英文:
Trying to make an if statement that checks if a text has ., ! or ? in it but it keeps giving errors
问题
以下是代码部分的翻译：
这是我的代码：导致错误的函数是 `count_sentences`。int count_letters(string text);int count_words(string text);int count_sentences(string text);int main(void){string text = get_string("Text: ");printf("%s\n", text);int letters = count_letters(text);printf("%i letters\n", letters);int words = count_words(text);printf("%i words\n", words);int sentences = count_sentences(text);printf("%i sentences\n", sentences);}int count_letters(string text){int letters = 0;for (int i = 0, len = strlen(text); i < len; i++){if (isalpha(text[i])){letters++;}}return letters;}int count_words(string text){int words = 0;for (int i = 0, len = strlen(text); i < len; i++){if (text[i] == ' '){words++;}}return words + 1;}int count_sentences(string text){int sentences = 0;for (int i = 0, len = strlen(text); i < len; i++){if (text[i] == '.' || text[i] == '!' || text[i] == '?'){sentences++;}}return sentences;}它给我报错：readability.c:66:28: 错误：使用带有常量操作数的逻辑‘||’ [-Werror，-Wconstant-logical-operand]if (text[i] == '.' || '!' || '?')^ ~~~readability.c:66:28: 注意：用‘|’进行位运算if (text[i] == '.' || '!' || '?')^~|致命错误：错误数过多，现在停止 [-ferror-limit=]有人知道如何修复吗？我知道我可以只需创建一个`else if`块并获得相同的结果，但我想知道是否可以这样做。它似乎更有效，并且看起来更好看。stackoverflow给我一个主要是代码的错误，所以请忽略这个。
英文:
Here is my code: the function that gives me errors is count_sentences.
int count_letters(string text);int count_words(string text);int count_sentences(string text);int main(void){string text = get_string("Text: ");printf("%s\n", text);int letters = count_letters(text);printf("%i letters\n", letters);int words = count_words(text);printf("%i words\n", words);int sentences = count_sentences(text);printf("%i sentences\n", sentences);}int count_letters(string text){int letters = 0;for (int i = 0, len = strlen(text); i < len; i++){if (isalpha(text[i])){letters++;}}return letters;}int count_words(string text){int words = 0;for (int i = 0, len = strlen(text); i < len; i++){if (text[i] == ' '){words++;}}return words + 1;}int count_sentences(string text){int sentences = 0;for (int i = 0, len = strlen(text); i < len; i++){if (text[i] == '.' || '!' || '?'){sentences++;}}return sentences;}
Its giving me:
readability.c:66:28: error: use of logical '||' with constant operand [-Werror,-Wconstant-logical-operand]if (text[i] == '.' || '!' || '?')^ ~~~readability.c:66:28: note: use '|' for a bitwise operationif (text[i] == '.' || '!' || '?')^~|fatal error: too many errors emitted, stopping now [-ferror-limit=]
Anyone know how to fix it? I know I can just make an else if block and get the same result but I would like to know if I could do it this way. it seems more effective and looks a lot nicer.
stackoverflow is giving me an error that is mostly code so ignore this please
答案1
得分: 3
if (text[i] == '.' || text[i] == '!' || text[i] == '?')
与
if (true)
具有相同的意思。这是因为 text[i] == '.' 会评估为 true 或 false，而 ! 和 ? 两者都会评估为 true，因为它们的数字值不为零。
正确的测试应该是
if (text[i] == '.' || text[i] == '!' || text[i] == '?')
或者更简单的方法是使用 strpbrk 来检查整个字符串，它会返回找到的字符的指针，如果没有找到要搜索的字符，则返回 NULL：
unsigned count_sentences(const char* text) {unsigned sentences = 0;for (; *text; ++text) {text = strpbrk(text, ".!?"); // 搜索任何一个字符'.', '!', '?'if (text == NULL) break; // 如果没有找到这些字符中的任何一个++sentences;}return sentences;}
<details><summary>英文:</summary>```cif (text[i] == '.' || '!' || '?')
has the same meaning as
if (true)
That's because text[i] == '.' evaluates to either true or false, but both '!' and '?' evaluate to true, since their numeric value is not zero.
The correct test would be
if (text[i] == '.' || text[i] == '!' || text[i] == '?')
or simpler, check the whole string with strpbrk which returns a pointer to the found char or NULL if none of the characters to search for is found:
unsigned count_sentences(const char* text) {unsigned sentences = 0;for (; *text; ++text) {text = strpbrk(text, ".!?"); // search for any of '.', '!', '?'if (text == NULL) break; // if none of the chars was found++sentences;}return sentences;}
答案2
得分: 2
|| 运算符比较两个布尔值，并在其中一个为 true 时停止评估。
在你的情况下，text[i] == '.' || '!' || '?' 这意味着：
- 评估
text[i] == '.'。如果为 true，则停止并返回 true。否则， - 评估
'!'。这是一个常量字符。因为它不等于 0，所以始终为 true。这不是你想要表达的逻辑，编译器检测到测试常量变量是奇怪的。这是一个警告，因为程序可以编译通过。很好地启用了-Werror以便早期检测到这个问题！
你想要做的是继续将 text[i] 与其他常量进行比较，如下所示：
text[i] == '.' || text[i] == '!' || text[i] == '?'。
由于这变得相当冗长，你可能希望稍后添加其他特殊字符，因此你会想使用更简洁的语法。请参考这个问题以获取更多信息：https://stackoverflow.com/questions/1071542/in-c-check-if-a-char-exists-in-a-char-array
英文:
The || operator compares two booleans and stops evaluating them as soon as one is true.
In your case, text[i] == '.' || '!' || '?', this means:
- evaluate
text[i] == '.'. If true, stop and return true. Otherwise, - evaluate
'!'. This is a constant character. Because it is not 0, it is always true. That's not the logic you want to express and the compiler detects that it is weird to test a constant variable. This is a warning because the program could compile like that. Good job on activating-Werrorto detect this early!
What you want to do is keep comparing text[i] to other constants like that:
text[i] == '.' || text[i] == '!' || text[i] == '?'.
Since this is getting quite verbose and you might want to add other special characters later, you will want to use a more condensed syntax. Refer to this question for that: https://stackoverflow.com/questions/1071542/in-c-check-if-a-char-exists-in-a-char-array
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