Comparison between pointer and integer in string class - C++

Question

I'm new to C++.

string str = "WWWHBBB";
if(str[mid] == "H" && str[0] != "W") return; // corrected after comments for second operand

The above line with if condition gives me an error.

Comparison between pointer and integer ('std::__1::basic_string, std::__1::allocator >::value_type' (aka 'char') and 'const char *')

I have searched the internet enough to know that array style access is fine in strings. Error basically is pointing out comparison about pointer and integer. Really? I thought I was comparing character H to another character in string str.

I tried if str[mid] really returns an iterator I should do *str[mid]. Nah! Didn't work either.

Double quotes mean it's a c-string (`const char*`), if you want a character literal, use single quotes. — flowit, Mar 31 '20 at 17:23
C++ isn't JavaScript. The types of quotes you use makes a difference. — PaulMcKenzie, Mar 31 '20 at 17:25
Gotcha... Learning on the fly and writing my programs. Please love me for it. — HalfWebDev, Mar 31 '20 at 17:26
Single quotes for single characters. Double quotes for text. — Thomas Matthews, Mar 31 '20 at 17:29

Cory Kramer · Answer 1 · 2020-03-31T17:24:20.610

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You'd want to compare with a single char

if (str[mid] == 'H' && str[mid] != 'W') return;

Note that double-quotes in this case refer to a const char[] whereas single-quotes refer to a single char.

The warning is telling you that the only way to make those comparisons in your case would be to compare the lhs char (result of str[mid]) to the rhs would be to let the array decay to const char* and compare the pointer address to the char.

edited Mar 31 '20 at 17:24

answered Mar 31 '20 at 17:22

Cory Kramer

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Can you please point me to a good but full fledged tutorial/reference for C++? – HalfWebDev Mar 31 '20 at 17:25
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@kushalvm https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list – Cory Kramer Mar 31 '20 at 17:27

Vlad from Moscow · Accepted Answer · 2020-03-31T17:33:00.827

3

In the expression of the if statement

 if(str[mid] == "H" && str[mid] != "W") return;

the string literals "H" and "W" that have the type const char[2] are implicitly converted to pointers to their first characters of the type const char *.

So you are trying to compare a character returned by the expression str[mid] with a pointer.

Instead of the string literals you need to use character literals like to compare characters

 if(str[mid] == 'H' && str[mid] != 'W') return;

You could also write like

 if(str[mid] == *"H" && str[mid] != *"W") return;

or

 if(str[mid] == "H"[0] && str[mid] != "W"[0]) return;

dereferencing the pointers but this will be confusing

Pay attention to that if str[mid] == 'H' then the second operand will always yield true. So it is enough to write

 if( str[mid] == 'H' ) return;

edited Mar 31 '20 at 17:33

answered Mar 31 '20 at 17:26

Vlad from Moscow

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As `'H' != 'W'`, we might even just write `if (str[mid] == 'H') return;`. – Jarod42 Mar 31 '20 at 17:30
@Jarod42 A good remark. I even did not look through the condition.:) – Vlad from Moscow Mar 31 '20 at 17:33
Hey Vlad, That's all good but I did not understand why would it say integer and not char for LHS. @Jarod42 the second operand was supposed to be str[0]. thanks :) – HalfWebDev Mar 31 '20 at 17:36
@kushalvm The type char belongs to standard integer types. – Vlad from Moscow Mar 31 '20 at 17:39
Oops! I should finish this formal reading first I guess for the language. Gracias! – HalfWebDev Mar 31 '20 at 17:41

Comparison between pointer and integer in string class - C++

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