指向C中的字符串數組的指針

Question

在C使用指針很有趣（不是真的）。

我有幾個字符串數組我想以簡單的方式聲明，最好是：

arrayOfStrings1 = {"word1", "word2", etc. };
arrayOfStrings2 = {"anotherword1", "anotherword2", etc. };
arrayOfStrings3 = etc.
etc.

類似於翻譯數組的東西（但不完全），所以我希望能夠在運行時間間隔這些。 為此，我想要一個指針指針pointerToArrayOfStrings ，我可以交換：

pointerToArrayOfStrings = arrayOfStrings1;
doStuff();
pointerToArrayOfStrings = arrayOfStrings2;
doSomeOtherStuff();

在我對字符串數組和指向這些字符串的指針的天真理解中，這就是我所嘗試的：

// Danish transforms
const unsigned char* da_DK[] = {"b","bb","c","c","cc","d","dd","e","f","ff","g","gg","h","hh","j","j","jj","k","k","kk","l","l","l","l","ll","m","mm","n","n","nn","p","pp","r","r","r","rr","s","s","s","ss","t","t","tt","v","v","vv","æ"};

// British english transforms
const unsigned char* en_GB[] = {"a","a","a","a","a","a","a","a","a","a","a","a","a","age","ai","aj","ay","b","cial","cian","cian","dj","dsj","ea","ee","ege","ei","ei","eigh","eigh","f","f","f","g","g","gs","i","i","i","j","j","k","ks","kw","l","m","n","n","o","r","s","s","sd","sdr","sion","sion","sj","sj","tial","tion","tion","tj","u","u","u","u","w","ye","ye","z"};

    // More languages....

const unsigned char** laguageStrings;

// Assign language
if (streq(language, "da-DK")){
    laguageStrings= da_DK;
}
else if (streq(language, "en-GB")){
    laguageStrings= en_GB;
}
else
         return 0;
}

語言是包含語言“en-GB”，“da-DK”等的char * ， streq()只是一個家庭釀造（比strcmp()稍快的字符串比較功能）。

長話短說，取決於編譯器這種方法可能有效，報告編譯器警告或編譯，但會給出意想不到的結果。

解決這個問題的正確方法是什么？

Answer 1

有與字符（串）的陣列的工作雙向C 。 它們如下：

char a[ROW][COL];
char *b[ROW];

圖形表示在代碼中作為內聯注釋提供。

根據您希望如何表示字符數組（字符串），您可以按如下方式定義指向該字符串的指針

    char (*ptr1)[COL] = a;
    char **ptr2 = b;

它們基本上是不同的類型（以微妙的方式），因此指向它們的指針也略有不同。

下面的示例演示了在C使用字符串的不同方法，我希望它可以幫助您更好地理解C的字符數組（字符串）。

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

#define ROW 5
#define COL 10

int main(void) 
{
    int i, j;
    char a[ROW][COL] = {"string1", "string2", "string3", "string4", "string5"};
    char *b[ROW];

    /*

    a[][]

      0   1   2   3   4   5   6     7    8   9
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 1 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 2 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 3 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 4 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 5 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+

    */  

    /* Now, lets work on b */    
    for (i=0 ; i<5; i++) {
        if ((b[i] = malloc(sizeof(char) * COL)) == NULL) {
            printf("unable to allocate memory \n");
            return -1;
        }
    }

    strcpy(b[0], "string1");
    strcpy(b[1], "string2");
    strcpy(b[2], "string3");
    strcpy(b[3], "string4");
    strcpy(b[4], "string5");

    /*

       b[]              0   1   2   3   4   5   6    7     8   9
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 1 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 2 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 3 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 4 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 5 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+

    */

    char (*ptr1)[COL] = a;
    printf("Contents of first array \n");
    for (i=0; i<ROW; i++)
        printf("%s \n", *ptr1++);


    char **ptr2 = b;
    printf("Contents of second array \n");
    for (i=0; i<ROW; i++)
        printf("%s \n", ptr2[i]);

    /* b should be free'd */
    for (i=0 ; i<5; i++)
        free(b[i]);

    return 0;
}

Answer 2

解決這個問題的正確方法是什么？

好吧，正確的方法是使用專門用於處理多語言接口的庫 - 例如gettext 。

另一種方法，雖然比較簡潔，但是在其他語言/技術中使用哈希表 （也稱為“字典”或“哈希映射”或“關聯映射”）： 在C中尋找一個好的哈希表實現

這可能不是你想要的答案，但是你問錯了問題到了正確的問題。