I am reading through a “Learn C” book right now and have come across a question I really don’t understand. The point of the exercise is to find the problem with this code:
char c;
c = 'a';
printf("c holds the character %c.",c);
..and then it gives the explanation that: “The text string “a” is composed of two characters, both ‘a’ and the terminating zero byte. The variable c is only a single byte in size. Even if c were 2 bytes long, you couldn’t copy a text string this way. Try copying the text one byte at a time into a variable large enough to hold the text string and its terminating zero byte.”
However, when I run the code above – it works perfectly fine. I thought I understood the theory behind why it is bad – the whole terminating 0 at the end of a string thing, so I rewrote the code like this to test:
char c[2];
*c = 'a';
printf("c holds the character %c.",c);
But this generates a problem. I am starting to get confused as to the problem. Wouldn’t this 2nd set of code pass the letter ‘a’ to the pointer at c[0] and then put the terminating 0 at c[1] – fully using the 2 spaces allotted for that array?
The text is wrong, or else you’re transcribing the code incorrectly.
In C, a double-quoted string is null-terminated. A single-quoted character is just one character.