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c - 为什么 '\\' 被认为是 int?

转载 作者:太空宇宙 更新时间:2023-11-04 08:36:57 25 4
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当试图比较 char* 和 '\' 以查看它们是否相同时,我只是使用了字符 == '\',但它给了我这个警告消息,当我运行它时我得到了段错误:11 .片段如下。

int strcompare(char *string1, char *string2) {
int i = 0;
while(string1[i] != 0 & string2[i] != 0) {
if (string1[i] != string2[i]) {
return false;
}
i++;
}
return true;
}


int special_character_is_in_pattern(char *pattern) {
int i = 0;
while(i<32) {
char *character = &pattern[i];
//int is_in = is_in_list(character, special_character_list);
return (strcompare(character, ".")|strcompare(character,"+")|strcompare(character,"?")|character == '\\') ? true : i++;
}
return false;
}

错误信息是这样的:

a.c:39:102: warning: comparison between pointer and integer ('char *' and 'int')
return (strcompare(character, ".")|strcompare(character,"+")|strcompare(character,"?")|character == '\\') ? true : i++;
~~~~~~~~~ ^ ~~~~
a.c:50:49: warning: incompatible integer to pointer conversion passing 'int' to parameter of type 'char *' [-Wint-conversion]
printf("%d\n",special_character_is_in_pattern('\\'));

需要一些帮助,谢谢

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