How can variable field width be implemented with printf()?

Question

The question is :

How can variable field width be implemented using printf()? That is, instead of %8d, the width should be s

Answer 1

First of all, let me tell you, the code you have shown is about controlling the precision, not the field width. For a shortened form^**

 %A.B<format specifier>

A denotes the field width and B makes the precision.

Now, quoting the C11 standard, chapter §7.21.6.1, fprintf() (emphasis mine)

Each conversion specification is introduced by the character %. After the %, the following appear in sequence:

[..]

• An optional precision that gives the minimum number of digits to appear for the d, i, o, u, x, and X conversions, the number of digits to appear after the decimal-point character for a, A, e, E, f, and F conversions, the maximum number of significant digits for the g and G conversions, or the maximum number of bytes to be written for s conversions. The precision takes the form of a period (.) followed either by an asterisk * (described later) or by an optional decimal integer; if only the period is specified, the precision is taken as zero. If a precision appears with any other conversion specifier, the behavior is undefined.

and

As noted above, a field width, or precision, or both, may be indicated by an asterisk. In this case, an int argument supplies the field width or precision. [...]

So, in your case,

printf("\"%.*s\"\n", i, text);

the precision will be supplied by i which can hold different values at run-time.

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The complete format ^{(broken down in separate lines for ease of readability)}

%
<Zero or more flags>
<optional minimum field width>
<optional precision>
<optional length modifier>
<A conversion specifier character>