c++ compile error: ISO C++ forbids comparison between pointer and integer

I am trying an example from Bjarne Stroustrup's C++ book, third edition. While implementing a rather simple function, I get the following compile time error:

error: ISO C++ forbids comparison between pointer and integer 

What could be causing this? Here is the code. The error is in the if line:

#include <iostream> #include <string> using namespace std; bool accept() { cout << "Do you want to proceed (y or n)?\n"; char answer; cin >> answer; if (answer == "y") return true; return false; } 

Thanks!

5 Answers

You have two ways to fix this. The preferred way is to use:

string answer; 

(instead of char). The other possible way to fix it is:

if (answer == 'y') ... 

(note single quotes instead of double, representing a char constant).

You need the change those double quotation marks into singles. ie. if (answer == 'y') returns true;

Here is some info on String Literals in C++:

A string literal is delimited by quotation marks and is of type char* not char.

Example: "hello"

So when you compare a char to a char* you will get that same compiling error.

char c = 'c'; char *p = "hello"; if(c==p)//compiling error { } 

To fix use a char literal which is delimited by single quotes.

Example: 'c'

"y" is a string/array/pointer. 'y' is a char/integral type

You must remember to use single quotes for char constants. So use

if (answer == 'y') return true;

Rather than

if (answer == "y") return true;

I tested this and it works