使用 USERNAME:PASSWORD 但不适用于 SERIAL:TOKEN 的字符串化? [英] Stringification working with USERNAME:PASSWORD but not for SERIAL:TOKEN?

问题描述

我有以下 Makefile(如果你问我为什么包含 \" 你可以参考我的 上一个问题)

I have the following Makefile (If you are asking me why there's \" included you can refer to my previous question)

BOARD_TAG    = mega2560
CPPFLAGS     = -DUSERNAME=\"$(USERNAME)\" -DPASSWORD=\"$(PASSWORD)\"
include $(ARDMK_DIR)/Arduino.mk

和代码:

void setup() {
  Serial.begin(9600);
  String auth_raw2(USERNAME ":" PASSWORD);
  Serial.println(auth_raw2);
}
void loop() {}

当我用 make USERNAME=hello PASSWORD=world 编译它时，一切正常，我看到 'hello:world' 被打印出来.

when I compile this with make USERNAME=hello PASSWORD=world, everything works and I see 'hello:world' being printed out.

但是，如果我将 USERNAME 替换为 SERIAL，将 PASSWORD 替换为 TOKEN:

However, if I substitute USERNAME to SERIAL and PASSWORD to TOKEN:

BOARD_TAG    = mega2560
CPPFLAGS     = -DSERIAL=\"$(SERIAL)\" -DTOKEN=\"$(TOKEN)\"
include $(ARDMK_DIR)/Arduino.mk

和

void setup() {
  Serial.begin(9600);
  String auth_raw2(SERIAL ":" TOKEN);
  Serial.println(auth_raw2);
}
void loop() {}

我收到错误，macro.ino:5:27: error: expected ‘)’ before string constant

请注意，$USERNAME 在我的 linux 机器上定义为 disappearedng 而 $PASSWORD、$SERIAL 和 $TOKEN 未定义.

Note that $USERNAME is defined on my linux box as disappearedng while $PASSWORD, $SERIAL, and $TOKEN are undefined.

为什么这适用于 USERNAME:PASSWORD 而不适用于 SERIAL:TOKEN?

推荐答案

所以看起来 Arduino 覆盖了 $SERIAL 参数.

So it seems like Arduino overrides the $SERIAL parameter.

将 SERIAL 切换到 DSERIAL 使编译正常

Switching SERIAL to DSERIAL makes compiling ok

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