It is known that both assignment = and conditional ?: operators have right associativity. In following code sample:
#include <stdio.h>
int main(void)
{
int a, b, c, d;
a = b = c = d = 1;
1 ? a++ : b ? c++ : d;
printf("%d %d %d %d\n", a, b, c, d);
return 0;
}
the assignment:
a = b = c = d = 1;
is equivalent to:
a = (b = (c = (d = 1)));
and correspondingly:
1 ? a++ : b ? c++ : d;
is the same as:
1 ? a++ : (b ? c++ : d);
What the Standard say about this last case? Does it guarantee that such combined expression is evaluated from left to right (so the c++ part is not evaluated), just as opposite to assignment?
The evaluation order of ?: is guaranteed: the first operand is evaluated first, then evaluate the second or the third operand depending on whether the first operand is true.
You are mainly confused about the relationship between operator precedence/associativity and order of evaluation. They play different roles. The former decides how operators are grouped, while the latter decides which sub-expression is evaluated first.
Consider the expression a * b + c * d, the precedence rule means it's equivalent to (a * b) + (c * d). But is it guaranteed that the compiler will evaluate a * b before c * d? The answer is no, in this example, the operator + doesn't guarantee the order of evaluation.
The conditional operator ?: is one of the few operators that do have a specified order of evaluation. (The rest are &&, ||, and ,).
In your example
1 ? a++ : b ? c++ : d;
1 ? a++ : (b ? c++ : d);
are always equivalent, in both expressions, 1 is evaluated first, and since it's true, a++ is evaluated next, the end.
Associativity and precedence do not define order of evaluation. These concepts are completely unrelated. Order of evaluation in C is defined by sequencing rules, not by precedence or associativity.
It is true that a = b = c = d = 1; is associated as a = (b = (c = (d = 1)));, but that does not mean that d = 1 should be evaluated first, especially in C language, where assignment operator evaluates to an rvalue.
Associtivity simply says that c should receive value 1 converted to the type of d ("as if" it was read from d). But that does not mean that d = 1 should be done first. In your example all variables have the same type, which means that the whole thing is equivalent to a = 1; b = 1; c = 1; d = 1; in absolutely any order. There's no sequencing inside a = b = c = d = 1 expression.
The same logic applies to ?: operator. Its associativity simply tells you which operand belongs to which operator. And the grouping is indeed 1 ? a++ : (b ? c++ : d);. But associativity does not tell you anything about the order of evaluation. Order of evaluation in ?: operator is defined separately and independently: the condition is always evaluated (sequenced) first, then one (and only one) of the branches is evaluated. In your example 1 is evaluated first, then a++ is evaluated and its result becomes the result of the entire expression. The (b ? c++ : d) part is not even touched.
1 ? a++ : b ? c++ : d;
is equivalent to
if (1) {
a++;
}
else {
if (b) {
c++;
}
else {
d;
}
}
So, the output will be
2 1 1 1
Related
I'm practicing to read C-code faster but I'm struggling to understand this portion:
int a = 7, b = 3, c = 2, d;
d = (b+c >a) || a > c ? a-- : a++;
When I enter this into my program and print out integer d, I get the result of 7. But I cannot really understand why. Can somebody please explain this to me?
You are assigning to d the value of either a-- or a++. Both of these expressions have a value of 7 (the original value of a); the difference between the operators is what they do to their operand (a) after evaluating the expression: one then increments it and the other decrements it.
That is why they are called post-increment and post-decrement operators.
Maybe you are getting confused with the pre-increment and pre-decrement operators, which are ++a and --a? (See also: Pre increment vs Post increment in array .)
As for the ternary operator (x ? y : z) and how to 'read' that, then you can take it as if x then y else z; so, if the result of the test (which has a logical OR, which means it will be true if either comparison is true) is true, then d = a--, otherwise d = a++. But, as mentioned already, both will give the same value to d, in this case (though they will have different effects on a). Note also that the ternary operator has a lower precedence than either the logical OR or the relational comparisons, so you can read as though everything between the = and the ? is in (another set of) brackets.
