我运行此代码时遇到段错误,我不确定原因。注释掉一个特定的行(标记如下)会删除段错误,这使我相信迭代器“i”的递归使用可能会造成麻烦,但即使将其更改为指针,我也会遇到段错误。
void executeCommands(string inputstream, linklist<linklist<transform> > trsMetastack)
{
int * i=new int;
(*i) = 0;
while((*i)<inputstream.length())
{
string command = getCommand((*i),inputstream);
string cmd = getArguments(command,0);
//cout << getArguments(command,0) << " " << endl;
if (cmd=="translate")
{
transform trs;
trs.type=1;
trs.arguments[0]=getValue(getArguments(command,2));
trs.arguments[1]=getValue(getArguments(command,3));
((trsMetastack.top)->value).push(trs);
executeCommands(getArguments(command,1),trsMetastack);
}
if (cmd=="group")
{
//make a NEW TRANSFORMS STACK, set CURRENT stack to that one
linklist<transform> transformStack;
trsMetastack.push(transformStack);
//cout << "|" << getAllArguments(command) << "|" << endl;
executeCommands(getAllArguments(command),trsMetastack); // COMMENTING THIS LINE OUT removes the segfault
}
if (cmd=="line")
{ //POP transforms off of the whole stack/metastack conglomeration and apply them.
while ((trsMetastack.isEmpty())==0)
{
while ((((trsMetastack.top)->value).isEmpty())==0) //this pops a single _stack_ in the metastack
{ transform tBA = ((trsMetastack.top)->value).pop();
cout << tBA.type << tBA.arguments[0] << tBA.arguments[1];
}
trsMetastack.pop();
}
}
“Metastack”是链接列表的链接列表,我必须在递归期间发送给函数,声明如下:
linklist<transform> transformStack;
linklist<linklist<transform> > trsMetastack;
trsMetastack.push(transformStack);
executeCommands(stdinstring,trsMetastack);
“Getallarguments”函数只是为了提取给定字符串的大部分字符串,如下所示:
string getAllArguments(string expr) // Gets the whole string of arguments
{
expr = expr.replace(0,1," ");
int space = expr.find_first_of(" ",1);
return expr.substr(space+1,expr.length()-space-1);
}
这是链表类定义。
template <class dataclass>
struct linkm {
dataclass value; //transform object, point object, string... you name it
linkm *next;
};
template <class dataclass>
class linklist
{
public:
linklist()
{top = NULL;}
~linklist()
{}
void push(dataclass num)
{
cout << "pushed";
linkm<dataclass> *temp = new linkm<dataclass>;
temp->value = num;
temp->next = top;
top = temp;
}
dataclass pop()
{
cout << "pop"<< endl;
//if (top == NULL) {return dataclass obj;}
linkm<dataclass> * temp;
temp = top;
dataclass value;
value = temp->value;
top = temp->next;
delete temp;
return value;
}
bool isEmpty()
{
if (top == NULL)
return 1;
return 0;
}
// private:
linkm<dataclass> *top;
};
感谢您花时间阅读本文。我知道这个问题很模糊,但我花了最后一小时尝试使用gdb进行调试,老实说,我不知道它可能是什么。
答案 0 :(得分:1)
它可能是任何东西,但我的猜测是,讽刺的是:堆栈溢出。 您可能希望尝试将数据结构作为引用传递,例如:
void executeCommands(string &inputstream, linklist<linklist<transform> > &trsMetastack)
但正如弗拉德指出的那样,你可能想熟悉gdb。