[英]C: Finding a specific character in a string
I am trying to write a program that searches for the first occurance of a specific character in a string.我正在尝试编写一个程序来搜索字符串中特定字符的第一次出现。 But no matter what character i type in the number the program gives back is the number of the last character.
但无论我在数字中输入什么字符,程序返回的都是最后一个字符的数字。
int where (char *str, char ltr);
int main () {
char word [80];
char letter;
printf("Type in a word: ");
scanf("%s", word);
printf("Type in a character: ");
scanf("%s", letter);
printf("%d", where(word, letter));
}
int where (char *str, char ltr){
int i = 0;
while(i < strlen(str)){
if(ltr == str[i]){
break;
}
i++;
}
return i;
}
The problem is that this here问题是这里
scanf("%s", letter);
Is undefined behavior because it's expecting a pointer to a character, but you give it a character.是未定义的行为,因为它需要一个指向字符的指针,但你给它一个字符。 The
%s
specifier is for reading strings, not single characters. %s
说明符用于读取字符串,而不是单个字符。 Your compiler should warn you about a type mismatch here.您的编译器应该在此处警告您类型不匹配。
Instead, change it to this:相反,将其更改为:
scanf(" %c", &letter);
%c
is for reading in single characters, and with the leading space you make it ignore any leading whitespace in the input (such as the newline entered after you word). %c
用于读取单个字符,并使用前导空格使其忽略输入中的任何前导空格(例如在单词后输入的换行符)。
You should also change scanf("%s", word);
您还应该更改
scanf("%s", word);
to scanf("%79s", word);
到
scanf("%79s", word);
in order to avoid undefined behavior when the user enters a very long word (this limits the word's length to your buffer's size).为了避免用户输入很长的单词时出现未定义的行为(这会将单词的长度限制为缓冲区的大小)。
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