I am new to C. I just want to know why initializing the array of int with int is working and why initializing an array of char with char is not working. Or am I wrong thinking that "1" is a char?
#include <stdio.h>
int main()
{
int incoming_message_test[2] = {1, 2}; // why does this work?
char incoming_message[2] = {"1", "2"}; // why does this not work?
return 0;
}
You must change "1", "2" with '1', '2'
This question was a bit deeper than I first saw.
The first works because char is an integer type, and this code is perfectly valid:
int x = 42;
char c = x;
However, string literals cannot be converted that way. Instead of "1", use '1'.
In C:
'c' is a char"c" is a string, meaning an array of N+1 chars. In this case char 'c' followed by string terminator '\0'.In C, a character literal contains one character that is surrounded with a single quotation ( ').
So instead of "1", "2", use '1', '2'.
Or am I wrong thinking that "1" is a char ?
Yes.
For single characters, use single quotes.
So, define you array like so:
char incoming_message[2] = {'1', '2'};
Single-quote character literals have the type char, while double-quote string literals have the type char * (a pointer (address) to a char).
"1"is not achar, it's an array of twochars.