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Home/ Questions/Q 886213
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T13:01:56+00:00 2026-05-15T13:01:56+00:00

I have a class symbol_table that has a vector of objects of another class

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I have a class symbol_table that has a vector of objects of another class row_st.also I have an enter method where inserts objects of row_st with a passed name into the vector of desired symbol_table.but when I call the enter to enter objects with name :
a;b;c;Iwill get the following result: a,b,c;b,c;c.the first element of vector gets the name of all the entered objects. and the second element also gets the name of the later entries.

  class row_st
  {
   public:
      char* name;
      type_u type;//int:0,flaot:1;char:2,bool:3,array:
      int offset;
      symbol_table *next;
      symbol_table *current;
  };
  class symbol_table
  {
   public:
    vector <row_st *> row;
     int type;
     int header;
     int starting_stmt;
     int index;
     int i;
     symbol_table *previous;
     symbol_table(){ header=0;
      previous=0; index=0;i=0;starting_stmt=0;}
  };

and here it is the enter method:

 int enter(symbol_table *table,char* name,type_u type){
     row_st *t=new row_st;
t->name=name;
t->type=type;
t->offset=table->index;
t->current=table;
table->index++;
t->next=0;
table->row.push_back(t);
table->header +=1;
return table->row.size()-1;
   }

the push_backed elements all points to the same address.the new call makes the same row_st every time it is called.what should I do?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T13:01:56+00:00Added an answer on May 15, 2026 at 1:01 pm

    As Neil Butterworth’s answer suggest, the trouble is probably not with this code, but the place where you call it. Using character pointers does not make it impossible to make things work, just harder.

    The problem in this case is definitely not with push_back. If you posted the method where you call this code it might be possible to see exactly what goes wrong.

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