The expression is parsed as
d = ((b+c >a) || a > c) ? a-- : a++;
so d gets the result of either a— or a++, which is 7 in both cases. The difference is what the value of a is after the expression is evaluated.
|| forces left to right evaluation, so b+c > a is evaluated first. Since b+c is 5, the result of the expression is false (0), so a > c is evaluated. The result of that expression is true (1), so the result of (b+c > a) || a > c is true (1), meaning we assign the result of a— to d. Thus d will be 7 and a will be 6.
This is how operator precedence works in C
Step 1
d = (3 + 2 > 7) || 7 > 2 ? a-- : a++;
Step 2
d = false || true ? a-- : a++;
Step 3
d = true ? a-- : a++;
Here value of 'a' will be changed but not in this statement so value of 'a' still be 7
But if you print a in other statement it will changed as 6.
To learn more about operator precedence https://en.cppreference.com/w/c/language/operator_precedence
#define MAX(a,b) ((a)>(b) ? (a) : (b))
int main(void) {
int a=2;
int b=3;
int c = MAX(a++,b++); // c=((a++)>(b++) ? (a++) : (b++));
printf("\na= %d", a);// a=3
printf("\nb= %d", b);//b=5
printf("\nc= %d", c);//c=4
a=3;
b=2;
cc = MAX(a++,b++); // c=((a++)>(b++) ? (a++) : (b++));
printf("\na= %d", a); // a=5
printf("\nb= %d", b); //b=3
printf("\nc= %d", c); //c=4
return 0;
}
I would like to know why c is not evaluating to 5.
It appears to me the evaluation order should be:
First both a and be get incremented in (a++)>(b++)
If for example the first one is greater, the ternary operator in
c=((a++)>(b++) ? (a++) : (b++)), goes to (a++), so a
increments again.
The result of the ternary expression, which is a twice-increment,
should be assigned to c, then c should have the greater value
twice-incremented, that is 5. However, I am getting 4. I suspect the
greater value's second increment is happening at the end, but I can't
explain why, since the parentheses seem to indicate that the
assignment comes at the end.
Any idea?
The postfix ++ operator has a result and a side effect. The result of a++ is the value of a before the increment - given
int a = 1;
int x = a++;
the value of x will be 1 and the value of a will be 2. Note that the side effect of adding 1 to a does not have to be applied immediately after evaluation - it only has to be applied before the next sequence point.
So, looking at
((a++) > (b++)) ? (a++) : (b++)
The ?: operator forces left-to-right evaluation, so the first thing that happens is that (a++) > (b++) is evaluated1. Since a is initially 2 and b is initially 3, the result of the expression is false (0). The ? operator introduces a sequence point, so the side effects to a and b are applied and a is now 3 and b is now 4.
Since the result of the condition expression was 0, we evaluate b++. Same thing, the result of the expression is the current value of b (4), and that value gets assigned to c. The side effect to b is applied at some point, and by the time everything's finished, a is 3, b is 5, and c is 4.
Although within that expression either a++ or b++ may be evaluated first, since the > operator doesn't force left-to-right evaluation.
Let's consider for example this declaration
int c = MAX(a++,b++);
after the macro substitution there will be
int c = (( a++) >( b++ ) ? (a++) : (b++));
The variables a and b are initialized like
int a=2;
int b=3;
As a is less than b then the third expression (b++) will be evaluated as a result of the conditional operator. In the first expression ( a++) >( b++ ) a and b were incremented. There is a sequence point after the evaluation of the first expression.
So a will be set to 3, b will be set to 4.
As it was already said the value of the conditional operator is the value of the third expression (b++) where there is used the post-increment. The value of the post-increment operator is the value of its operand before incrementing.
From the C Standard (6.5.2.4 Postfix increment and decrement operators)
2 The result of the postfix ++ operator is the value of the
operand. As a side effect, the value of the operand object is
incremented (that is, the value 1 of the appropriate type is added to
it).
So the value of the conditional operator is 4. This value will be assigned to the variable c. But as a side effect the value of b will be incremented.
Thus after this declaration a will be equal to 3, b will be equal to 5 and c will be equal to 4.
