logEvent.tijd only has eight elements, so accessing the ninth element is pointless.
Even if you meant to use strcpy, you need nine elements to store an eight character string.
Use strcpy to fill up the memory buffer of your struct with the string
You can't initialize your buffer the way you do it. It is not the correct way to do an array or cstring assignment in C
logEvent.tijd[8] is the 9th element (which is beyond your array size as AWOL pointed to give you a hint) of your buffer and is of type character (char). If that memory was legit what you can store in there is a char or anything that can be converted to a char. logEvent.tijd[5] = 'A'; would be OK because 'A' is a char.
{"13:51:00"} is an array of 1 element of type pointer to a constant string (const char *)
So when you do logEvent.tijd[8] = {"13:51:00"}; you are trying to store In a (overflown) memory location designed for a char a pointer to a constant string which obviously makes no sense at all, hence the compiler telling you invalid conversion from 'const char*' to 'char'
baukepluggee:
error message invalid conversion from 'const char*' to 'char'
Regardless of your code, the error you are getting occurs when you try to place the contents of one variable type into another that doesn't match. Here's an example:
void send (const char *buffer)
{
char *bufptr;
bufptr = buffer; // make a movable buffer pointer
// ...............
}
That code will give you the "invalid conversion from 'const char*' to 'char'" error because the compiler can't make a non-constant variable (bufptr) into a constant variable (buffer).
The solution is to typecast the variable - like this: