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法学检查给定字符串是否为C语言关键字的程序
关键字是在C++库中预定义或保留的词，具有固定的含义，用于执行内部操作。C++语言支持60多个关键字。
所有关键字都小写，例如auto、break、case、const、continue、int等。
C++语言中32个关键字也存在于C语言中。
| 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 |
以下30个保留字在C语言中不存在，而是C++语言新增的：
| 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
解释
关键字是保留字，不能用作程序中的变量名。
C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，则该字符串为关键字。
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
更新于：2019年10月7日
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