The simplest way of printing a portion of a char[] in C

You can use printf(), and a special format string:

char *str = "0123456789";
printf("%.6s\n", str + 1);

The precision in the %s conversion specifier specifies the maximum number of characters to print. You can use a variable to specify the precision at runtime as well:

int length = 6;
char *str = "0123456789";    
printf("%.*s\n", length, str + 1);

In this example, the * is used to indicate that the next argument (length) will contain the precision for the %s conversion, the corresponding argument must be an int.

Pointer arithmetic can be used to specify the starting position as I did above.

[EDIT]

One more point, if your string is shorter than your precision specifier, less characters will be printed, for example:

int length = 10;
char *str = "0123456789";
printf("%.*s\n", length, str + 5);

Will print “56789“. If you always want to print a certain number of characters, specify both a minimum field width and a precision:

printf("%10.10s\n", str + 5);

or

printf("%*.*s\n", length, length, str + 5);

which will print:

"     56789"

You can use the minus sign to left-justify the output in the field:

printf("%-10.10s\n", str + 5);

Finally, the minimum field width and the precision can be different, i.e.

printf("%8.5s\n", str);

will print at most 5 characters right-justified in an 8 character field.