I saw this code:
if (cond) {
perror("an error occurred"), exit(1);
}
Why would you do that? Why not just:
if (cond) {
perror("an error occurred");
exit(1);
}
In your example it serves no reason at all. It is on occasion useful when written as
if(cond)
perror("an error occured"), exit(1) ;
-- then you don't need curly braces. But it's an invitation to disaster.
The comma operator is to put two or more expressions in a position where the reference only allows one. In your case, there is no need to use it; in other cases, such as in a while loop, it may be useful:
while (a = b, c < d)
...
where the actual "evaluation" of the while loop is governed solely on the last expression.
Legitimate cases of the comma operator are rare, but they do exist. One example is when you want to have something happen inside of a conditional evaluation. For instance:
std::wstring example;
auto it = example.begin();
while (it = std::find(it, example.end(), L'\\'), it != example.end())
{
// Do something to each backslash in `example`
}
It can also be used in places where you can only place a single expression, but want two things to happen. For instance, the following loop increments x and decrements y in the for loop's third component:
int x = 0;
int y = some_number;
for(; x < y; ++x, --y)
{
// Do something which uses a converging x and y
}
Don't go looking for uses of it, but if it is appropriate, don't be afraid to use it, and don't be thrown for a loop if you see someone else using it. If you have two things which have no reason not to be separate statements, make them separate statements instead of using the comma operator.
The main use of the comma operator is obfuscation; it permits doing two
things where the reader only expects one. One of the most frequent
uses—adding side effects to a condition, falls under this
category. There are a few cases which might be considered valid,
however:
The one which was used to present it in K&R: incrementing two
variables in a for loop. In modern code, this might occur in a
function like std::transform, or std::copy, where an output iterator
is incremented symultaneously with the input iterator. (More often, of
course, these functions will contain a while loop, with the
incrementations in separate statements at the end of the loop. In such
cases, there's no point in using a comma rather than two statements.)
Another case which comes to mind is data validation of input parameters
in an initializer list:
MyClass::MyClass( T const& param )
: member( (validate( param ), param) )
{
}
(This assumes that validate( param ) will throw an exception if
something is wrong.) This use isn't particularly attractive, especially
as it needs the extra parentheses, but there aren't many alternatives.
Finally, I've sometimes seen the convention:
ScopedLock( myMutex ), protectedFunction();
, which avoids having to invent a name for the ScopedLock. To tell
the truth, I don't like it, but I have seen it used, and the alternative
of adding extra braces to ensure that the ScopedLock is immediately
destructed isn't very pretty either.
This can be better understood by taking some examples:
First:
Consider an expression:
x = ++j;
But for time being, if we need to assign a temporarily debug value, then we can write.
x = DEBUG_VALUE, ++j;
Second:
Comma , operators are frequently used in for() -loop e.g.:
for(i = 0, j = 10; i < N; j--, i++)
// ^ ^ here we can't use ;
Third:
One more example(actually one may find doing this interesting):
if (x = 16 / 4), if remainder is zero then print x = x - 1;
if (x = 16 / 5), if remainder is zero then print x = x + 1;
It can also be done in a single step;
if(x = n / d, n % d) // == x = n / d; if(n % d)
printf("Remainder not zero, x + 1 = %d", (x + 1));
else
printf("Remainder is zero, x - 1 = %d", (x - 1));
PS: It may also be interesting to know that sometimes it is disastrous to use , operator. For example in the question Strtok usage, code not working, by mistake, OP forgot to write name of the function and instead of writing tokens = strtok(NULL, ",'");, he wrote tokens = (NULL, ",'"); and he was not getting compilation error --but its a valid expression that tokens = ",'"; caused an infinite loop in his program.
The comma operator allows grouping expression where one is expected.
For example it can be useful in some case :
// In a loop
while ( a--, a < d ) ...
But in you case there is no reason to use it. It will be confusing... that's it...
In your case, it is just to avoid curly braces :
if(cond)
perror("an error occurred"), exit(1);
// =>
if (cond)
{
perror("an error occurred");
exit(1);
}
A link to a comma operator documentation.
There appear to be few practical uses of operator,().
Bjarne Stroustrup, The Design and Evolution of C++
Most of the oft usage of comma can be found out in the wikipedia article Comma_operator#Uses.
One interesting usage I have found out when using the boost::assign, where it had judiciously overloaded the operator to make it behave as a comma separated list of values which can be pushed to the end of a vector object
#include <boost/assign/std/vector.hpp> // for 'operator+=()'
using namespace std;
using namespace boost::assign; // bring 'operator+=()' into scope
{
vector<int> values;
values += 1,2,3,4,5,6,7,8,9; // insert values at the end of the container
}
Unfortunately, the above usage which was popular for prototyping would now look archaic once compilers start supporting Uniform Initialization
So that leaves us back to
There appear to be few practical uses of operator,().
Bjarne Stroustrup, The Design and Evolution of C++
In your case, the comma operator is useless since it could have been used to avoid curly braces, but it's not the case since the writer has already put them. Therefore it's useless and may be confusing.
