检查给定字符串是否是关键字的C程序？

关键字是在C++库中预定义或保留的单词，具有固定的含义，并用于执行内部操作。C++语言支持超过64个关键字。

每个关键字都以小写字母形式存在，如auto、break、case、const、continue、int等。

C++语言中的32个关键字也可在C语言中使用。

auto	double	int	struct
break	else	long	switch
case	enum	register	typedef
char	extern	return	union
const	float	short	unsigned
continue	for	signed	void
default	goto	sizeof	volatile
do	if	static	while

这是C++中新增的30个保留字，不在C语言中。

asm	dynamic_cast	namespace	reinterpret_cast
bool	explicit	new	static_cast
catch	false	operator	template
class	friend	private	this
const_cast	inline	public	throw
delete	mutable	protected	true
try	typeid	typename	using
using	using	wchar_t

Input: str=”for”
Output: for is a keyword

Explanation

• 关键字是程序中不能用作变量名的保留字。

• C编程语言中有32个关键字。

将字符串与每个关键字进行比较，如果字符串相同，则字符串是关键字。

Example

 示例

#include <stdio.h>
#include <string.h>
int main() {
   char keyword[32][10]={
      "auto","double","int","struct","break","else","long",
      "switch","case","enum","register","typedef","char",
      "extern","return","union","const","float","short",
      "unsigned","continue","for","signed","void","default",
      "goto","sizeof","voltile","do","if","static","while"
   } ;
   char str[]="which";
   int flag=0,i;
   for(i = 0; i < 32; i++) {
      if(strcmp(str,keyword[i])==0) {
         flag=1;
      }
   }
   if(flag==1)
      printf("%s is a keyword",str);
   else
      printf("%s is not a keyword",str);
}

输出

which is a keyword

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