I'm trying to use switch case with char[2] in C, but it only supports integers.
int main() {
char c[2] = "h";
switch (c) {
case "h":
printf("done!");
}
return 1;
}
For better understanding, what I'm trying to do is:
if "h" in ")($+#&@+)#"
Basically I want to make a condition that matches a character with a group of characters, but efficiently, the first thing that came to mind is to use switch case but it only supports int. But when I use (int)c which is recommended in other stackoverflow answers, it doesn't return the ascii value.
Using the switch statement in this case does not make a sense.
Just use the standard function strchr declared in <string.h>.
#include <string.h>
//...
if ( strchr( c, 'h' ) != NULL ) puts( "done!" );
You can't compare arrays of characters (strings) using a switch statement directly, because switch statements only work with fundamental types; not arrays of fundamental types.
The reason using (int)c isn't returning the ASCII value is because you're casting char[]->int instead of char->int.
The difference between single and double quotes is an important one in C:
'h' is of type char, and is equal to 104 (Lowercase 'h' in ASCII)"h" is of type const char*, and points to 2 characters: h, & the null terminating character \0To check if a character is present in a string, you can use the strchr function as mentioned in Vlad from Moscow's answer, or if you want to do it yourself using switch statements you could try this:
int main()
{
char c[2] = "h";
bool found = false;
for (int i = 0; i < 2; ++i) {
switch (c[i]) {
case 'h': // compare c[i] == 'h'
found = true;
break;
default:
break;
}
}
if (found)
printf("Done!");
return 0;
}