如何将 char** 传递给需要 const char 指针数组的函数

Question

如下所示，我想将

char** x

定义的变量以只读方式传递给函数。

参考书显示了线性搜索源代码，通过将其作为由

char *x

定义的临时参数传递给

search

函数，对由

const int a[]

定义的变量执行线性搜索。

所以我想到了一个想法，如果是字符串呢？然后我写了下面的代码。

gcc、clang 和 MSVC 分析可以在这里找到。

// Linear Search

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

int scanf_s(const char *format ,...); // for gcc and clang environment
// int search(const char **a, int n, char *key) {       //※[1]not allowed in c lang specification⇒NG
// int search(char* const *a, int n, char *key) {       //※[2]⇒NG
// int search(const char* const *a, int n, char *key) { //※[3]warning occured in gcc and clang！！
// int search(char const * const *a, int n, char *key) {//※[4]same as above
// int search(const char* const a[], int n, char* key) {//※[5]same as above
// int search(const char* a[], int n, char* key) {      //※[6]I thought this was ok, but warning occured in gcc and clang！！
int search(char** a, int n, char* key) {            //in conclusion, gcc,clang and MSVC only allowing this!!
    int i = 0;

    for (i = 0; i < n; i++) {
        if (strcmp(a[i], key) == 0)
            return i;
    }
    return -1;

    /* or while style!! the reference book shows this style!!
    while (1) {
        if (i == n)
            return -1;
        if (strcmp(a[i], key) == 0)
            return i;
        i++;
    }
    */
}

int main(void) {
    char** x;
    char* ky;
    int nx;
    int idx;
    int i;

    puts("Linear Search");
    printf("How many Elements??：");
    scanf_s("%d", &nx);
    
    x = malloc(sizeof(char*) * nx);
    if (x == NULL) {
        printf("Pointer to Pointer x malloc failed!!\n");
        exit(EXIT_FAILURE);
    }

    for (i = 0; i < nx; i++) {
        printf("x[%d]：", i);
        x[i] = malloc(sizeof(char) * 35);
        if (x[i] == NULL) {
            printf("Pointer Array x[] malloc failed!!\n", i);
            exit(EXIT_FAILURE);
        }
        scanf_s("%s", x[i], 35);
    }

    printf("Target Value：");
    ky = malloc(sizeof(char) * 35);
    if (x == NULL) {
        printf("target value malloc failed!!\n");
        exit(EXIT_FAILURE);
    }
    // Or
    // ky = calloc(35, sizeof(char));
    scanf_s("%s", ky, 35);

    idx = search(x, nx, ky);

    if (idx == -1)
        puts("no target value.");
    else
        printf("%s is in x[%d]！！\n", ky, idx);

    free(ky);
    for (i = 0; i < nx; i++) {
        free(x[i]);
    }
    free(x);
    system("pause");
    return 0;
}

带有

const

的代码会使 gcc 和 clang 显示警告，尽管 MSVC 可以编译而没有任何警告，如注释（※[1]~[6]）所示。

那么，在

search

函数必须接受

char** x

定义的变量作为只读参数的情况下，如何编写代码呢？

Answer 1

编译器希望数组中的字符串是

const char*

类型。

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

#define STRING_LENGTH 36

int scanf_s(const char *format ,...);

unsigned int search(const char* const stringArray[], unsigned int stringCount, const char* stringToSearch)
{
   for(unsigned int i = 0; i < stringCount; i++)
   {
      if(strcmp(stringArray[i], stringToSearch) == 0)
      {
         return i;
      }
   }

   return UINT_MAX;
}

int main(void)
{
   printf("Enter the count of strings: ");
   unsigned int strCount;
   scanf_s("%u", &strCount);
   char* *strArray = malloc(sizeof(char*) * strCount);
   if(strArray == NULL)
   {
       printf("Could not get memory for the string array!\n");
       return EXIT_FAILURE;
   }

   for(unsigned int i = 0; i < strCount; i++)
   {
       strArray[i] = calloc(sizeof(char), STRING_LENGTH);
       if(strArray[i] == NULL)
       {
          printf("Could not get memory for the next string!\n");
          return EXIT_FAILURE;
       }
       printf("Enter %ith string (%i characters at most): ", (i + 1), (STRING_LENGTH - 1));
       scanf_s("%s", strArray[i], STRING_LENGTH);
   }

   char* strToSearch = calloc(sizeof(char), STRING_LENGTH);
   if(strToSearch == NULL)
   {
      printf("Could not get memory for the string to be searched!\n");
      return EXIT_FAILURE;
   }
   printf("Enter string to be searched (%i characters at most) : ", (STRING_LENGTH - 1));
   scanf_s("%s", strToSearch, STRING_LENGTH);

   unsigned int result = search((const char* const*)strArray, strCount, strToSearch);
   if(result != UINT_MAX) {
      printf("String found at: %u\n", result);
   } else {
      printf("String was not found!\n");
   }

   free(strToSearch);
   for(unsigned int i = 0; i < strCount; i++)
   {
       free(strArray[i]);
   }
   free(strArray);

   return EXIT_SUCCESS;
}

我在您提供的链接中尝试了此代码。 当我用 gcc 编译并运行它时，示例输出如下：

Enter the count of strings: 3
Enter 1th string (35 characters at most): Goodbye
Enter 2th string (35 characters at most): Cruel
Enter 3th string (35 characters at most): World
Enter string to be searched (35 characters at most) : Cruel
String found at: 1

注意：最后一个版本更好。 :)

检查这里