Possible Duplicate:
In C arrays why is this true? a[5] == 5[a]
How is it possible that this is valid C++?
void main()
{
int x = 1["WTF?"];
}
On VC++10 this compiles and in debug mode the value of x is 84 after the statement.
What's going on?
3 Answers
Array subscript operator is commutative. It's equivalent to int x = "WTF?"[1]; Here, "WTF?" is an array of 5 chars (it includes null terminator), and [1] gives us the second char, which is 'T' - implicitly converted to int it gives value 84.
Offtopic: The code snippet isn't valid C++, actually - main must return int.
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3 Comments
default
+1 for the main must return int :)
ehremo
ok it's valid Microsoft C++ in that it compiles :)
puppydrum64
I'm surprised that it doesn't also grab the letters after it and give you
0x003F4654, since you declared the variable x as an int rather than a charint x = 1["WTF?"];
equals to
int x = "WTF?"[1];
84 is "T" ascii code
Comments
The reason why this works is that when the built-in operator [] is applied to a pointer and an int, a[b] is equivalent to *(a+b). Which (addition being commutative) is equivalent to *(b+a), which, by definition of [], is equivalent to b[a].
answered Jan 17, 2013 at 12:29
Angew is no longer proud of SO
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operator [](int)would this work:5[myType()]int*. It does exactly what you think it might. Good times.i++[arName]and such. Keeps them on their toes =P