如何检查字符串是否包含某个字符?

问题描述

对于C编程我还是很陌生，如果有的话，我如何能够检查字符串是否包含某个字符?

I'm fairly new to C programming, how would I be able to check that a string contains a certain character for instance, if we had:

void main(int argc, char* argv[]){

  char checkThisLineForExclamation[20] = "Hi, I'm odd!"
  int exclamationCheck;
}

因此，我如何将 exclamationCheck 设置为1(如果为！")呢?存在，如果不存在则为0?非常感谢您提供的任何帮助.

So with this, how would I set exclamationCheck with a 1 if "!" is present and 0 if it's not? Many thanks for any assistance given.

推荐答案

通过使用 strchr()，例如:

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

int main(void)
{
  char str[] = "Hi, I'm odd!";
  int exclamationCheck = 0;
  if(strchr(str, '!') != NULL)
  {
    exclamationCheck = 1;
  }
  printf("exclamationCheck = %d\n", exclamationCheck);
  return 0;
}

输出:

如果您正在寻找一种简洁的班轮，那么您可以按照@melpomene的方法进行操作:

If you are looking for a laconic one liner, then you could follow @melpomene's approach:

int exclamationCheck = strchr(str, '!') != NULL;

---

如果不允许使用C字符串库中的方法，则按照@SomeProgrammerDude的建议，您可以简单地对字符串进行迭代，并且如果有任何字符是感叹号，如本示例所示:

---

If you are not allowed to use methods from the C String Library, then, as @SomeProgrammerDude suggested, you could simply iterate over the string, and if any character is the exclamation mark, as demonstrated in this example:

#include <stdio.h>

int main(void)
{
  char str[] = "Hi, I'm odd";
  int exclamationCheck = 0;
  for(int i = 0; str[i] != '\0'; ++i)
  {
    if(str[i] == '!')
    {
      exclamationCheck = 1;
      break;
    }
  }
  printf("exclamationCheck = %d\n", exclamationCheck);
  return 0;
}

输出:

请注意，当找到至少一个感叹号时，您可以中断循环，这样就无需遍历整个字符串.

Notice that you could break the loop when at least one exclamation mark is found, so that you don't need to iterate over the whole string.

PS: main()在C和C ++中应该返回什么? int ，而不是 void .

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