如何在C中使用2D数组指针来存储字符串？

Lets assume that I have a file, that contains the following strings:

让我们假设我有一个包含以下字符串的文件：

option0 value0
option1 value1
....... ......
optionX valueX

I am trying to make a C program to read the file. I decided to create a 2D array of pointers to strings and store all options in the pointee(s) of the first row and all values in the pointee(s) of the second row. This way:

我正在尝试制作一个C程序来读取该文件。我决定创建一个指向字符串的指针的2D数组，并将所有选项存储在第一行的指针对象中，并将所有值存储在第二行的指针对象中。这条路：

pointer[0][0] should point to option0
pointer[1][0] should point to value0
pointer[0][1] should point to option1
pointer[1][1] should point to value1

It seems that the program reads the file successfully, but I can not access the strings to print them out. Here is a stripped down version of the code, which causes the problem:

程序似乎成功读取文件，但我无法访问字符串以将其打印出来。这是代码的精简版本，导致问题：

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

int main (void) {

#define BUFF 1024

FILE *fd;
int i = 0;
char buff[BUFF];
char *options [2][20];

fd = fopen("foo.txt", "r");

while (!feof(fd)) {

if (fgets(buff, sizeof(buff), fd) != NULL) {

        (options[0][i] = malloc(sizeof (char) * 512)
        (options[1][i] = malloc(sizeof (char) * 512)

                sscanf(buff, "%s %s", options[0][i], options[1][i]);
                i++;

}
}

while (i >= 0) {

printf ("Option %i is %s its value is %s", i, *option[0][i], *option[1][i]);

i--;
}
fclose(fd);
return 0;
}

I played a few days with that, but I can not seem to make it work. I checked various * questions and other online resources, but even experimenting did not help me figure out that.

我玩了几天，但我似乎无法使它工作。我检查了各种*问题和其他在线资源，但即使是试验也没有帮我弄明白。

What I want to know is: Are the strings stored properly with malloc and sscanf? If 'no', why? Why am I not able to print the string with the printf statement? How to fix that?

我想知道的是：字符串是否与malloc和sscanf一起正确存储？如果'不'，为什么？为什么我无法使用printf语句打印字符串？如何解决？

Please be aware that the code shown is only for testing purposes, so I am aware that there is no error checking, etc..

请注意，显示的代码仅用于测试目的，因此我知道没有错误检查等。

2 个解决方案

#1

You are dereferencing the pointers when giving them to printf. It expects char*, not char for %s.

在将它们提供给printf时，您正在取消引用指针。它期望char *，而不是％s的char。

printf ("Option %i is %s its value is %s", i, *option[0][i], *option[1][i]);

This should be

这应该是

printf ("Option %i is %s its value is %s", i, option[0][i], option[1][i]);

Another issue is you have to decrement i before printing in the while loop. You increment it after each memory allocation, so in the end it points past the last element. Move the i--; to be the first inside the while loop.

另一个问题是你必须在while循环中打印之前减少i。你在每次内存分配后递增它，所以最后它指向最后一个元素。移动i--;成为while循环中的第一个。

Also sizeof(char) is always 1, so you don't need that in calculating sizes for malloc().

sizeof（char）也始终为1，因此在计算malloc（）的大小时不需要它。

Please also adjust your compiler's warnings to a higher level. For example gcc with -Wall will say:

还请将编译器的警告调整到更高级别。例如gcc with -Wall会说：

warning: format '%s' expects argument of type 'char *', but argument 2 has type 'int' [-Wformat=]

警告：格式'％s'需要类型为'char *'的参数，但参数2的类型为'int'[-Wformat =]

#2

1) i-- has to come before the printf

1）我必须在printf之前来

2) As in the answer of Sami, dont dereference the pointers in printf

2）正如在Sami的答案中，不要取消引用printf中的指针

while (0 <=-- i)
    printf ("Option %i is %s its value is %s", i, option[0][i], option[1][i]);

#1