printf(“%d”, 1.0) 是否未定义？
According to section 4.9.6.1 of the C89 draft, %d is a character that specifies the type of conversion to be applied.
The word conversion implies, in my opinion, that printf("%d", 1.0) is defined.
Please confirm or refute this.
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转换是将语言值转换为该值的词汇表示形式。
你的理论是错误的；行为未定义。规范说（7.19.6.1p8 和 9，使用 C99 TC2）：
和
The conversion is the conversion of a language value to a lexical representation of that value.
Your theory is wrong; behavior is undefined. The spec says (7.19.6.1p8 and 9, using C99 TC2):
And
Printf 是一个可变参数函数，因此无法进行转换。编译器只是安排将双精度数推入参数列表。 Printf 无法确定它是 double 与 int 与大象。结果？混乱。
Printf is a varargs function, so no conversion is possible. The compiler just arranges to push a double onto the arguments list. Printf has no way to find out that it's a double versus an int versus an elephant. Result? Chaos.
这里的“转换”一词是指将
int（这是这里唯一可接受的参数类型）转换为由该int的十进制表示形式组成的字符串。代码>.它与从其他类型（例如double）到int的转换无关。The word "conversion" here is referring to the conversion of an
int(which is the only acceptable argument type here) to a string of characters that make of the decimal representation of thatint. It has nothing to do with conversion from other types (such asdouble) toint.不确定它是正式未定义还是错误 - 但它是错误的！
Not sure if it's officially undefined or an error - but it's wrong!