I have three variables: a, b, c. Let's say they are integers. I want to find the first non-zero value among them, in that particular order, without looping. The following seems to work, but I am not sure if that is because I am lucky, or because the language guarantees it:
int main(int argc, char *argv[]) {
int a = 0;
int b = 3;
int c = 5;
int test;
if ((test = a) != 0 || (test = b) != 0 || (test = c) != 0) {
printf("First non-zero: %d\n", test);
} else {
printf("All zero!\n");
}
return 0;
}
Is the repeated assignment with short-circuiting shown here guaranteed to work as intended, or am I missing something?
This might be one place where a three-letter answer would be acceptable, but a two-letter answer might require more explanation.
It would!
Because of the nature of the OR operator if any of the condition is
true then the test stops.
Thus i think what you did was basically equivalent to:
test = a != 0 ? a : b != 0 ? b : c != 0 ? c : 0;
printf("%d\n",test);
but heck yours looks good.
[update]
As per what chqrlie mentioned it can be further simplified to:
test = a ? a : b ? b : c;
Yes, your expression is fully defined because there is a sequence point at each || operator and the short circuit evaluation guarantees that the first non zero value assigned to test completes the expression.
Here is a crazy alternative without sequence points that may produce branchless code:
int test = a + !!a * (b + !!b * c);
printf("%d\n", test);
The code is very bad practice but it is guaranteed to work fine.
This is because the || and && operators have special characteristics - unlike most operators in C, they guarantee that the evaluation of the left operand is sequenced (executed) before the evaluation of the right operand. This is the reason that the code works. There's also a guarantee that the right operand will not be evaluated if it is sufficient to evaluate the left one ("short circuit"). Summarized in C17 6.5.14/4:
Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the
second operand is evaluated, there is a sequence point between the evaluations of the first
and second operands. If the first operand compares unequal to 0, the second operand is
not evaluated.
"Sequence point" being the key here, which is what gives the expression a deterministic outcome.
Had you used pretty much any other operator (like for example bitwise |), then the result would be undefined, because you have multiple side effects (assignments) on the same variable test in the same expression.
A more sound version of the same algorithm would involve storing the data in an array and loop through it.
I am writing a small test app in C with GCC 4.8.4 pre-installed on my Ubuntu 14.04. And I got confused for the fact that the expression a=(b++); behaves in the same way as a=b++; does. The following simple code is used:
#include <stdint.h>
#include <stdio.h>
int main(int argc, char* argv[]){
uint8_t a1, a2, b1=10, b2=10;
a1=(b1++);
a2=b2++;
printf("a1=%u, a2=%u, b1=%u, b2=%u.\n", a1, a2, b1, b2);
}
The result after gcc compilation is a1=a2=10, while b1=b2=11. However, I expected the parentheses to have b1 incremented before its value is assigned to a1.
Namely, a1 should be 11 while a2 equals 10.
Does anyone get an idea about this issue?
However, I expected the parentheses to have b1 incremented before its value is assigned to a1
You should not have expected that: placing parentheses around an increment expression does not alter the application of its side effects.
Side effects (in this case, it means writing 11 into b1) get applied some time after retrieving the current value of b1. This could happen before or after the full assignment expression is evaluated completely. That is why a post-increment will remain a post-increment, with or without parentheses around it. If you wanted a pre-increment, place ++ before the variable:
a1 = ++b1;
Quoting from the C99:6.5.2.4:
The result of the postfix ++ operator is the value of the operand.
After the result is obtained, the value of the operand is incremented.
(That is, the value 1 of the appropriate type is added to it.) See the
discussions of additive operators and compound assignment for
information on constraints, types, and conversions and the effects of
operations on pointers. The side effect of updating the stored value
of the operand shall occur between the previous and the next sequence
point.
You can look up the C99: annex C to understand what the valid sequence points are.
In your question, just adding a parentheses doesn't change the sequence points, only the ; character does that.
Or in other words, you can view it like there's a temporary copy of b and the side-effect is original b incremented. But, until a sequence point is reached, all evaluation is done on the temporary copy of b. The temporary copy of b is then discarded, the side effect i.e. increment operation is committed to the storage,when a sequence point is reached.
Parentheses can be tricky to think about. But they do not mean, "make sure that everything inside happens first".
Suppose we have
a = b + c * d;
The higher precedence of multiplication over addition tells us that the compiler will arrange to multiply c by d, and then add the result to b. If we want the other interpretation, we can use parentheses:
a = (b + c) * d;
But suppose that we have some function calls thrown into the mix. That is, suppose we write
a = x() + y() * z();
Now, while it's clear that the return value of y() will be multiplied by the return value of z(), can we say anything about the order that x(), y(), and z() will be called in? The answer is, no, we absolutely cannot! If you're at all unsure, I invite you to try it, using x, y, and z functions like this:
int x() { printf("this is x()\n"); return 2; }
int y() { printf("this is y()\n"); return 3; }
int z() { printf("this is z()\n"); return 4; }
The first time I tried this, using the compiler in front of me, I discovered that function x() was called first, even though its result is needed last. When I changed the calling code to
a = (x() + y()) * z();
the order of the calls to x, y, and z stayed exactly the same, the compiler just arranged to combine their results differently.
Finally, it's important to realize that expressions like i++ do two things: they take i's value and add 1 to it, and then they store the new value back into i. But the store back into i doesn't necessarily happen right away, it can happen later. And the question of "when exactly does the store back into i happen?" is sort of like the question of "when does function x get called?". You can't really tell, it's up to the compiler, it usually doesn't matter, it will differ from compiler to compiler, if you really care, you're going to have to do something else to force the order.
