Why does cout print the ascii value of a char when used with a ternary and int?

Question

Why does this code produce 42012 instead of *012? I can see that it is converting the asterisk to its ASCII value but why?

    vector<int> numbers = {-1,0,1,2};
    for(int num: numbers){
        cout << (num == -1 ? '*' : num); //42012
    }

    for(int num: numbers){
        if(num == -1) cout << '*'; //*012
        else cout << num;
    }

If I use a normal if else statement it works. Why?

See: https://stackoverflow.com/questions/56634198/why-does-stdstring-view-create-a-dangling-view-in-a-ternary-expression , the issue is different, the cause is the same — CuriouslyRecurringThoughts, Jun 22 '19 at 18:15

score 3 · Accepted Answer · answered Jun 22 '19 at 19:17

3

The ternary expression returns the "common type" of its true part and false part, and the common type between char and int is int, so '*' is promoted to int.

answered Jun 22 '19 at 19:17

Jarod42

203,559
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score 1 · Answer 2 · answered Jun 22 '19 at 20:47

1

Found a geeks for geeks article on it here

answered Jun 22 '19 at 20:47

Andrew Boyley

37
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Why does cout print the ascii value of a char when used with a ternary and int?

2 Answers2