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。通过这种方式，您可以读取任何文件，无论字符串长度如何，它都会动态构建一个数组，并用文件填充它。我把它作为一个练习来练习C，希望能把它推广到其他类型和结构中。
然而，由于某种原因，我的第一个条目不匹配，导致了意外的行为。我知道，使用C，你需要基本上对所有内存进行微观管理，在使用代码时，我试图为每个条目分配内存，但这是正确的方法吗？我在脑海中运行了代码，从逻辑上讲，从0个条目开始是有道理的，但我不明白为什么第一个条目失败，而其余条目有效。
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 %dn",s,size,i);
printf("Using MALLOC with %d bytesn",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 %dn",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] = ' ';
//Print it out and increase i
printf("a[%3d] is now %sn",i,a[i]);
i++;
}
printf("I is now %dnnn",i);
a[i] = NULL;
//print out array
for(j=0; j<i; j++) printf("%3d. %-40sn",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<numbers.txt
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
代码中有很多冗余。首先，您只为字符串s分配一个字节，并向其中读取一个字符串，这将导致undefined behavior（主要是问题的原因）。a也是如此。
必须是
char **a = (char**)malloc(sizeof(char*) * SOME_CONSTANT);
接下来，您将为数组中的每个字符一次分配一个字符，这可以在一行中完成（您的代码效率太低。对malloc进行了大量函数调用）。此处
for(j=0; j<(size+1); j++) a[i][j] = (char)(malloc(sizeof(char)));
哪个可以是
a[i] = (char*)(malloc(sizeof(char) * (size+1)));
您没有分配足够的内存。
char *s = (char *) malloc(sizeof(char));
...
while( scanf("%79s",s) == 1){
您分配1个字节sizeof char，然后将79个字节读入该地址。对于其他78个字节来说，它不会很好。任何事情都能奏效，这纯粹是运气使然；只是碰巧你没有使用任何被无意覆盖的内存。