#17 strings not working in a 'printf' command

[Ignacio Diaz-Emparanza]

It seems that the susbtitution of strings is not working in 'printf' commands. In the script below, the first loop does not use the string @varname so the printf command works ok, but the second loop uses that string and gretl gives an error.

<script>
nulldata 50
genr x = normal()
string varname = "x"
loop for i=1..50 --quiet
scalar ti=i
printf "%s\t %10.0f\n", date(ti), x[i]
endloop

loop for i=1..50 --quiet
scalar ti=i
printf "%s\t %10.0f\n", date(ti), @varname[i]
endloop
</script>

[Riccardo "Jack" Lucchetti]

Logged In: YES
user_id=1368575
Originator: NO

This is due to the fact that in a (s)printf statement string substitution ended with the format line; I think this was meant to optimise things. The following patch fixes it: Allin, please check I haven't overlooked something.

Index: lib/src/gretl_string_table.c

RCS file: /cvsroot/gretl/gretl/lib/src/gretl_string_table.c,v
retrieving revision 1.36
diff -u -r1.36 gretl_string_table.c
--- lib/src/gretl_string_table.c 14 Nov 2007 22:44:52 -0000 1.36
+++ lib/src/gretl_string_table.c 11 Dec 2007 20:29:10 -0000
@@ -992,10 +992,6 @@

while (*s && !err) {
if (pf) {
- if (*s == '"' && (bs % 2 == 0)) {
- /* reached end of format string: stop substituting */
- break;
- }
if (*s == '\\') {
bs++;
} else {

[Allin Cottrell]

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user_id=330339
Originator: NO

It is by design that string substitution does not operate for the
_arguments_ to (s)printf: in that context @xxx string variables are
treated as variables, not macros. See section 11.2 of the User's
Guide.

The proposed patch would break this sort of usage:

string foo = "avocado"
printd "foo = %s\n", @avocado

[Nobody/Anonymous]

Logged In: NO

I see. I hadn't thought of this.

Besides, the problem is easy to circumvent by assigning the value to be printed to a temporary scalar, eg

<script>
loop for i=1..50 --quiet
scalar ti=i
scalar tmp = @varname[i]
printf "%s\t %10.0f\n", date(ti), x
endloop
</script>

I guess this bug can be closed, then.