For clarity this declaration
int c = (( a++) >( b++ ) ? (a++) : (b++));
in fact can be rewritten in the logically equivalent way.
int result = a > b;
++a;
++b;
int c;
if ( result != 0 )
{
c = a++;
}
else
{
c = b++;
}
For a++ it's equivalent as saying
int temp = a;
a = a + 1;
return temp;
so in the case of a=2, b=3, we can go to the condition (a++)>(b++) and we can have a rewritten as
int temp1 = a;
a = a + 1;
return temp1;
and b rewritten as
int temp2 = b;
b = b + 1;
return temp2;
now since it's only a conditional statement, we're really just evaluating the old values of a and b before the increment which is temp1 = 2 and temp2 = 3 while at the same time a and b values have changed a = 3, b = 4. Since temp1 < temp2 from the old value, we go to the false clause of the ternary operator (b++) and do the same thing as we did before
int temp3 = b;
b = b + 1;
return temp3;
so now b is 5 while the returned temp3 is the previous value of b which is 4. Hope this helps!
I have a question, how the compiler operate on the following code:
#include<stdio.h>
int main(void)
{
int b=12, c=11;
int d = (b == c++) ? (c+1) : (c-1);
printf("d = %i\n", d);
}
I am not sure why the result is d = 11.
In int d = (b == c++) ? (c+1) : (c-1);:
The value of c++ is the current value of c, 11. Separately, c is incremented to 12.
b == 11 is false, since b is 12.
Since (b == c++) is false, (c-1) is used. Also, the increment of c to 12 must be completed by this point.
Since c is 12, c-1 is 11.
d is initialized to that value, 11.
According to the C Standard (6.5.15 Conditional operator)
4 The first operand is evaluated; there is a sequence point between
its evaluation and the evaluation of the second or third operand
(whichever is evaluated). The second operand is evaluated only if the
first compares unequal to 0; the third operand is evaluated only if
the first compares equal to 0; the result is the value of the second
or third operand (whichever is evaluated), converted to the type
described below.110)
So in the initializing expression of this declaration
int d = (b == c++) ? (c+1) : (c-1);
the variable b is compared with the value of the variable c because the post-increment operator returns the value of its operand before incrementing it.
As the values are not equal each other (b is set to 12 while c is set to 11) then the sub-expression (c-1) is evaluated.
According to the quote there is a sequence point after evaluation of the condition of the operator. It means that after evaluation of the condition c has the value 12 after applying the post-increment operator to the variable c. As a result the variable d is initialized by the value 1 (12 - 1).
Translated to a regular if-statement your code would look like this:
int b=12, c=11;
int d;
if (b == c++)
d = c+1;
else
d = c-1;
The clue here is that c is incremented after the condition is checked. So you enter the else state but c already has the value 12 there.
Beacuse the condition is false, therefore the false case will happen: c-1, but since you incremented c in the condition by c++, therefore c is now 12. The result thus 12 - 1 which is 11.
EDIT:
What OP misunderstood was the post increment.
So what actually happen is like this:
#include<stdio.h>
int main(void)
{
int b=12, c=11;
int d;
if (b == c) { // 12 == 11 ? -> false
c = c + 1;
d = c + 1;
} else { // this executes since condition is false
c = c + 1; // post increment -> c++ -> c = 12 now
d = c - 1; // 12 - 1 = 11 -> d = 11
}
printf("d = %i\n", d);
}
Refer to Ternary Operator.
Syntax
condition ? value_if_true : value_if_false
So, you wrote
int d = (b == c++) ? (c+1) : (c-1);
In this situation, the result will be 11 because, after if checks, 'c' value is increased(c+1=12) and only after that it sets 'd' value as c(12)-1 which is 11.
If you used, for example:
int d = (b == ++c) ? (c+1) : (c-1);
"c" value would be increased before checking the statement, so it would be true and "d" value would be c(12)+1 which is 13.
#include <stdio.h>
int main() {
int a = 1;
int b = a || (a | a) && a++;
printf("%d %d\n", a, b);
return 0;
}
when I ran this code the results were 1 and 1.
