c-将令牌分配给结构节点类型变量后，令牌为null

Question

我正在解决这个奇怪的问题(至少对我来说(我正在使用lex和yacc创建AST虽然所有的标记都有效，语法也有效(我只用print测试了它(当我试图将一个新节点分配给一个变量时，只是为了测试，我试图打印节点的令牌，每次尝试时，令牌都是(null(例如，这是我代码的一部分：

ifelse: IF LEFTBRACKET cond RIGHTBRACKET LEFTBLOCK body RIGHTBLOCK {$$ = mknode("IF",$3,$6);printf("token is %sn",$$->token);} 

mknode函数是这样工作的：

node* mknode(char* token, node* left, node* right){
node* newnode = (node*)malloc(sizeof(node));
char* newstr = (char*)malloc(sizeof(token)+1);
strcpy(newstr, token);
newnode->left = left;
newnode->right = right;
return newnode;
}

因此，对于这个例子，这是输出：

token is (null)

知道为什么它保持为空吗？如果很重要的话，我将VMWare与Ubuntu 20.04一起使用

完整的lex文件：

%{
#include "y.tab.h"
#include <stdio.h>
#include <string.h>
%}
%%
""" {printf("LEX: double quote heren");return DQUOTE;}
"'" {printf("LEX: single quote heren");return SQUOTE;}
"}" {printf("LEX: } heren");return RIGHTBLOCK;}
"{" {printf("LEX: { heren");return LEFTBLOCK;}
";" {printf("LEX: ; heren");return SEMICOLON;}
"," {printf("LEX: comma heren");return COMMA;}
"(" {printf("LEX: opening bracket heren");return LEFTBRACKET;}
")" {printf("LEX: closing bracket heren");return RIGHTBRACKET;}

&& {printf("LEX: and heren");return AND;}
"||" {printf("LEX: || heren");return OR;}
"=" {printf("LEX: assign heren");return ASSIGN;}
== {printf("LEX: == heren");return EQ;}
">" {printf("LEX: > heren");return GT;}
">=" {printf("LEX: >= heren");return GTEQ;}
"<" {printf("LEX: < heren");return LT;}
"<=" {printf("LEX: <= heren");return LTEQ;}
"-" {printf("LEX: - heren");return SUB;}
"!" {printf("LEX: ! heren");return NOT;}
"!=" {printf("LEX: != heren");return NOTEQ;}
"/" {printf("LEX: div heren");return DIV;}
"+" {printf("LEX: add heren");return ADD;}
"*" {printf("LEX: mul heren");return MUL;}
"&" {printf("LEX: & heren");return ADRS;}
if {printf("LEX: if heren");return IF;}
else {printf("LEX: else heren");return ELSE;}
do {printf("LEX: do heren");return DO;}
while {printf("LEX: while heren");return WHILE;}
for {printf("LEX: for heren");return FOR;}
var {printf("LEX: var heren");return VAR;}
return {printf("LEX: return heren");return RETURN;}
null {printf("LEX: nullval heren");return NULLVAL;}
void {printf("LEX: func return type heren");return VOID;}
"int*" {printf("LEX: int* type heren");return INTPOINT;}
"char*" {printf("LEX: char* type heren");return CHARPOINT;}
"real*" {printf("LEX: real* type heren");return REALPOINT;}
int {printf("LEX: int type heren");return INT;}
real {printf("LEX: real type heren");return REAL;}
char {printf("LEX: char type heren");return CHAR;}
bool {printf("LEX: bool type heren");return BOOL;}
"true"|"false" {printf("LEX: boolval heren");return BOOLVAL;}
[0-9]+ {yylval.string = yytext;printf("LEX: int val heren");return INTVAL;}
"-"|0|[1-9][0-9]+"."[0-9]+|[1-9][0-9]+'.'[0-9]['E'|'e']['+'|'-'][0-9]+ {printf("LEX: realval heren");yylval.string = strdup(yytext);return REALVAL;}
[a-zA-Z][0-9]*"_"[a-zA-Z]* {printf("LEX: ID heren");yylval.string = strdup(yytext);return ID;}
[a-zA-Z] {printf("LEX: char heren");yylval.string = strdup(yytext);return CHARVAL;}
[a-zAZ]*[0-9]*[a-zAZ]+[0-9]*[a-zAZ]* {printf("LEX: string heren");yylval.string = strdup(yytext);return STRING;}
. ;
%%

