C++ Help (ISO C++ forbids comparison between pointer and integer.) -

i'm trying make simple program stacks seem getting error when try run it. error says "iso c++ forbids comparison between pointer , integer.". appreciated.

my code:

#include <iostream> #include <string> using namespace std;  const int maxstack = 5;  struct stacktype{     string name[maxstack];     int top; };  void createstack(stacktype &stack); void destroystack(stacktype &stack); bool fullstack(stacktype stack); void push(stacktype &stack, string &newelement); bool emptystack(stacktype stack); void pop(stacktype &stack, string &poppedelement);  int main(){     stacktype stack;     string newelement, poppedelement;     char quest;      createstack(stack);     cout<<"do want enter data? (y/n)";     cin>>quest;      while((quest == "y" || quest == "y") && !(fullstack(stack))){ //i error on line         cout<<"please enter name";         cin>>newelement;         push(stack, newelement);         cout<<"do want enter data? (y/n)";         cin>>quest;     }      cout<<endl<<endl;      while(!emptystack(stack)){         pop(stack, poppedelement);         cout<<poppedelement<<endl;     }      destroystack(stack);     system("pause");     return 0; }  void createstack(stacktype &stack){     stack.top = -1; } void destroystack(stacktype &stack){     stack.top = -1;; } bool fullstack(stacktype stack){     if(stack.top == maxstack - 1){         return true;     }else{         return false;     } } void push(stacktype &stack, string &newelement){     stack.top++;     stack.name[stack.top] = newelement; } bool emptystack(stacktype stack){     if(stack.top == -1){             return true;     }else{         return false;     } } void pop(stacktype &stack, string &poppedelement){     poppedelement = stack.name[stack.top];     stack.top--; } 

quest char, yet "y" string literal type const char[2]. when try compare these quest == "y", string literal converted pointer first element, , attempting compare char pointer. that's error telling you.

instead of string literal, want character literal 'y', has type char.