It could be useful for the itinerary operator if you want to execute two or more instructions when the condition is true or false. but keep in mind that the return value will be the most right expression due to the comma operator left to right evalutaion rule (I mean inside the parentheses)
For instance:
a<b?(x=5,b=6,d=i):exit(1);
The boost::assign overloads the comma operator heavily to achieve this kind of syntax:
vector<int> v;
v += 1,2,3,4,5,6,7,8,9;
Conditional operator in C is used like this:
condition ? value_if_true : value_if_false
What does 0 mean when it's used in the value_if_false?
I've seen some people using it like this, for example.
a == b ? i++ : 0
It seems like it does nothing. Does this work like return 0 in other functions?
In C language, ternary is shorter version of if statement and it requires both statements, if_true and if_false. It would be like this (in fact it can have multiple statements for one case, separated with comma):
Short:
condition ? if_true : if_false;
Long:
if (condition) {
if_true;
} else {
if_false;
}
You can also assign the value if you put something infront of condition.
Short:
result = condition ? if_true : if_false;
Long:
if (condition) {
result = if_true;
} else {
result = if_false;
}
Now here is the trick. In C language, writing 0; is a valid statement, so your ternary becomes in longer version same as code below:
if (a == b) {
i++;
} else {
0; /* This is valid C statement */
}
Or if you have assignment too, it would be:
if (a == b) {
result = i++;
} else {
result = 0;
}
You can also do this:
int a;
/* Code here ... */
condition ? a = 5: 0;
That is effectively the same as:
if (condition) {
a = 5;
} else {
/* DO not touch a */
}
The ?: operator is a ternary operator, but it is not called "ternary" as some answers and/or comments here suggest. It just is the arity of the operator, just as + is a binary operator or as & is unary. If it has a name at all, it is called "Conditional Expression"-operator
It is not quite equivalent to if/else, because it is a conditional value (with the consequence, that both expressions must have the same type) in the first place, not a conditional execution. Of course, both types can be cast to make them equal.
In the case of what the OP does, a better option (if if shall not be used) is in my opinion:
a == b && i++;
which resembles a bit more logical what happens. But of course it is a matter of style.
The reason why someone might want to write a == b ? i++ : 0; is that s/he probably wants to have an (Caution! You are now entering an opinion-based area) easier and faster alternative to if (a == b) i++; - although this is of course opinion-based and I personally not share the same opinion.
One thing I can think of as a "blocker" at the if statement is the requirement to write the parentheses () which can be omitted by using the conditional operator instead.
"But why the 0?"
The C syntax requires a third operand for the conditional operator. Else if you would want to compile for example:
a == b ? i++;
you will get an error from the compiler:
"error: expected ':' before ';' token"
Or respectively, doing so:
a == b ? i++ : ;
would raise:
"error: expected expression before ';' token"
So they use 0 as kind of "syntax satisfier" to be able to use the conditional operator as replacement for the if statement. You could use any other numeral value here as well, but 0 is the most readable value, which signifies that it has no use otherwise.
To showcase the use at an example:
#include <stdio.h>
int main (void)
{
int a, b, c = 4;
a = 2;
b = 2;
a == b ? c++ : 0;
printf("%d",c);
return 0;
}
The output for c will be 5, because a == b.
Note that a == b ? i++ : 0 is different when used f.e. inside of an assignment like f.e.:
int c = a == b ? i++ : 0;
Here c is either getting assigned by i or 0, dependent upon a == b is true or not. If a == b is true, c is assigned by i. If a == b is wrong, c is assigned by 0.
Side Notes:
To view it from a technical point, ?= is called the "conditional operator". The conditional operator is one of the group of ternary operators.
If you want to learn more about the conditional operator ?=, look at ISO:IEC 9899:2018 (C18), §6.5.15 - "Conditional operator" for more information.
There's nothing special about 0 one could write
a == b ? i++ : 1
And it would behave the same way.
Only difference is when you assign the expression to say another variable:
int c = a == b ? i++ : 1;
// a == b -> c will be i++
// a != b -> c will be 1
However it's much cleaner to write
if (a == b) i++;
It helps to think of the ternary operator as a shorthand way or writing an if-else statement.
If(a == b){
i++;
}else{
//if assigned, return 0. Else do nothing.
}
What is the "hanging else" problem? (Is that the right name?)
Following a C++ coding standard (forgot which one) I always
use brackets (block) with control structures. So I don't
normally have this problem (to which "if" does the last(?)
else belong), but for understanding possible problems in
foreign code it would be nice with a firm understanding of
this problem. I remember reading about it in a book about
Pascal many years ago, but I can't find that book.
Ambiguous else.
Some info here: http://theory.stanford.edu/~amitp/yapps/yapps-doc/node3.html
But the classic example is:
if a then
if b then
x = 1;
else
y = 1;
vs.
if a then
if b then
x = 1;
else
y = 1;
Which if does the else belong to?
if (a < b)
if (c < d)
a = b + d;
else
b = a + c;
(Obviously you should ignore the indentation.)
