Dereferencing string iterator yields int

Question

I get this error

comparison between pointer and integer ('int' and 'const char *')

For the following code

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int main()
{
    std::string s("test string");
    for(auto i = s.begin(); i != s.end(); ++i)
    {
        cout << ((*i) != "s") << endl;
    }
}

Why does dereferencing the string iterator yield an int and not std::string?

You should write `cout << ((*i) != 's') << endl;` (note the single quotes). — Thomas Sparber, Sep 15 '15 at 07:26
You are comparing a character (`*i`) with a pointer to a string (`"s"`). That can't work. — David Schwartz, Sep 15 '15 at 07:27
if you use `auto` already, why not use range-for as in `for (auto const& c : s) { cout << (c != 's') << '\n'; }`? (note that you rarely want to use `endl` and most often `\n`) — TemplateRex, Sep 15 '15 at 08:04

score 9 · Accepted Answer · answered Sep 15 '15 at 07:26

Actually, it does not yield an int, it yields a char (because a string iterator iterates over the characters in the string). Since the other operand of != is not a char (it's a const char[2]), standard promotions and conversions are applied to the arguments:

char is promoted to int via integral promotion
const char[2] is converted to const char* via array-to-pointer conversion,

This is how you arrive at the int and const char* operands the compiler is complaining about.

You should compare the dereferenced iterator to a character, not to a string:

cout << ((*i) != 's') << endl;

"" encloses a string literal (type const char[N]), '' encloses a character literal (type char).

Dereferencing string iterator yields int

1 Answers1