A C++ program. Why does this code work?

Question

int main()
{
  int a=1;
  int b=2;
  int c=a+++b;
  cout<<"c"<<c<<endl;
}

c's value turns out to be 3. While, it gives me an error for c=a++b. What is happening here? Why does c=a+++b work?

It is parsed as `a++ + b` which is an increment and an addition. — Bo Persson, Jan 24 '16 at 09:36
Although C, [this question](http://stackoverflow.com/questions/5341202/why-doesnt-ab-work-in-c) is a similar concept. One key phrase relating to this is *maximal munch*. That is, `a+++b` is `a++ + b`, not `a + ++b`. — chris, Jan 24 '16 at 09:44

score 2 · Answer 1 · answered Jan 24 '16 at 09:47

A key part of why a+++b "works", and a++b doesn't is the way the C and C++ language parsing is defined. It is what is called a 'greedy' parser. It will combine as many elements as possible to produce a valid token.

So, given that it's a greedy parser, a++b becomes a++ b, which is not valid. a+++b becomes a++ + b, which is syntactically valid - whether that is what you WANT is another matter. If you want to write a + +b, then you need spaces to separate the tokens.

Devin Wall · Answer 2 · 2016-01-24T09:44:09.403

1

Looks like a is post increment then added to b with bad spacing. For example a++ + b. The variable a is evaluated then incremented. That being said a++b is invalid syntax.

edited Jan 24 '16 at 09:44

answered Jan 24 '16 at 09:38

Devin Wall

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1
16

score 1 · Answer 3 · answered Jan 24 '16 at 09:42

1

Have a look at the C++ operator precedence.

(++) post-increment has the highest precedence, sou you end up with (a++) + b.

answered Jan 24 '16 at 09:42

Derlin

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A C++ program. Why does this code work?

3 Answers3