The code below compiles well
int a=5,b=4,c;
a>b?30:40;
Also does,
int a=5,b=4,c;
a>b?c=30:40;
But why this does not work?
int a=5,b=4,c;
a>b?c=30:c=40;
You are being bitten by precedence. ?: has very low precedence, but not as low as = or , (see the operator precedence table).
Your code is parsed as:
(a>b ? c=30 : c) = 40;
Rather than:
a>b ? c=30 : (c=40);
You don't need parenthesis around c=30 because ? and : act like parentheses to the expression within.
Believe it or not, (a>b ? c=30 : c) = 40 is valid C++ (but not valid C). The expression (a>b ? c=30 : c) is an lvalue referencing the variable c, to which 40 is assigned.
You've run into a precedence problem with the = operator. If you insist on assignment inside of your ternary operator, merely wrap the sub expressions in parentheticals:
int d = a > b ? (c = 30) : (c = 40); // explicit precedence
The last one:
int a=5,b=4,c;
a>b?c=30:c=40;
fails because it's trying to assign 40 to a>b?c=30:c, which obviously won't work. The = has lower precedence, and a>b?c=30:c is a valid expression (though you can't assign to it). The = in the c=30 part is sort of an exception because it's in the middle of the ternary operator, between the ? and the :. To fix it you'd simply need to add parentheses around the c=40 so that it's evaluated as a single value for the 'else' part of the ternary operator, i.e. a>b?c=30:(c=40);
The second example
a>b?c=30:40;
doesn't assign anything to c unless a is greater than b... which it is when a is 5 and b is 4, as in this case; but note that if a were not greater than b, no assignment would occur.
From the first example
a>b?30:40
is a valid expression, with a value of 30 or 40, but you're not doing anything with that value so of course it serves no purpose there.
Of course, you'd normally use something more like:
c = a>b ? 30 : 40;
where a>b ? 30 : 40 will evaluate to 30 or 40, which is then assigned to c. But I suspect you know that and simply want to know why the c=40 is not treated as a single value for the 'else' part of the ternary operator in the last example.
Related
I'm starting to learn to program in c, and I thought I was already pretty confident with the precedence of operators, until I did this:
a > b ? c = a : (c = b);
Of course at the first time I didn't use parenthesis on the last sentence, but since that ended up causing compiling issues I searched how to solve that problem on this forum and I read that adding parenthesis could do the job. However, I thought that the expressions inside parenthesis get executed before anything else written in the same line, which would mean that the c = b sentence was executed first and then the ternary operator. I did something similar but easier to read in order to get a better idea of what was happening with this operator precedence thing and tried executing this line:
printf("Number is %d", i) + (i = 5);
I know this expression returns returns a value, but since I don't need it and this isn't a line that I will keep for more than 5 seconds, I won't store it in any variable. What gets my attention in this case is that, when I execute the code, I doesn't show up on the screen with the value 5, but instead it uses the previous value, which means that the computer is just reading it from left to right. When I do:
(i = 5) + printf(Numer is %d, i);
it first does the assignment of i and only after that the printf function is executed. My question is: how does the computer execute an expression that uses operators of different orders of precedence? It clearly doesn't run first the operator with the highest precedence, because in the first printf the value stored wasn't the one assigned on the parenthesis, but it also doesn't just read from left to right because in that case there would be no operator precedence. How does it work?
Parenthesis and operator precedence only dictate how operands are grouped. It does not dictate the order of evaluation.
In this expression:
a > b ? c = a : (c = b);
The three parts of the ternary operator are a > b, c = a, and c = b respectively. This operator also has the property that only one of the second and third clause are evaluated, based on the result of the first. Formally speaking, there is a sequence point between the evaluation of the first clause and of either the second or third. So a > b is first evaluated. If it is nonzero, c = a is evaluated, otherwise c = b is evaluated.
In this expression:
printf("Number is %d", i) + (i = 5);
There is nothing that dictates whether printf("Number is %d", i) or i = 5 is evaluated first. Unlike the ternary operator, there is no sequence point between the evaluation of the operands of the + operator. This expression also has a problem: i is both read and written in the same expression without a sequence point. Doing so triggers undefined behavior. This is also true for:
(i = 5) + printf(Numer is %d, i);
On a side note, this:
a > b ? c = a : (c = b);
Can be more clearly written as:
c = a > b ? a : b;
why does a<=20? b=10 : c=30; give lvalue error?
error: lvalue required as left operand of assignment
is it because b=10 and c=30 are statements and not expressions?
It is because conditional operator has higher precedence than assignment operator and the expression is interpreted as
((a<=20)? (b=10) : c)=30;
What is returned from conditional operator is not a lvalue (N15706.5.15, footnote 110), so the program will emit compile error.
You can use parentheses to overcome this problem of precedence.
a<=20? (b=10) : (c=30);
Using normal if statement should be better unless you have some reasons like:
Doing code golf
Trying to make your program hard to read
Want to use what is returned from the expression
if (a <= 20) {
b = 10;
} else {
c = 30;
}
Your code is missing parentheses around assignments.
