解释一下C代码片段：预处理器+打印=？
此代码片段的输出是 %s is a string is a string。请解释一下。
#include <stdio.h>
#define scanf "%s is a string"
int main()
{
printf(scanf, scanf);
}
The output for this code snippet is %s is a string is a string. Please explain.
#include <stdio.h>
#define scanf "%s is a string"
int main()
{
printf(scanf, scanf);
}
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评论（5）
您究竟希望我们解释什么？替换宏并获取
剩下的就是
printf的预期正常日常行为。PS
#define scanf ...???What exactly do you want us to explain? Subsititute the macro and get
The rest is the expected normal everyday behavior of
printf.P.S.
#define scanf ...???预处理器进行盲替换以给出：
第一个参数中的
%s是字符串的格式说明符，并被第二个参数替换。第二个参数中的%s没有什么特别的。The preprocessor does a blind substitution to give:
The
%sin the first argument is the format specifier for a string and is replaced with the 2nd argument. There is nothing special about the%sin the 2nd argument.这是一些相当奇怪的代码，但输出将是“%s 是一个字符串是一个字符串”，因为在这两种情况下 scanf 都扩展为“%s 是一个字符串”，然后 printf 将其替换为 %s。
This is some rather bizarre code, but the output would be "%s is a string is a string" because scanf is expanded to "%s is a string" in both cases and then printf substitutes that in for the %s.
printf("%s 是一个字符串","%s 是一个字符串");
我想令人困惑的是 printf 将如何处理第二个 %s。
要澄清这一点，printf 不是递归函数。如果您正在打印字符串并且该字符串具有任何格式标识符，则它不会被视为格式标识符。它被视为普通字符串。因此，在这种情况下，第二个“%s 是字符串”只是一个普通字符串。该字符串中的 %s 不是格式标识符。
如果你有类似 printf("%s %s 是一个字符串", "%s %s 是一个字符串"); 的东西
然后，是的，您将收到运行时错误，指出 printf 缺少某些参数。
printf("%s is a string","%s is a string");
I guess the confusion is what the printf will do with the second %s.
To clear this, printf is not a recursive function. If you are printing a string and that string has any format-identifier it is not considered as format identifier. It is considered as plain string. So in this case the second "%s is a string" is just a plain string. %s in this string is not format-identifier.
If you had something like printf("%s %s is a string", "%s %s is a string");
Then yes you will get runtime error saying that printf is missing some argument.
要理解这段代码，请一一运行以下语句：-
to understand this code,run the following statements one by one:-