According to the C language Operator Precedence the operation && is supposed to happen before the operation ||. So shouldn't the result be 2 1 ? (a = 2, b = 1)
when OR'ing expressions in C, a shortcut is taken, I.E. as soon as an expression is evaluated to TRUE, the rest of the OR'd expressions are not evaluated
The first expression a evaluates to TRUE, so all the rest of the expressions are not evaluated, so a is never incremented
When applying the operator precedence rules, the expression is equivalent to:
int b = a || ((a | a) && a++);
The evaluation or the operands to || and && is performed from left to right and shortcut evaluation prevents evaluating the right operand if the left operand can determine the result: since a is non zero, a || anything evaluates to 1 without evaluating the right operand, hence bypassing the a++ side effect.
Therefore both a and b have value 1 and the program prints 1 1.
Conversely, if you had written int b = (a || ((a | a)) && a++;, the left operand of && would have value 1, so the right operand need to be evaluated. a++ evaluates to 1 but increments a, so b have final value 1 and a is set to 2, producing your expected result.
The confusion comes from equating operator precedence with order of evaluation. These are two separate notions: operator precedence determines the order in which to apply the operators, but does not determine the order of evaluation of their operands.
Only four operators have a specified order of evaluation of their operands: &&, ||, ? : and , and the first 2 may skip evaluation of one, depending on the value of the other operand and the third only evaluates the first and only one among the second and third operands. For other operators, the order of evaluation of the operands is unspecified, and it may differ from one compiler to another, one expression to another or even one run to another, although unlikely.
When I try to put a || (a | a) into a parenthesis. The result was as you expected.
So I guess that when C compiler executes an OR operator and it get a True value to the OR, the execution will finish immediately.
In your case, when the compiler executed the a || (a | a) (1 || something) operation. Value of b will be declared to 1 right away and a++ operator won't be execute.
Consider the following C code.
#include <stdio.h>
int main(void)
{
int test = 0;
int a= 10,b = 20;
test ? a*2 : b*3;
printf("a = %d, b = %d\n",(test ? a = 200 : b = 300),(test ? a =2 : b = 3));
return 0;
}
When trying to compile it throws the following error.
file1.c:11:50: error: lvalue required as left operand of assignment
printf("a = %d, b = %d\n",(test ? a = 200 : b = 300),(test ? a =2 : b = 3));
^
file1.c:11:74: error: lvalue required as left operand of assignment
printf("a = %d, b = %d\n",(test ? a = 200 : b = 300),(test ? a =2 : b = 3));
^
I have already provided an l value as left operand for the assignment operations and besides why does it not throw an error in the line
test ? a*2 : b*3;
and it produces error only in the line
printf("a = %d, b = %d\n",(test ? a = 200 : b = 300),(test ? a =2 : b = 3));
Please explain.
You're probably missing parenthesis and the compiler thinks you mean:
(test ? a = 200 : b) = 300
when you actually meant:
test ? (a = 200) : (b = 300)
The ternary operator(?:) has higher precedence(not strictly) as compared to assignment(=). So, putting assignment statement inside round brackets can have more precedence as shown:
#include <stdio.h>
int main(void)
{
int test = 0, ret;
int a= 10,b = 20;
test ? a*2 : b*3;
printf("a = %d, b = %d\n",(test ? (a = 200) : (b = 300)), (test ? (a =2) : (b = 3)));
return 0;
}
Strictly speaking, it's not the matter of Operator precedence, but of language grammar.
According to C Operator Precedence:
The C language standard doesn't specify operator precedence. It specifies the language grammar, and the precedence table is derived from it to simplify understanding. There is a part of the grammar that cannot be represented by a precedence table: an assignment-expression is not allowed as the right hand operand of a conditional operator, so e = a < d ? a++ : a = d is an expression that cannot be parsed, and therefore relative precedence of conditional and assignment operators cannot be described easily.
However, many C compilers use non-standard expression grammar where ?: is designated higher precedence than =, which parses that expression as e = ( ((a < d) ? (a++) : a) = d ), which then fails to compile due to semantic constraints: ?: is never lvalue and = requires a modifiable lvalue on the left. This is the table presented on this page.