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

Question

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;
}
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.

Answer 1

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

Answer 2

You can use plain search for ! character with
Code
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[] = "Hi, I'm odd!";
int exclamationCheck = 0;
int i=0;
while (str[i]!='\0'){
if (str[i]=='!'){
exclamationCheck = 1;
break;
}
i++;
}
printf("exclamationCheck = %d\n", exclamationCheck);
return 0;
}