"warning: operation of ... may be undefined" for ternary operation -- not if/else block

Question

Here's my code:

int main() {
    static int test = 0;
    const int anotherInt = 1;
    test = anotherInt > test ? test++ : 0;
    if (anotherInt > test)
        test++;
    else
        test = 0;
    return 0;
}

Here's the warning produced when I build it:

../main.cpp:15:40: warning: operation on ‘test’ may be undefined [-Wsequence-point]
  test=     anotherInt>test ? test++ : 0;
                                        ^

Why does C++ give me a warning on the ternary operation, but not the regular if..else statement?

Your `if` isn't equivalent. Your ternary is doing `test = test++;` — chris, Feb 14 '16 at 07:00
You changed `test++` to `test=test++`, which makes a big difference. — Alan Stokes, Feb 14 '16 at 07:01
Note that the _undefined_ refers to the output of the expression, not that of any variable. Different compilers will evaluate "text = text++" differently; thus, it gets flagged. — Oldest Software Guy, Feb 14 '16 at 07:06

score 4 · Accepted Answer · answered Feb 14 '16 at 07:01

4

They are not equivalent. Note that in the ternary operator expression, you assigned the result to test

Change the if condition to:

if(anotherInt > test)
    test = test++;  // undefined!

You would probably see the same warning here as well.

answered Feb 14 '16 at 07:01

Yu Hao

119,891
44
235
294

Yep, I got the same warning. If I want to use a ternary operator, what must I do? – Username Feb 14 '16 at 07:03
3

@Username Replace `test++` with `test + 1`. – Yu Hao Feb 14 '16 at 07:05
`test = anotherInt > test ? test+1 : 0;` – raz Apr 29 '20 at 11:08

score -3 · Answer 2 · answered Feb 14 '16 at 07:02

-3

u may know that in this code: anotherInt>test ? test++ : 0; the computer maybe run test++ first,maybe run anotherInt>test ? first.so in a expression,if u use a variable ,u should not change it in some other place in this expression .U can change test++ to test+1.

answered Feb 14 '16 at 07:02

shadow work

1
1

3

No, that's not the problem. The condition of the ternary operator is guaranteed to be evaluated before either expression. – 5gon12eder Feb 14 '16 at 07:04

"warning: operation of ... may be undefined" for ternary operation -- not if/else block

2 Answers2