fix(Serial): Uart::write() must not spin when begin() failed

If Serial.begin(baud) fails during hardware init (e.g. an LPUART
clocked from LSE can't reach the requested baud rate), uart_init()
returns false and _ready is set to false -- but Uart::write() never
checks _ready before entering its transmit path.

Uart::write(const uint8_t*, size_t) contains:

    while (!availableForWrite()) {
        // nop, the interrupt handler will free up space for us
    }

Since the hardware was never actually brought up, the TX interrupt
this loop waits on never fires, so once the 63-byte TX ring buffer
fills up (a few Serial.print() calls after a failed begin()), this
spins forever and the MCU deadlocks permanently.

Fix: return 0 immediately if !_ready, before touching the buffer or
entering the wait loop, using the same _ready accessor already used
elsewhere in this file (see Uart::begin(), which sets it, and the
existing operator bool()-style accessor in Serial.h). write(uint8_t)
needs no separate guard since it already delegates to this overload.

Fixes #3071

Author: saikumar-mandaji

cores/arduino/Serial.cpp

@@ -583,6 +583,14 @@ void Uart::flush(uint32_t timeout)
 
 size_t Uart::write(const uint8_t *buffer, size_t size)
 {
+  // If begin() never brought the hardware up (e.g. the requested baud
+  // rate can't be reached from the configured kernel clock), the TX
+  // interrupt this function waits on below will never fire. Bail out
+  // instead of spinning forever once the TX buffer fills up.
+  if (!_ready) {
+    return 0;
+  }
+
   size_t size_intermediate;
   size_t ret = size;
   size_t available = availableForWrite();