linux-user: Return void from queue_signal()
The linux-user queue_signal() function always returns 1, and none of
its callers check the return value. Give it a void return type
instead.
The return value is a leftover from the old pre-2016 linux-user
signal handling code, which really did have a queue of signals and so
might return a failure indication if too many signals were queued at
once. The current design avoids having to ever have more than one
signal queued via queue_signal() at once, so it can never fail.
Signed-off-by: Peter Maydell <peter.maydell@linaro.org>
Reviewed-by: Philippe Mathieu-Daudé <f4bug@amsat.org>
Message-Id: <20220114153732.3767229-4-peter.maydell@linaro.org>
Signed-off-by: Laurent Vivier <laurent@vivier.eu>
diff --git a/linux-user/signal-common.h b/linux-user/signal-common.h
index 42aa479..2113165 100644
--- a/linux-user/signal-common.h
+++ b/linux-user/signal-common.h
@@ -59,8 +59,8 @@
void process_pending_signals(CPUArchState *cpu_env);
void signal_init(void);
-int queue_signal(CPUArchState *env, int sig, int si_type,
- target_siginfo_t *info);
+void queue_signal(CPUArchState *env, int sig, int si_type,
+ target_siginfo_t *info);
void host_to_target_siginfo(target_siginfo_t *tinfo, const siginfo_t *info);
void target_to_host_siginfo(siginfo_t *info, const target_siginfo_t *tinfo);
int target_to_host_signal(int sig);
