编译原理中的词法分析需要进行错误报告吗?还是应该一并在语法分析中给出。
如果需要,那么需要判断哪几类错误?
编译原理中的词法分析需要进行错误报告吗
答案:2 悬赏:70 手机版
解决时间 2021-12-18 19:35
- 提问者网友:萌卜娃娃
- 2021-12-18 08:12
最佳答案
- 五星知识达人网友:独行浪子会拥风
- 2022-01-05 20:17
词法分析中需要进行处理和报告的错误为:源程序中出现的不符合文法的字符,标识符等等。
这些一定要在词法分析中完成,搁到后面会很麻烦。
其余错误,如类型不匹配,返回值出错,语句语法错误等,在语法分析过程中进行。
这些一定要在词法分析中完成,搁到后面会很麻烦。
其余错误,如类型不匹配,返回值出错,语句语法错误等,在语法分析过程中进行。
全部回答
- 1楼网友:独钓一江月
- 2022-01-05 20:30
编译的词法分析,一般是先画一个状态转换图,一般是有多少分支,就有多少if语句,分支里面再分(可能有循环语句)。注意记住词的类别和词的字符串,请以以下代码为例,理会一下词法分析的大致过程。
while(s[i]!='#')
{
while(s[i]==' '||s[i]=='\t'||s[i]=='\n')
{
if(s[i]=='\n')
line++;
i++;
}
if(s[i]=='#')
break;
j=i;
if(s[i]>='a'&&s[i]<='z'||s[i]>='a'&&s[i]<='z')
{
i++;
while(s[i]>='a'&&s[i]<='z'||s[i]>='a'&&s[i]<='z'||s[i]>='0'&&s[i]<='9')
i++;
if((i-j)==2&&s[j]=='i'&&s[j+1]=='f')
{
strcpy(dancishuzu[dancigeshu].name,"if");
dancishuzu[dancigeshu].bianhao=4;
dancigeshu++;
}
else if((i-j)==3&&s[j]=='i'&&s[j+1]=='n'&&s[j+2]=='t')
{
strcpy(dancishuzu[dancigeshu].name,"int");
dancishuzu[dancigeshu].bianhao=2;
dancigeshu++;
}
else if((i-j)==3&&s[j]=='f'&&s[j+1]=='o'&&s[j+2]=='r')
{
strcpy(dancishuzu[dancigeshu].name,"for");
dancishuzu[dancigeshu].bianhao=6;
dancigeshu++;
}
else if((i-j)==4&&s[j]=='m'&&s[j+1]=='a'&&s[j+2]=='i'&&s[j+3]=='n')
{
strcpy(dancishuzu[dancigeshu].name,"main");
dancishuzu[dancigeshu].bianhao=1;
dancigeshu++;
}
else if ((i-j)==4&&s[j]=='c'&&s[j+1]=='h'&&s[j+2]=='a'&&s[j+3]=='r')
{
strcpy(dancishuzu[dancigeshu].name,"char");
dancishuzu[dancigeshu].bianhao=3;
dancigeshu++;
}
else if ((i-j)==4&&s[j]=='e'&&s[j+1]=='l'&&s[j+2]=='s'&&s[j+3]=='e')
{
strcpy(dancishuzu[dancigeshu].name,"else");
dancishuzu[dancigeshu].bianhao=5;
dancigeshu++;
}
else if ((i-j)==5&&s[j]=='w'&&s[j+1]=='h'&&s[j+2]=='i'&&s[j+3]=='l'&&s[j+4]=='e')
{
strcpy(dancishuzu[dancigeshu].name,"while");
dancishuzu[dancigeshu].bianhao=7;
dancigeshu++;
}
else{
dancishuzu[dancigeshu].bianhao=10;
count=0;
while(j='0'&&s[i]<='9')
{
while(s[i]>='0'&&s[i]<='9')
i++;
dancishuzu[dancigeshu].bianhao=11;
count=0;
while(j')
{
if(s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=28;
strcpy(dancishuzu[dancigeshu].name,">=");
dancigeshu++;
i+=2;
}
else
{
i++;
dancishuzu[dancigeshu].bianhao=26;
strcpy(dancishuzu[dancigeshu].name,">");
dancigeshu++;
}
}
else if(s[i]=='<')
{
if(s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=29;
strcpy(dancishuzu[dancigeshu].name,"<=");
dancigeshu++;
i+=2;
}
else
{
i++;
dancishuzu[dancigeshu].bianhao=27;
strcpy(dancishuzu[dancigeshu].name,"<");
dancigeshu++;
}
}
else if(s[i]=='!'&&s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=31;
strcpy(dancishuzu[dancigeshu].name,"!=");
dancigeshu++;
i+=2;
}
else
{
printf("\nline:%derror!",line);
i++;
return;
}
}
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