Question

Let’s say I have a char* str = '0123456789' and I want to cut the first and the last three letters and print just the middle, what is the simplest, and safest, way of doing it?

Now the trick: The portion to cut and the portion to print are of variable size, so I could have a very long char*, or a very small one.

Answer 1

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.