C++ literal followed by a bracket

Question

I came across this code I have no clue what it is doing

// ptr is a pointer to an int
if (2[ptr] == 5){
    // do something
}

I did some of my tests but no matter what value I pass into ptr, 0, negatives, positives, 2[ptr] always evaluates to 0. I am surprised it actually compiles. Can someone explain what this expression is ?

Where did you "come across" this code? Who wrote it? Why don't you ask them what it does, and why? — Lightness Races in Orbit, Apr 18 '17 at 15:57
Whoever wrote code using that syntax should not be allowed to deliver code to the codebase anymore :) — Cory Kramer, Apr 18 '17 at 15:58
@BoundaryImposition You don't have to ask them what it does or why they did it: simply fire them on the spot :-) — Sergey Kalinichenko, Apr 18 '17 at 16:00
lol it's more like a code snippet for fun in school to test c++ knowledge. This is absolutely not readable for any production code haha — Bobby, Apr 18 '17 at 16:00
This kind of code only exists in puzzles and intentionally obfuscated code. — Barmar, Apr 18 '17 at 16:01
also, http://stackoverflow.com/questions/41273307/c11-swapping-array-name-and-index for more — Swift - Friday Pie, Apr 18 '17 at 16:51
@Barmar I saw only once when this syntax was used legitimately, but there both index and pointer were results of function calls and braces wre used like scope braces — Swift - Friday Pie, Apr 18 '17 at 16:53

score 2 · Accepted Answer · answered Apr 18 '17 at 15:59

2

2[ptr] is the same as ptr[2] (more or less). Somebody is playing a little joke on you.

As for why it always evaluates to 0, that depends on the rest of the program, which we cannot see.

answered Apr 18 '17 at 15:59

Lightness Races in Orbit

378,754
76
643
1,055

C++ literal followed by a bracket

1 Answers1