This compiles and runs correctly:
int a = 20, b = -1, c = -1;
a<=20? (b=10) : (c=30);
printf("b=%d, c=%d\n", b, c);
Demo.
Note: It goes without saying that such (mis)use of ternary operator has negative impact on readability of your code, and should be avoided in favor of a regular if statement.
Can anybody please tell me why this statement is giving an error - Lvalue Required
(a>b?g=a:g=b);
but this one is correct
(a>b?g=a:(g=b));
where a , b and g are integer variables , and a and b are taken as input from keyboard.
In the expression,
(a > b ? g = a : g = b);
the relational operator > has the highest precedence, so a > b is grouped as an operand. The conditional-expression operator ? : has the next-highest precedence. Its first operand is a>b, and its second operand is g = a. However, the last operand of the conditional-expression operator is considered to be g rather than g = b, since this occurrence of g binds more closely to the conditional-expression operator than it does to the assignment operator. A syntax error occurs because = b does not have a left-hand operand (l-value).
You should use parentheses to prevent errors of this kind and produce more readable code which has been done in your second statement
(a > b ? g = a : (g = b));
in which last operand g = b of : ? has an l-value g and thats why it is correct.
Alternatively you can do
g = a > b ? a : b
The expression:
(a>b?g=a:g=b)
parsed as:
(a>b?g=a:g)=b
And we can't assign to an expression so its l-value error.
Read: Conditional operator differences between C and C++ Charles Bailey's answer:
Grammar for ?: is as follows:
conditional-expression:
logical-OR-expression
logical-OR-expression ? expression : conditional-expression
This means that a ? b : c = d parses as (a ? b : c) = d even though (due to the 'not an l-value' rule) this can't result in a valid expression.
One side note:
Please keep space in you expression so that it become readable for example.
(a>b?g=a:g=b);
Should be written as:
(a > b? g = a: g = b);
similarly, you should add space after ; and ,.
The problem is operator precedence: In C the ternary conditional operator (?:) has a higher precedence than the assignment operator (=).
Without parenthesis (which don't do anything here) your expression would be this:
a > b ? g = a : g = b;
The operator with the highest precedence in there would be the comparison >, so this is where you'll get your first logical grouping:
(a > b) ? g = a : g = b;
The next highest expression is the ternary conditional, which results in the following expression:
((a > b) ? (g = a) : (g)) = b;
As you can see, you'll now end up with an lvalue (i.e. a value; not a variable) on the left side of your assignment operator, something that won't work.
As you already noticed, the solution to this is to simply group the expressions on your own. I'd even consider this good practice, especially if you're unsure how your precedence might play out. If you don't want to think about it, add parenthesis. Just keep code readability in mind, so if you can, resolve the operator precedence on your own, to ensure you've got everything right and readable.
As for readability: I'd probably use a classic if() here or move the assignment operator outside the ternary conditional, which is how you usually define max():
g = a > b ? a : b;
Or more general as a macro or inline function:
#define max(a, b) ((a) > (b) ? (a) : (b))
inline int max(int a, int b) {
return a > b ? a : b;
}
if(a>b)
{
g = a;
}
else
{
g = b;
}
that can be replaced with this
g = a > b ? a : b; //if a>b use the first (a) else use the second (b)
Your expression (a>b?g=a:g=b) is parsed as :
(a>b?g=a:g)=b
// ^^^
From the Microsoft documentation :
conditional-expression:
logical-or-expression
logical-or-expression ? expression : conditional-expression
In C, the operator ?: has an higher precedence that the operator =. Then it means that ( a ? b : c = d ) will be parsed as ( a ? b : c ) = d. Due to l-value's rule, the first expression is also valid but is not doing what you think.
To avoid this error, you can do also :
g = ( a > b ) ? a : b;
This question usually triggers a barrage of answers trying to explain the situation through the concept of operator precedence. In reality it cannot be explained that way, since this is a typical example of an input, on which surrogate concepts like "operator precedence" break down. As you probably know, there's really no "operator precedence" in C. There are only grammatical groupings, which generally cannot be expressed precisely through any linear ordering of operators.
Let's take a look at what the language specification says about it. The relevant portions of C grammar in this case are the grammars of ?: operator and = operator. For ?: operator it is
conditional-expression:
logical-OR-expression
logical-OR-expression ? expression : conditional-expression
and for the = operator it is
assignment-expression:
conditional-expression
unary-expression assignment-operator assignment-expression
In the first case the critical part is the last operand of ?: operator: it is not an expression, but rather a conditional-expression. The conditional-expression is a different entry point into the grammar of C expression: it "enters" the grammar at the point where it is no longer possible to include a top-level = operator into a conditional-expression. The only way to "smuggle" a = operator into a conditional-expression is to descend the grammar all the way to the very bottom
primary-expression:
identifier
constant
string-literal
( expression )
generic-selection
and then wrap around all the way to the top using the ( expression ) branch. This means that a conditional-expression can contain a = operator only when it is explicitly wrapped in (...). E.g. the grammar prohibits you from having g = b as the last operand of ?: operator. If you want something like that, you have to explicitly parenthesize it: <smth> ? <smth> : (g = b).
