Question

I’ve faced with problem while compiling lambda-function:

... (int level = 3) ...
QString str = [level] {QString s;for(int i=0;i++<level;s.append(" "));return s;};

Content of the error:

error: conversion from 'GainStatistic::getWidgetAndProps(QObject*, int)::<lambda()>' to non-scalar type 'QString' requested

I’ve tried this variant:

... (int level = 3) ...
QString str ([level] {QString s;for(int i=0;i++<level;s.append(" "));return s;});

error: no matching function for call to 'QString::QString(GainStatistic::getWidgetAndProps(QObject*, int)::<lambda()>)'

But lambda-expression in a function is simply value of some type? Is that right? Thus, QString(lambda-that-returns-QString) must call the QString::QString(const QString& ref) constructor and this must work:

... (int level = 3) ...
QString str([level] {const QString& ref = "123";return ref;}); //leads to the same error

Another variant:

QString str = [level]->QString {QString s;for(int i=0;i++<level;s.append(" "));return s;};

error: expected token ';' got 'str'

MinGW 4.6.1

Answer 1

You try to assign a lambda to a QString. What do you expect to happen? A lambda taking no arguments is a nullary function. You need to call it to get its return value.

e.g.

int x = [] { return 23; }();
                         ^^
                        call

Also, thanks for showing me the syntax for a no-argument lambda. I didn’t know this was possible. I’m also a little unsure if it is really legal.

Edit: It is legal. 5.1.2

lambda-expression:
    lambda-introducer lambda-declarator{opt} compound-statement

lambda-declarator:
    (parameter-declaration-clause) mutable{opt}