这将导致未定义的行为(主要是导致问题的原因)。 a 也是如此。

必须是

char **a = (char**)malloc(sizeof(char*) * SOME_CONSTANT);

接下来，您一次为数组中的每个字符分配一个字符，这可以在一行中完成(您的代码效率太低。对 malloc 的大量函数调用)。这里

for(j=0; j<(size+1); j++)   a[i][j] = (char)(malloc(sizeof(char)));

可以是

a[i] = (char*)(malloc(sizeof(char)  * (size+1)));