C++ Comparison of String Literals

Question

I\'m a c++ newbie (just oldschool c). My son asked for help with this and I\'m unable to explain it. If he had asked me \"how do I compare strings\" I would have told hi

Answer 1

In C++, the results are unspecified. I will be using N3337 for C++11.

First, we have to look at what the type of a string literal is.

§2.14.5

9 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).

Arrays are colloquially said to decay to pointers.

§4.2

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 result is a pointer to the first element of the array.

Since your string literals both contain one character, they're the same type (char[2], including the null character.)

Therefore the following paragraph applies:

§5.9

2 [...]

Pointers to objects or functions of the same type (after pointer conversions) can be compared, with a result defined as follows:

[...]

— If two pointers p and q of the same type point to different objects that are not members of the same object or elements of the same array or to different functions, or if only one of them is null, the results of p<q, p>q, p<=q, and p>=q are unspecified.

Unspecified means that the behavior depends on the implementation. We can see that GCC gives a warning about this:

warning: comparison with string literal results in unspecified behaviour [-Waddress]
     std::cout << ("Z" < "A");

The behavior may change across compilers or compiler settings but in practice for what happens, see Wintermute's answer.

Answer 2

String literals have static storage duration. In all these comparisons there are compared addresses of memory allocated by the compiler for string literals. It seems that the first string literal that is encountered by the compiler is stored in memory with a lower address compared with the next encountered string literal.

Thus in this program

int main() 
{ 
  cout << ("Z"< "A");
  cout << ("A"< "Z");
}

string literal "Z" was alllocated with a lower address than string literal "A" because it was found first by the compiler.

Take into account that comparison

  cout << ("A"< "A");

can give different results depending on the options of the compiler because the compiler may either allocate two extents of memory for the string literals or use only one copy of the string literals that are the same.

From the C++ Standard (2.14.5 String literals)

12 Whether all string literals are distinct (that is, are stored in nonoverlapping objects) is implementation defined. The effect of attempting to modify a string literal is undefined.

The same is valid for C.

Answer 3

The string constants ("A" and "Z") in C++ are represented by the C concept - array of characters where the last character is '\0'. Such constants have to be compared with strcmp() type of function.

If you would like to use the C++ std::string comparison you have to explicitly state it:

cout << (std::string( "A") < "Z");

Answer 4

You are comparing memory addresses. The example that follow explains how to compare 2 strings:

#include "stdafx.h"
#include <iostream>
#include <cstring> //prototype for strcmp()

int _tmain(int argc, _TCHAR* argv[])
{
 using namespace std;

 cout << strcmp("A", "Z"); // will print -1
 cout << strcmp("Z", "A"); // will print 1

 return 0;
}

Answer 5

If you want to compare actual C++ strings, you need to declare C++ strings:

int main() 
{
  const std::string a("A");
  const std::string z("Z");

  cout << (z < a) << endl; // false
  cout << (a < z) << endl; // true
}

Answer 6

You are comparing memory addresses. Apparently your compiler places the string literals in memory in the order it encounters them, so the first is "lesser" than the second.

Since in the first snippet it sees "A" first and "Z" second, "A" is lesser. Since it sees "Z" first in the second, "Z" is lesser. In the last snippet, it already has literals "A" and "Z" placed when the second command rolls around.