The following code snippet gives unexpected output in Turbo C++ compiler:
char a[]={'a','b','c'};
printf("%s",a);
Why doesn't this print abc? In my understanding, strings are implemented as one dimensional character arrays in C.
Secondly, what is the difference between %s and %2s?
This is because your string is not zero-terminated. This will work:
char a[]={'a','b','c', '\0'};
The %2s specifies the minimum width of the printout. Since you are printing a 3-character string, this will be ignored. If you used %5s, however, your string would be padded on the left with two spaces.
char a[]={'a','b','c'};
Well one problem is that strings need to be null terminated:
char a[]={'a','b','c', 0};
Without change the original char-array you can also use
char a[]={'a','b','c'};
printf("%.3s",a);
or
char a[]={'a','b','c'};
printf("%.*s",sizeof(a),a);
or
char a[]={'a','b','c'};
fwrite(a,3,1,stdout);
or
char a[]={'a','b','c'};
fwrite(a,sizeof(a),1,stdout);
Because you aren't using a string. To be considered as a string you need the 'null termination': '\0' or 0 (yes, without quotes).
You can achieve this by two forms of initializations:
char a[] = {'a', 'b', 'c', '\0'};
or using the compiler at your side:
char a[] = "abc";
Whenever we store a string in c programming, we always have one extra character at the end to identify the end of the string.
The Extra character used is the null character '\0'.
In your above program you are missing the null character.
You can define your string as
char a[] = "abc";
to get the desired result.