A very similar situation exists with the second piece of grammar: assignment operator. The left-hand side (LHS) of assignment is unary-expression. And unary-expression "enters" the general grammar of C expression at the point where it is too late to include a top level ?: operator. The only way to reach the ?: operator from unary-expression is to descend all the way down to primary-expression and take the ( expression ) branch. This means that grammar prohibits you from having a > b ? g = a : g as the LHS operand of = operator. If you want something like that, you have to explicitly parentesize it: (a > b ? g = a : g) = <smth>.
For this reason "popular" answers claiming that "operator precedence" makes the language to parse your expression as
(a > b ? g = a : g) = b
are actually completely incorrect. In reality, there's no derivation tree in formal C grammar that would make your input fit the syntax of C language. Your input is not parsable at all. It is not an expression. It is simply syntactically invalid. C language sees it as a syntactic gibberish.
Now, in practice you might see some implementations to respond with a "lvalue required as left operand of assignment" diagnostic message. Formally, this is a misleading diagnostic message. Since the above input does not satisfy the grammar of C language expression, there's no "assignment" in it, there's no "left operand" and there's no meaningful "lvalue" requirement.
Why do compilers issue this strange message? Most likely they do indeed parse this input as a valid C expression
(a > b ? g = a : g) = b
The result of ?: is never an lvalue in C, hence the error. However, this interpretation of your input is non-standard (extended?) behavior, which has no basis in formal C language. This behavior of specific implementations might be caused by their attempts to reconcile C and C++ grammars (which are quite different in this area), by their attempts to produce a more readable (albeit "fake") error message or by some other reason.
Typically, in such implementations a similar issue also would pop up in case of inputs like
a + b = 5
The same error would be issued, suggesting a (a + b) = 5 parse, while from the pedantic point of view a + b = 5 is not parsable as an expression at all (for the same reasons as described above).
Again, formally, this is not enough to say that the compiler is "broken": the compiler is required to detect a constraint violation and issue some diagnostic message, which is exactly what happens here. The fact that the text of the diagnostic message does not correctly reflect the nature of the problem is inconsequential (the compiler can simply say "Ha ha ha!"). But one undesirable consequence of such misleading diagnostics is that it misleads users into misinterpreting the problem, which is BTW painfully evident from the barrage of formally incorrect answers posted to this question.
Hey thank you very much for your time! I'm having trouble understanding the syntax of a statement in my audio coding textbook. In one example there is a print function that goes like this
printf("%d semitones up or %d semitones down\n", interval,
interval ? 12-interval : 0 );
The part I don't understand is the conditional operator, or "?". It seems like I should just read it as "if interval does not equal 0, interval = 12 - interval" but the syntax here seems strange. I'm used to the conditional operator being a more fleshed out statement, like:
a = b > c ? b : c;
"If b is bigger than c, than a = b; else a = c"
Could someone point me to any other reference for this, or explain more about this syntax? I can't find similar examples.
You're almost right, but there's no assignment taking place. It's saying "if interval is non-zero, pass 12 - interval to the printf statement, otherwise pass 0".
In general the ternary operator looks like this:
a ? b : c
Where a, b, and c are all expressions. If a evaluates to non-zero, the ternary operator evaluates as if it were b, and if a evaluates to zero, the ternary operator's result is the result of evaluating c.
Your second example is a combination of the ternary operator and the assignment operator. The ternary operator itself doesn't perform any assignments.
Any expression that results in a boolean will do. In the case of C, where integers can be used as booleans, the value 0 is considered false and anything else is considered true.
So, in your case, interval ? 12-interval : 0, means: if interval is nonzero, use 12-interval, otherwise, use 0. To be extra verbose, you could rewrite it to:
interval != 0 ? 12-interval : 0
Following program gives error
#include<stdio.h>
int main ()
{
int a=10,b;
a>=5?b=100:b=200;
printf("\n%d",b);
}
the error is
ka1.c: In function ‘main’:
ka1.c:5: error: lvalue required as left operand of assignment
now if I replace the line
a>=5?b=100:b=200;
by
a>=5?b=100:(b=200);
and then compile then there is no error.
So I wanted to know what is wrong with
a>=5?b=100:b=200;
The ternary operator (?:) has higher precedence than the assignment operator (=). So your original statement is interpreted as:
((a >= 5) ? (b = 100) : b) = 200;
Write it like this instead:
b = (a >= 5) ? 100 : 200;
This is idiomatic C. (The brackets around the condition are not really necessary, but they aid readability.)
You're using the ternary operator incorrectly. Both of your examples are wrong, even though one compiles. The expression evaluates to either the second or third sub-expression depending upon the truth value of the first.
So a ? b : c will be the same thing as b if a is true, or c if a is false.
The proper way of using this operator is to assign the result to a variable:
b = a>= 5 ? 100 : 200;
Because it tries to do: (a>=5?b=100:b)=200
But the thing in parentheses is not lvalue.