That's the "hanging else problem".
C/C++ gets rid of the ambiguity by having a rule that says you can't have an-if-without-an-else as the if-body of an-if-with-an-else.
Looking at this from a langauge design point of view.
The standard BNF-like grammar for if-else:
Statement :- .. STUFF..
| IfStatement
IfStatement :- IF_TOKEN '(' BoolExpression ')' Statement IfElseOpt
IfElseOpt :- /* Empty */
| ELSE_TOKEN Statement
Now from a parsers point of view:
if (cond1) Statement1
if (cond2) Statement2
else Statement3
When you get to the ELSE_TOKEN the parser has two options, SHIFT or REDUCE. The problem is that which to choose requires another rule that the parser must follow. Most parsers generators default to SHIFT when given this option.
I don't see the problem for Pascal?
This one is incorrectly indented.
if a then
if b then
x = 1;
else
y = 1;
Removing the semi-colon from after x = 1 would make it correctly indented.
This one correctly indented
if a then
if b then
x = 1;
else
y = 1;
From this answer, I saw this:
unsigned long
hash(unsigned char *str)
{
unsigned long hash = 5381;
int c;
while (c = *str++)
hash = ((hash << 5) + hash) + c; /* hash * 33 + c */
return hash;
}
The problem is that I don't see how the while stops from executing. Generally an assignment isn't resulted in true? I mean if(a = 5) will be true.
Also the compiler gave me this warning:
warning: suggest parentheses around assignment used as truth value [-Wparentheses]
Can anyone step me trough on what exactly goes on with the loop? I would strongly suggests that an example would be the best approach here.
Also where should the parentheses go to make this more clear?
The value of an assignment is the value that gets assigned. For example, this code:
int a;
printf("%d\n", a=5);
… will print out 5.
So, this loop runs until c is assigned something false.
And in C, 0 is false, and strings are null-terminated, so at the end of the string, the NUL character, 0, will end the loop.
The warning is because if (a = 5) is a typo for if (a == 5) more often than it's intentional code. Putting extra parentheses around an assignment, like if ((a == 5)), is an idiomatic way to say "I meant to use the value of this assignment". With some compilers, you may get a different, stronger warning in the case where the value is constant, as in my example, than in examples like yours, but the issue is the same: if you meant to test the value of an assignment, and you don't want to disable the warning globally, the way to convince GCC to leave you along is with extra parentheses: while ((c = *str++)).
See this question for more details. You may also want to search the GCC mailing lists for threads like this one discussing the warning.
In my search for an example of a software phase lock loop I came across the following question
Software Phase Locked Loop example code needed
In the answer by Adam Davis a site is given that is broken and I have tried the new link that is given in a comment but I cant get that to work either.
The answer from Kragen Javier Sitaker gave the following code as a simple example of a software phase locked loop.
main(a,b){for(;;)a+=((b+=16+a/1024)&256?1:-1)*getchar()-a/512,putchar(b);}
Also included in his answer was a link to what should be a much more readable example but this link is also broken. Thus I have been trying to translate the above code into simpler and more readable code.
I have come this far:
main(a,b){
for(;;){
// here I need to break up the code somehow into a if() statement('s).
if(here I get lost){
a = a+1;
if(here i get lost some more){
b = b+1;
}
}
}
Thanks to the SO question What does y -= m < 3 mean?
I know it is possible to break up the a+= and b+= into if statements.
But the (&256? 1 : -1)*getchar()-a/512,putchar(b); part in the code is killing me. I have been looking on Google and on SO to the meaning of the symbols and functions that are used.
I know the & sign indicates an address in memory.
I know that the : sign declares a bit field OR can be used in combination with the ? sign which is a conditional operator. The combination of the two I can use like sirgeorge answer in what does the colon do in c?
Theory Behind getchar() and putchar() Functions
I know that getchar() reads one character
I know that putchar() displays the character
But the combination of all these in the example code is not readable for me and . I can not make it readable for my self even do I know what they all separately do.
So my question: How do I read this software phase lock loop code?
main(a,b){for(;;)a+=((b+=16+a/1024)&256?1:-1)*getchar()-a/512,putchar(b);}
What I get is:
main (a, b)
{
char c;
for (;;)
{
c = getchar();
b = (b + 16 + (a / 1024));
if(!(b & 256))
{
c = c * -1;
}
a = a + c - (a/512);
putchar(b);
}
}
I had to add a c variable to not get lost.
What the program does:
Take a and b.
Infinite loop:
get a char input in c
calculate b
If (b bitwise AND 256)
c = -c
Calculate a
Print b
It seems it translate input into something else, I have to see the code in action to understand better myself.
Hope it helped!
Hint:
https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B
a+= => a = a +
a?b:c => if(a){return b;} else {return c;} (As a function itself, it don t truly return)
Add parentheses, it help.
a & b is bitwise AND:
a/b |0|1|
0|0|0|
1|0|1|