GCC does not recognize escape character '\' on multi-line input

Question

The below will raise compilation warning warning: unknown conversion type character ‘w’ in format [-Wformat=] on a new version of gcc: gcc (Debian 12.2.0-14) 12.2.0

#include <stdio.h>


int main() {
    char query[4096];
    snprintf(query, sizeof(query), "SELECT e.expression_id, e.name, e.channel, e.subsidiary_of, ds.input_channel FROM expression e "
            " JOIN shoe s ON s.shoe_id = e.shoe_id JOIN device_serial ds ON s.device_id = ds.device_id " 
            "WHERE e.name NOT LIKE '\%walker\%' ORDER BY e.channel");
    printf("%s \n", query);
    return 0;
}

It works fine on a shorter input and also on older versions of gcc. How can I make it recognize the backslash as an escape character?

I want to include `LIKE '%walker%'` in the char array. The slashes are to escape the percent symbol from being treated like a format string — Cheetaiean, Mar 29 '23 at 20:22
To print a literal `%` sign you need to double it as in `"%%"` — Eugene Sh., Mar 29 '23 at 20:23
I don't understand the point of calling snprintf when you aren't formatting any variables. That's just a very expensive string copy with extra escaping rules you don't need. — Paul Dempsey, Mar 30 '23 at 02:21
the query will be used to execute a PSQL command, I simplified the code for the purposes of the question — Cheetaiean, Mar 31 '23 at 03:09

Vlad from Moscow · Accepted Answer · 2023-03-29T20:28:59.287

4

Just write

"WHERE e.name NOT LIKE '%%walker%%' ORDER BY e.channel"

From the C Standard (7.21.6.1 The fprintf function)

8 The conversion specifiers and their meanings are:

% A % character is written. No argument is converted. The complete conversion specification shall be %%.

edited Mar 29 '23 at 20:28

answered Mar 29 '23 at 20:23

Vlad from Moscow

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GCC does not recognize escape character '\' on multi-line input

1 Answers1