Hello, a user reported <https://bugzilla.redhat.com/show_bug.cgi?id=1199959> that indent does not support compound literals (see <http://www.drdobbs.com/the-new-c-compound-literals/184401404>) and that it can cause indent to print errors and misindent the code.
Input:
#include <stdio.h>
int main(int argc, char **argv)
{
typedef struct {
char *capital;
char *country;
} t_association;
t_association a = { "foo", "bar" };
t_association m;
m = (t_association){"Paris", "France"};
m = (t_association){"Rome", "Italy"};
if (1)
m = (t_association){"Paris", "France"};
else
m = (t_association){"Rome", "Italy"};
return 0;
}
Output:
$ indent -br < test.c
#include <stdio.h>
int
main (int argc, char **argv)
{
typedef struct
{
char *capital;
char *country;
} t_association;
t_association a = { "foo", "bar" };
t_association m;
m = (t_association) {
"Paris", "France"};
m = (t_association) {
"Rome", "Italy"};
if (1)
m = (t_association) {
"Paris", "France"};
indent: Standard input:16: Error:Unmatched 'else'
else
m = (t_association) {
"Rome", "Italy"};
return 0;
}
As you can see, the `m = (t_association){"Paris", "France"};' is recgnized as
function call (or just a new statement block) and it's broken at the left
brace.
Then the real if-else lines are thought to be misballanced and an error is
printed.
I believe indent should handle the assignement the same way as the definition
of `a' above. Also please note that a coumpound statement can be used as part
of bigger expression or to be simply void expression. Any idea how to fix it?
-- Petr
pgp7vJfPBZTeE.pgp
Description: PGP signature
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