f-string evaluation of conditional expressions with != operator seems to fail

[JD-Veiga]

Bug report

Bug description:

Hi,

I really feel puzzled by this behaviour:

>>> print(f'{True == True=}')
True == True=True
>>> print(f'{True != True=}')
True False

I expected that the output of print(f'{True != True=}') would be True != True=False.

According to docs, "[t]o display both the expression text and its value after evaluation, (useful in debugging), an equal sign '=' may be added after the expression". So print(f'{True != True=}') fails because the expression text is not fully displayed -- even "=" is omitted.

Sorry if this is a duplicate -- I was not able to find any other report here and in Cython issue tracker.

Thank you.

CPython versions tested on:

3.12, 3.13

Operating systems tested on:

macOS

Linked PRs

• gh-129093: Fix f-string debug text sometimes getting cut off when expression contains ! #129159
• [3.13] gh-129093: Fix f-string debug text sometimes getting cut off when expression contains ! (GH-129159) #129163
• [3.12] gh-129093: Fix f-string debug text sometimes getting cut off when expression contains ! #129164

[StanFromIreland]

Tested on Windows 11 with 3.13 and same bug.

[ericvsmith]

Also:

>>> f'{1==0=}'
'1==0=False'
>>> f'{1!=0=}'
'1True'

[JD-Veiga]

Yeah! I think that any conditional expression containing != behaves in the same manner -- whatever the result of evaluating the expression is.

[ericvsmith]

Yeah! I think that any conditional expression containing != behaves in the same manner -- whatever the result of evaluating the expression is.

Yes, I was just trying to point out that the True in the last evaluated expression comes not from the source code of the expression, but rather the value. Which is also why I removed the spaces around !=.

[JelleZijlstra]

The AST looks like this:

>>> print(ast.dump(ast.parse("f'{1!=0=}'"), indent=True))
Module(
 body=[
  Expr(
   value=JoinedStr(
    values=[
     Constant(value='1'),
     FormattedValue(
      value=Compare(
       left=Constant(value=1),
       ops=[
        NotEq()],
       comparators=[
        Constant(value=0)]),
      conversion=114)]))])

So it does evaluate 1 != 0, but instead of outputting a constant 1!=0= (as it should) it outputs just 1. I suspect something is looking for the ! that introduces conversion specifiers and gets confused by the ! in the != operator.

A ! inside a string literal does work correctly though:

>>> f"{'a!b'=}"
"'a!b'='a!b'"

[sobolevn]

This is a parsing issue:

>>> ast.dump(ast.parse("f'{1 != 1=}'"))
"Module(body=[Expr(value=JoinedStr(values=[Constant(value='1 '), FormattedValue(value=Compare(left=Constant(value=1), ops=[NotEq()], comparators=[Constant(value=1)]), conversion=114)]))])"

>>> ast.dump(ast.parse("f'{1 == 1=}'"))
"Module(body=[Expr(value=JoinedStr(values=[Constant(value='1 == 1='), FormattedValue(value=Compare(left=Constant(value=1), ops=[Eq()], comparators=[Constant(value=1)]), conversion=114)]))])"

[tomasr8]

This seems to fix it (with test_fstring passing), but I don't fully understand why yet:

diff --git a/Parser/lexer/lexer.c b/Parser/lexer/lexer.c
index dbbb94a407c..3b433addbe1 100644
--- a/Parser/lexer/lexer.c
+++ b/Parser/lexer/lexer.c
@@ -212,7 +212,7 @@ _PyLexer_update_fstring_expr(struct tok_state *tok, char cur)
         case '}':
         case '!':
         case ':':
-            if (tok_mode->last_expr_end == -1) {
+            if (tok_mode->last_expr_end == -1 || cur == '}') {
                 tok_mode->last_expr_end = strlen(tok->start);
             }
             break;

I'll try to add some tests and perhaps open a PR tomorrow

[picnixz]

Could this be related to #124363?

[JD-Veiga]

Could be, @picnixz, but I do not think so. As @JelleZijlstra pointed out:

I suspect something is looking for the ! that introduces conversion specifiers and gets confused by the ! in the != operator.

while #124363, as long as I understand, is related to raw strings and escape characters, which is not the case here.

The problem here seems to be that the expression text is printed up to ! while the expression itself is correctly evaluated, such as in the following example:

>> print(f'{(2 - 1) != 1=}')
(2 - 1) False

Notice that:

>> print(f'{(2 - 1) == 1=}')
(2 - 1) == 1=True

Even = sign is omitted in != expressions.

[pablogsal]

Thanks everyone for the investigation and thanks @tomasr8 for the analysis. Fantastic job!

[JD-Veiga]

Thanks to everybody. This was awesome.