CODEWITHSUNDEEP
c++
Levente Mészáros
Levente Mészáros

Reputation: 960

Isn't C++ operator== supposed to be left associative?

Why does this C++ code print what it prints?

#include <vector>
#include <stdio.h>

int p(const char *x) {
   printf("%s\n", x);
   return 0;
}

int main()
{
   if (p("LEFT") == p("RIGHT")) ;

   std::vector<int> v1;
   v1.push_back(1);
   std::vector<int>::iterator it = v1.erase(v1.begin());

   if (it == v1.end())
      printf("OK\n");
   else
      printf("FAIL\n");

   std::vector<int> v2;
   v2.push_back(1);
   if (v2.erase(v2.begin()) == v2.end())
      printf("OK\n");
   else
      printf("FAIL\n");

   return 0;
}

I'm confused about the OK/FAIL part. This is in contradiction with the LEFT/RIGHT part. What's happening here?

The output for me is:

LEFT
RIGHT
OK
FAIL

Upvotes: 1

Views: 127

Answers (2)

Luchian Grigore
Luchian Grigore

Reputation: 258618

Associativity has nothing to do with order of evaluation.

In fact, you're invoking unspecified behavior. You can't tell which part of == will evaluate first.

Think about this:

int x;  // x is 0 initially
int foo()
{
   x++;
   return x;
}
int goo()
{
   return x;
}

int main()
{
   bool b = foo() == goo();
}

If foo (returns 1) evaluates first, b will be true (goo will return 1).

If goo (returns 0) evaluates first, b will be false (foo will return 1).

Upvotes: 7

jcoder
jcoder

Reputation: 30035

It being left associative does not affect the order in which its parameters are evaluated which is unspecifed by the language. It simply means that if you write "a == b == c" then it will interpret it as ((a==b) == c). But it can calculate a, b and c in advance and in any order it likes.

Upvotes: 3

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