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/* This checks various ways of dead code inside if statements
where there are non-obvious ways of how the code is actually
not dead due to reachable by labels. */
extern int printf (const char *, ...);
static void kb_wait_1(void)
{
unsigned long timeout = 2;
do {
/* Here the else arm is a statement expression that's supposed
to be suppressed. The label inside the while would unsuppress
code generation again if not handled correctly. And that
would wreak havoc to the cond-expression because there's no
jump-around emitted, the whole statement expression really
needs to not generate code (perhaps except useless forward jumps). */
(1 ?
printf("timeout=%ld\n", timeout) :
({
int i = 1;
while (1)
while (i--)
some_label:
printf("error\n");
goto some_label;
})
);
timeout--;
} while (timeout);
}
int main (void)
{
int i = 1;
kb_wait_1();
/* Simple test of dead code at first sight which isn't actually dead. */
if (0) {
yeah:
printf ("yeah\n");
} else {
printf ("boo\n");
}
if (i--)
goto yeah;
/* Some more non-obvious uses where the problems are loops, so that even
the first loop statements aren't actually dead. */
i = 1;
if (0) {
while (i--) {
printf ("once\n");
enterloop:
printf ("twice\n");
}
}
if (i >= 0)
goto enterloop;
/* The same with statement expressions. One might be tempted to
handle them specially by counting if inside statement exprs and
not unsuppressing code at loops at all then.
See kb_wait_1 for the other side of the medal where that wouldn't work. */
i = ({
int j = 1;
if (0) {
while (j--) {
printf ("SEonce\n");
enterexprloop:
printf ("SEtwice\n");
}
}
if (j >= 0)
goto enterexprloop;
j; });
/* The other two loop forms: */
i = 1;
if (0) {
for (i = 1; i--;) {
printf ("once2\n");
enterloop2:
printf ("twice2\n");
}
}
if (i > 0)
goto enterloop2;
i = 1;
if (0) {
do {
printf ("once3\n");
enterloop3:
printf ("twice3\n");
} while (i--);
}
if (i > 0)
goto enterloop3;
/* And check that case and default labels have the same effect
of disabling code suppression. */
i = 41;
switch (i) {
if (0) {
printf ("error\n");
case 42:
printf ("error2\n");
case 41:
printf ("caseok\n");
}
}
i = 41;
switch (i) {
if (0) {
printf ("error3\n");
default:
printf ("caseok2\n");
break;
case 42:
printf ("error4\n");
}
}
return 0;
}
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