Check If a String Contains a String in C++

Check substring exists in a string in C

if (strstr(sent, word) != NULL) {
    /* ... */
}

Note that strstr returns a pointer to the start of the word in sent if the word word is found.

How do I check if a string contains a certain character?

By using strchr(), like this for example:

#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;
}

Output:

exclamationCheck = 1

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

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

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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;
}

Output:

exclamationCheck = 0

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.

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PS: What should main() return in C and C++? int, not void.

Seeing if a string contains a substring in C

Remove semicolon after the if condition. Like

if(strstr(outputArr[j], "@") != NULL)

Because

 if(strstr(outputArr[j], "@") != NULL);

is equivalent to

if(strstr(outputArr[j], "@") != NULL)
{

}

C99-6.8.3 paragraph 3:

A null statement (consisting of just a semicolon) performs no operations.

Simple way to check if a string contains another string in C?

if (strstr(request, "favicon") != NULL) {
    // contains
}

How to check if string contains a substring at the end?

Your function has problems:

• you omit the return type, this is no longer supported by Standard C.
• The first for loop effectively erases the string pointed to by str2. There is no guarantee it even has a null terminator.
• the while loop invokes undefined behavior if the str1 does not contain a . or if it is shorter than str2.
• do not hardcode ASCII values such as 97 or 122. It is neither portable not even readable. Use 'a' and 'z' or preferably the functions defined in <ctype.h>.

Here is a simple string function for your purpose:

int str_ends_with(const char *s, const char *suffix) {
    size_t slen = strlen(s);
    size_t suffix_len = strlen(suffix);

    return suffix_len <= slen && !strcmp(s + slen - suffix_len, suffix);
}

Checking if a string contains a substring C++

Wouldn't it be easier for strings to have two std::string instead of two char[20]?

Then you would be able to say this with no problem:

if (strings.string1.find(strings.string2) != std::string::npos) {
    std::cout << "found!" << '\n';
}

char is not a class like std::string, it doesn't have any member functions. This means char[20].find doesn't exist. std::string is a class, however. And it does have a member function called std::string::find(), so doing this to a std::string is legal.

Check if a string contains a string in C++

Use std::string::find as follows:

if (s1.find(s2) != std::string::npos) {
    std::cout << "found!" << '\n';
}

Note: "found!" will be printed if s2 is a substring of s1, both s1 and s2 are of type std::string.

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