This question already has answers here:
Are compound statements (blocks) surrounded by parens expressions in ANSI C?
(2 answers)
Warning "Use of GNU statement expression extension"
(4 answers)
Closed 7 years ago.
I found strange macros in a driver implementation that I can't explain to myself.
Simplified example is:
cout << ({int i=0; while(i<10) {++i;} i;}) << endl;
It will output 10.
But why this does expression become an rvalue at all? It seems to work in C and in C++.
Can someone explain me? Pointing to keywords and to reference will be great.
The is a GCC extension:
A compound statement enclosed in parentheses may appear as an
expression in GNU C.
The last thing in the compound statement should be an expression
followed by a semicolon; the value of this subexpression serves as the
value of the entire construct.
Related
This question already has answers here:
When were the 'and' and 'or' alternative tokens introduced in C++?
(8 answers)
Closed 5 years ago.
Is the word and equivalent to the && operator?
if (inner > 10 and !id)
{
std::cout << "idle" << std::endl;
}
This code was originally translated from Python.
I was sure that the 'if' line would result in a compilation error. But it does pass.
Visual studio (2015) marks it as an error, but it does compile with g++ (and also on this site https://www.onlinegdb.com/online_c++_compiler) and seems to run as expected.
Is this correct syntax or did I miss something?
Yes, according to:
https://en.wikipedia.org/wiki/Digraphs_and_trigraphs#C++
and is equivalent to && ... another issue is that it is not really widely used and it is one character longer than && ... and we know that C++ programmers try to optimize everything. Even the length of their source code, so don't expect to find it widespread in production code.
This question already has answers here:
Why can't I use a "break" statement inside a ternary conditional statement in C++?
(2 answers)
Closed 7 years ago.
I like to use shorthanded "if statements" of the format
if-condition?then-statement:else-statement
Whys is it that this works...
if (num==0)break;else continue;
And this doesn't?
num==4?break:continue;
Seeing as the two statements are logically equivalent.
The first is a statement, the second is an expression. You cannot use statements in expressions.
The syntax of if is:
if (<expression>) <statement> [else <statement>]
The syntax of the expression is:
<Boolean expression> ? <expression> : <expression>
The syntax does not allow statements in an expression.
This question already has answers here:
Uses of C comma operator [duplicate]
(20 answers)
Closed 9 years ago.
Good day, everyone.
I've come across a peculiar piece of code today, which I don't quite understand.
I don't even know how to search for this particular problem.
In this code, which works, a variable assignment is done like this:
if(condition) {
Var1 = false, Var2 = false;
}
Now, I was under the impression, that ALL commands need to be terminated by a semicolon instead of a comma. I am familiar with the syntax
Var1 = Var2 = false;
but not with the one posted above. The compiler (g++) doesn't even throw me a warning or anything...am I missing something from the specification here?
Or is the compiler generous with me and just replaces the , with a ; internally? If so, shouldn't he at least throw a warning?
Thank you for your time.
am I missing something from the specification here?
Yes, it's the "comma operator", specified by C++11 5.18. It evaluates the sub-expression to the left, then the one to the right, and the overall result is that of the right-hand one.
In this case, it's equivalent to two expression statements separated by ;
It's useful in places like if/while/for where you're only allowed one expression, but might want to do more than one thing:
while (++i, --j != 0)
and also if you like to jam multiple statements together to make life difficult for whoever has to read your code.
In the C and C++ programming languages, the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type). (read more)
As Alexandru Barbarosie pointed out, there's quite a thorough explanation on what's happening at https://stackoverflow.com/questions/1613230/uses-of-c-comma-operator
To quickly summarize it for whomever stumbles across this post: When used outside of for loops and stuff, the , actually has the same effect as the ;.
For more information, please visit the link.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
C++ Comma Operator
that probably a trivial question, but I don't know the answer. And this has been troubling me this afternoon.
I was just writing a function to convert RVB to YUV. Nothing really special, but mistakenly used the comma (,) instead of a dot in my numbers.
It compiles but the result was not what I expected, for example "-3713796" instead of a 0-255 range number.
(0,615*(double) 61) - (0,51498*(double) 61) - (0,10001*(double) 61)
So what does it mean ?
If it's not a compilation error, it's probably usefull for something but what?
Ps: I was using C++ with Qt.
You've accidentally used the comma operator. It evaluates its first operand and discards the result, and then evaluates the second operand and returns its value and type.
In my experience, it's most commonly used in loop statements. For other uses, see Uses of C comma operator
It's the comma operator which evaluates each operand and returns the rightmost. In this case it evaluated 0, then 615*(double) 61 and resulted in a value of that product.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
C extension: <? and >? operators
Take a look at the top answer (by sclo) to problem D of this Google Code Jam. It's C++ code, it must have compiled, and it contains statements such as this one:
double& ret = F[mask][cur][b];
if(j==cur) {
ret<?=f(tmp,j,b||bad[i])+M[cur][i]; // WTF is <?= ???
}
This doesn't compile in my Visual Studio 2008. What does the <?= mean?
It's a gcc extension: C extension: <? and >? operators
Recent manuals say:
The G++ minimum and maximum operators (‘<?’ and ‘>?’) and their compound forms (‘<?=’) and ‘>?=’) have been deprecated and are now removed from G++. Code using these operators should be modified to use std::min and std::max instead...
It's simply not valid C++. < Might be less than, an open angle bracket for a template argument list, or the start of a digraph however non of those can be followed by ?, then =.
It's a now deprecated g++ extension to the c++ language.
a <? b
is the minimum, returning the smaller of the numeric values a and b;
a >? b
is the maximum, returning the larger of the numeric values a and b.
There are also compound versions
<?=
and
>?=
that do assignment as well.
It doesn't compile also with GCC, and I've never heard of an operator <?=.
Anyway I would hazard a guess that a<?=b could have a semantics like: a = (a<b) ? b : a, but again, this is just a guess.