完整的yacc文件：

%{
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string.h>
#include "lex.yy.c"
int yyerror();
typedef struct node
{
char* token;
struct node *left;
struct node *right;
}node;
node* mknode(char* token, node* left, node* right);
void printtree(node* tree);
%}
%union
{
struct node *node;
char* string;
}    
%token <string> DIV ADD MUL SUB AND NOT OR RETURN ASSIGN
%token <string> EQ GT GTEQ LT LTEQ NOTEQ SEMICOLON COMMA LEFTBRACKET RIGHTBRACKET RIGHTBLOCK LEFTBLOCK
%token <string> ID CHARVAL INTVAL REALVAL BOOLVAL STRING ADRS DQUOTE SQUOTE 
%token <node> INT REAL BOOL CHAR VOID INTPOINT CHARPOINT REALPOINT VAR NULLVAL 
%token <node> IF ELSE WHILE DO FOR 
%type <string> name oper type rettype ret
%type <node> code ifelse body action args argnum math cond params block valvar
%left SEMICOLON COMMA RIGHTBRACKET RIGHTBLOCK
%right LEFTBLOCK 
%%
st: code {printf("YACC: Code done!n");}
code: rettype name params block code {
$$ = mknode("(FUNC",mknode($2,mknode("(ARGS",$3,NULL),mknode("(RET",mknode($1,NULL,NULL),NULL)),mknode("(BODY",$4, NULL));
printf("YACC: func readyn");}| {}
params: LEFTBRACKET args RIGHTBRACKET {$$ = $2;}

block: LEFTBLOCK code RIGHTBLOCK {$$ = $2;}| 
LEFTBLOCK body RIGHTBLOCK {$$ = $2;}| 
LEFTBLOCK code body RIGHTBLOCK {$$ = mknode("",$2,$3);}
ifelse: IF LEFTBRACKET cond RIGHTBRACKET LEFTBLOCK body RIGHTBLOCK {$$ = mknode("IF",$3,$6);printf("token is %sn",$$->token);printf("YACC: if readyn");}|
IF LEFTBRACKET cond RIGHTBRACKET LEFTBLOCK body RIGHTBLOCK ELSE LEFTBLOCK body RIGHTBLOCK {$$ = mknode("IF",$3,mknode("",$6,mknode("ELSE",$10,NULL)));
printf("YACC: if else readyn");}
rettype:VOID {$$ = "VOID";}|
type {$$ = $1;}
name: STRING {$$ = $1;}|
CHARVAL {$$ = $1;}
args: type argnum {$$ = mknode($1,mknode(" ",NULL,NULL),$2);}| 
{$$ = mknode("",NULL,NULL);printf("YACC: args readyn");}
argnum: name argnum {$$ = mknode($1,mknode(" ",NULL,NULL),$2);}| 
COMMA argnum {$$ = mknode(" ",$2,NULL);}| 
SEMICOLON args {$$ = mknode(" ",$2,NULL);}|  
{$$ = NULL;printf("YACC: args num readyn");}
type: INT {$$ = "INT";}|
REAL {$$ = "REAL";}|
CHAR {$$ = "CHAR";}|
INTPOINT {$$ = "INT*";}|
CHARPOINT {$$ = "CHAR*";}|
REALPOINT {$$ = "REAL*";};
body: action body {$$ = mknode(" ",$1,$2);}|
action {$$ = mknode(" ",$1,NULL);}| 
ifelse  {$$ = mknode("(IF-ELSE",mknode("n",NULL,NULL),$1);printf("YACC: block readyn");}
action: name ASSIGN math {$$ = mknode($2,mknode($1,NULL,NULL),mknode(" ",$3,NULL));printf("YACC: action readyn");}| 
RETURN ret SEMICOLON {$$ = mknode("(RET",mknode($2,NULL,NULL),NULL);printf("YACC: return action readyn");}
ret: INTVAL {$$ = $1;}| 
SQUOTE CHARVAL SQUOTE {$$ = $2;}| 
REALVAL {$$ = $1;}| 
DQUOTE STRING DQUOTE  {$$ = $2;}| 
name {$$ = $1;}| 
ADRS name {$$ = $2;}
math: valvar oper math {$$ = mknode($2, $1,$3);}| 
valvar SEMICOLON {$$ = mknode(" ",$1,NULL);}| 
valvar math {$$ = mknode(" ",$1,$2);}
oper: ADD {$$ = "+";}|
DIV {$$ = "/";}|
SUB {$$ = "-";}|
MUL {$$ = "*";}
cond: valvar EQ valvar {$$ = mknode($2,$1,$3);}| 
valvar GT valvar {$$ = mknode($2,$1,$3);}| 
valvar GTEQ valvar {$$ = mknode($2,$1,$3);}| 
valvar LT valvar {$$ = mknode($2,$1,$3);}| 
valvar LTEQ valvar {$$ = mknode($2,$1,$3);}| 
valvar NOTEQ valvar {$$ = mknode($2,$1,$3);}
valvar: name {$$ = mknode($1,NULL,NULL);}|
INTVAL {$$ = mknode($1,NULL,NULL);}
%%
int main(){
return yyparse();
}
void printtree(node* tree){
printf("%sn", tree->token);
if(tree->left)
printtree(tree->left);
if(tree->right)
printtree(tree->right);
}
node* mknode(char* token, node* left, node* right){
node* newnode = (node*)malloc(sizeof(node));
char* newstr = (char*)malloc(strlen(token)+1);
strcpy(newstr, token);
newnode->left = left;
newnode->right = right;
return newnode;
}

int yyerror(){
printf("language errorn");
return 0;
}

目前的输入测试代码是：

void foo(int x){
if (x==5){
return 'a';
}
}

Answer 1

在许多其他问题中，mknode函数从未设置newnode->token。所以它是未定义的——你很幸运，它包含一个null，而不是一个会崩溃的无效指针。