C编程警告: array subscript has type 'char' [-Wchar-subscripts]

Question

我似乎无法解决这个问题。下面是我的代码:

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

_Bool are_anagrams (const char *word1, const char *word2);

int main (void)
{
    char an1[30], an2[30];
    int j;
    printf("Enter first word: ");
    scanf("%s", an1);
    printf("Enter second word: ");
    scanf("%s", an2);
    printf("The words are");

    j = are_anagrams (an1, an2);

    if (j == 0)
    {
        printf(" not anagrams. \n");
    }else
        printf(" anagrams. \n");

    return 0;
}

_Bool are_anagrams (const char *word1, const char *word2)
{
    int i;
    int check[26] = {0};
    for(i=0; i<30; i++)
        if(word1[i] == '\0')
            i=40;
        else
        {
            word1[i] = toupper(word1[i]);
            check[word1[i]-65]++;
        }

    for(i=0; i<30; i++)
        if(word2[i] == '\0')
            i=40;
        else
        {
            word2[i] = toupper(word2[i]);
            check[word2[i]-65]--;
        }

    for(i=0; i<26; i++)
        if(check[i] != 0)
        {
            return 0;
        }

    return 1;
}

这些是错误信息:

anagram1.c:38:3: warning: array subscript has type ‘char’ [-Wchar-subscripts]
   word1[i] = toupper(word1[i]);
   ^
anagram1.c:38:3: error: assignment of read-only location ‘*(word1 + (sizetype)((long unsigned int)i * 1ul))’
anagram1.c:46:4: warning: array subscript has type ‘char’ [-Wchar-subscripts]
    word2[i] = toupper(word2[i]);
    ^
anagram1.c:46:4: error: assignment of read-only location ‘*(word2 + (sizetype)((long unsigned int)i * 1ul))’

Answer 1

警告:

warning: array subscript has type ‘char’

是“toupper()”需要“int”类型作为参数的结果，而问题代码提供的是“char”类型。

word1[i] = toupper(word1[i]);
...
word2[i] = toupper(word2[i]);

要消除警告，给 toupper() 'int' 值:

word1[i] = toupper((unsigned char)word1[i]);
...
word2[i] = toupper((unsigned char)word2[i]);

为了彻底，您可以将“toupper()”返回的值从“int”转换回“char”:

word1[i] = (char)toupper((unsigned char)word1[i]);
...
word2[i] = (char)toupper((unsigned char)word2[i]);

---

错误:

error: assignment of read-only location

是尝试使用“const”标志修改值的结果:

_Bool are_anagrams (const char *word1, const char *word2)

如果合适，您可以通过消除“const”标志来消除错误:

_Bool are_anagrams (char *word1, char *word2)

或者，您可以制作“const”字符串的本地工作副本:

_Bool are_anagrams (const char *I__word1, const char *I__word2)
   {
   int rCode = 0;
   int i;
   int check[26] = {0};
   char *word1 = strdup(I__word1);
   char *word2 = strdup(I__word2);

   for(i=0; i<30; i++)
      if(word1[i] == '\0')
         i=40;
      else
         {
         word1[i] = toupper(word1[i]);
         check[word1[i]-65]++;
         }

   for(i=0; i<30; i++)
      if(word2[i] == '\0')
         i=40;
      else
         {
         word2[i] = toupper(word2[i]);
         check[word2[i]-65]--;
         }

   for(i=0; i<26; i++)
      if(check[i] != 0)
        goto CLEANUP;

   rCode=1;

CLEANUP:
   free(word2);
   free(word1);

   return(rCode);
   }

注意:以上代码使用问题代码正文，可能准确也可能不准确。本回答无意修复问题代码中的其他问题；只是为了演示通过创建参数的非“const”副本来绕过参数上的“const”标志的正确方法