Why in Visual C++ STL: "C" not bigger than "b"

Question

Why C not bigger than b, when B bigger than a?

Code in Visual C++ 2017 STL:

void main ()
{       
    cout << ("A" > "A") << endl;
    cout << ("a" > "A")<< endl;
    cout << ("B" > "a")<< endl;
    cout << ("b" > "B")<< endl;
    cout << ("C" > "b")<< endl; 
    cout << ("c" > "C")<< endl;
    cout << ("AA" > "c")<< endl;
    cout << ("Aa" > "AA")<< endl;
    cout << ("aA" > "AA")<< endl;
    cout << ("aa" > "aA")<< endl;
    cout << ("BA" > "aa")<< endl;
    cout << ("Ba" > "BA")<< endl;
    cout << ("D" > "c")<< endl;
    string pause;
    cin >> pause;
}

OutPut: 0 1 1 1 0 1 1 1 1 1 1 1 1

Do you want to compare strings or characters? – doctorlove Jul 12 '17 at 11:20 — doctorlove, Jul 12 '17 at 11:20
You are comparing pointers, not strings or characters. – Pontifex Jul 12 '17 at 11:21 — Pontifex, Jul 12 '17 at 11:21
`void main` isnt c++ – 463035818_is_not_an_ai Jul 12 '17 at 11:23 — 463035818_is_not_an_ai, Jul 12 '17 at 11:23
strings, because I am comparing "BB" or "AA" – justromagod Jul 12 '17 at 13:11 — justromagod, Jul 12 '17 at 13:11
#include #include #pragma once using namespace std; – justromagod Jul 12 '17 at 17:59 — justromagod, Jul 12 '17 at 17:59

score 6 · Accepted Answer · answered Jul 12 '17 at 11:27

6

You have made an assumption that > and < perform lexicographic comparison on string literals.

Unfortunately, that assumption does not hold. You are just comparing pointers.

Use strcmp instead and, next time, read the documentation instead of making assumptions.

Also, main returns int, not void.

answered Jul 12 '17 at 11:27

Lightness Races in Orbit

378,754
76
643
1,055

Relational operators for string [link](http://www.cplusplus.com/reference/string/string/operators/) http://www.cplusplus.com/reference/string/string/operators/ Performs the appropriate comparison operation between the string objects lhs and rhs. The functions use string::compare for the comparison. These operators are overloaded in header – justromagod Jul 12 '17 at 13:07
@justromagod: You are not using `std::string` anywhere in the code. – Lightness Races in Orbit Jul 12 '17 at 13:55
(well, except for the unrelated `pause` at the end... which, by the way, [you should not do](https://stackoverflow.com/a/36374595/560648)) – Lightness Races in Orbit Jul 12 '17 at 13:59
https://www.joelonsoftware.com/2002/11/11/the-law-of-leaky-abstractions/. I found answer about issue, I belive it is leaky abstraction in C++ – justromagod Jul 25 '17 at 21:15
@justromagod I don't entirely disagree – Lightness Races in Orbit Aug 30 '18 at 15:41

doctorlove · Answer 2 · 2017-07-12T11:32:45.333

3

You are comparing pointers.

"A" is a pointer to chararaters. (Actaully const char[2] as pointed out in comments) 'A' is a character.

Try this:

int main ()
{
    cout << ('A' > 'A') << endl;
    cout << ('a' > 'A')<< endl;
    cout << ('B' > 'a')<< endl;
    cout << ('b' > 'B')<< endl;
    cout << ('C' > 'b')<< endl; 
    cout << ('c' > 'C')<< endl;
}

You now need to think about what you want from a string (or char array).

Is "AA" less or greater than "c"? It comes first in a dictionary, but it's shorter.

If you want "dictionary" (lexographical) order you need to compare strings; strcmp or look at string comparision operators (if you decide to use std::string instead)

edited Jul 12 '17 at 11:32

answered Jul 12 '17 at 11:27

doctorlove

18,872
2
46
62

`"A"` is `const char[2]`, not `char*`. – Baum mit Augen Jul 12 '17 at 11:30
OK, true. I was trying to emphasis the pointer v characters, but what I've said in words is not correct... thanks. – doctorlove Jul 12 '17 at 11:31
By the time it is passed to any functions or operators, including the built-in ones, the expression is a `const char*`, and that's why you get pointer comparisons. Array comparisons aren't even defined – Lightness Races in Orbit Aug 30 '18 at 15:42
@LightnessRacesinOrbit is that a comment on my aside about the comments? – doctorlove Aug 30 '18 at 15:44
It's a continuation of the comments discussion yes – Lightness Races in Orbit Aug 30 '18 at 16:07
You are of course right. I was driving at it decays to a const char * - and what would an array comp mean... bu this is an old answer! – doctorlove Aug 30 '18 at 22:38

Vlad from Moscow · Answer 3 · 2017-07-12T11:51:16.263

In the statements like these

cout << ("A" > "A") << endl;
cout << ("a" > "A")<< endl;
cout << ("B" > "a")<< endl;
//...

there are used string literals that have the static storage duration. In C++ string literals have types of constant character arrays. For example the string literal "A" has type const char[2] because in memory it is stored like { 'A', '\0' }.

Execute this statement

std::cout << sizeof( "A" ) << std::endl;

and you will see that the result will be 2.

From the C++ Standard (2.13.5 String literals)

8 Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration (3.7).

In expressions like this "A" > "A" arrays are implicitly converted to pointers to their first characters. So in this expression there are compared two pointers.

From the C++ STandard (4.2 Array-to-pointer conversion)

1 An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The temporary materialization conversion (4.4) is applied. The result is a pointer to the first element of the array.

In general it is not necessary that string literals with equal contents will be stored as one character array. So in fact this expression

"A" == "A"

can yield either true or false depending on compiler options that is whether these two string literals stored as one array or as two distinct arrays by the compiler.

It seems your compiler stores equal string literals as one array. In this case this statement

cout << ("A" > "A") << endl;

outputs 0.

Also it seems that the compiler placed the array that stores the string literal "a" after the array that stores string literal "A". So the address of the first character of the string literal "a", that is of the character 'a', is higher than the address of the first character of the string literal "A". So this statement

cout << ("a" > "A")<< endl;

outputs 1.

Also the compiler placed the array that stores the string literal "B" after the array that stores the string literal "a". So again the address of the first character of the string literal "B" is higher than the address of the first character of the string literal "a" and in this statement

cout << ("B" > "a")<< endl;

there are compared these addresses. So the statement outputs 1.

And so on.

It seems what you need to do is to compare character literals instead of the string literals. For example

cout << ( 'A' > 'A') << endl;
cout << ('a' > 'A')<< endl;
cout << ('B' > 'a')<< endl;
//...

Why in Visual C++ STL: "C" not bigger than "b"

3 Answers3