And in any case, remember that the definition of i++ is that it gives the old value of i out to the surrounding expression. That's a pretty absolute rule, and it can not be changed just by adding some parentheses! That's not what parentheses do.
Let's go back to the previous example involving functions x, y, and z. I noticed that function x was called first. Suppose I didn't want that, suppose I wanted functions y and z to be called first. Could I achieve that by writing
x = z() + ((y() * z())?
I could write that, but it doesn't change anything. Remember, the parentheses don't mean "do everything inside first". They do cause the multiplication to happen before the addition, but the compiler was already going to do it that way anyway, based on the higher precedence of multiplication over addition.
Up above I said, "if you really care, you're going to have to do something else to force the order". What you generally have to do is use some temporary variables and some extra statements. (The technical term is "insert some sequence points.") For example, to cause y and z to get called first, I could write
c = y();
d = z();
b = x();
a = b + c * d;
In your case, if you wanted to make sure that the new value of b got assigned to a, you could write
c = b++;
a = b;
But of course that's silly -- if all you want to do is increment b and have its new value assigned to a, that's what prefix ++ is for:
a = ++b;
Your expectations are completely unfounded.
Parentheses have no direct effect on the order of execution. They don't introduce sequence points into the expression and thus they don't force any side-effects to materialize earlier than they would've materialized without parentheses.
Moreover, by definition, post-increment expression b++ evaluates to the original value of b. This requirement will remain in place regardless of how many pair of parentheses you add around b++. Even if parentheses somehow "forced" an instant increment, the language would still require (((b++))) to evaluate to the old value of b, meaning that a would still be guaranteed to receive the non-incremented value of b.
Parentheses only affects the syntactic grouping between operators and their operands. For example, in your original expression a = b++ one might immediately ask whether the ++ apples to b alone or to the result of a = b. In your case, by adding the parentheses you simply explicitly forced the ++ operator to apply to (to group with) b operand. However, according to the language syntax (and the operator precedence and associativity derived from it), ++ already applies to b, i.e. unary ++ has higher precedence than binary =. Your parentheses did not change anything, it only reiterated the grouping that was already there implicitly. Hence no change in the behavior.
Parentheses are entirely syntactic. They just group expressions and they are useful if you want to override the precedence or associativity of operators. For example, if you use parentheses here:
a = 2*(b+1);
you mean that the result of b+1 should be doubled, whereas if you omit the parentheses:
a = 2*b+1;
you mean that just b should be doubled and then the result should be incremented. The two syntax trees for these assignments are:
= =
/ \ / \
a * a +
/ \ / \
2 + * 1
/ \ / \
b 1 2 b
a = 2*(b+1); a = 2*b+1;
By using parentheses, you can therefore change the syntax tree that corresponds to your program and (of course) different syntax may correspond to different semantics.
On the other hand, in your program:
a1 = (b1++);
a2 = b2++;
parentheses are redundant because the assignment operator has lower precedence than the postfix increment (++). The two assignments are equivalent; in both cases, the corresponding syntax tree is the following:
=
/ \
a ++ (postfix)
|
b
Now that we're done with the syntax, let's go to semantics. This statement means: evaluate b++ and assign the result to a. Evaluating b++ returns the current value of b (which is 10 in your program) and, as a side effect, increments b (which now becomes 11). The returned value (that is, 10) is assigned to a. This is what you observe, and this is the correct behaviour.
However, I expected the parentheses to have b1 incremented before its value is assigned to a1.
You aren't assigning b1 to a1: you're assigning the result of the postincrement expression.
Consider the following program, which prints the value of b when executing assignment:
#include <iostream>
using namespace std;
int b;
struct verbose
{
int x;
void operator=(int y) {
cout << "b is " << b << " when operator= is executed" << endl;
x = y;
}
};
int main() {
// your code goes here
verbose a;
b = 10;
a = b++;
cout << "a is " << a.x << endl;
return 0;
}
I suspect this is undefined behavior, but nonetheless when using ideone.com I get the output shown below
b is 11 when operator= is executed
a is 10
OK, in a nutshell: b++ is a unary expression, and parentheses around it won't ever take influence on precedence of arithmetic operations, because the ++ increment operator has one of the highest (if not the highest) precedence in C. Whilst in a * (b + c), the (b + c) is a binary expression (not to be confused with binary numbering system!) because of a variable b and its addend c. So it can easily be remembered like this: parentheses put around binary, ternary, quaternary...+INF expressions will almost always have influence on precedence(*); parentheses around unary ones NEVER will - because these are "strong enough" to "withstand" grouping by parentheses.
(*)As usual, there are some exceptions to the rule, if only a handful: e. g. -> (to access members of pointers on structures) has a very strong binding despite being a binary operator. However, C beginners are very likely to take quite awhile until they can write a -> in their code, as they will need an advanced understanding of both pointers and structures.
The parentheses will not change the post-increment behaviour itself.
a1=(b1++); //b1=10
It equals to,
uint8_t mid_value = b1++; //10
a1 = (mid_value); //10
Placing ++ at the end of a statement (known as post-increment), means that the increment is to be done after the statement.
Even enclosing the variable in parenthesis doesn't change the fact that it will be incremented after the statement is done.
From learn.geekinterview.com:
In the postfix form, the increment or decrement takes place after the value is used in expression evaluation.
In prefix increment or decrement operation the increment or decrement takes place before the value is used in expression evaluation.
That's why a = (b++) and a = b++ are the same in terms of behavior.
In your case, if you want to increment b first, you should use pre-increment, ++b instead of b++ or (b++).
Change
a1 = (b1++);
to
a1 = ++b1; // b will be incremented before it is assigned to a.
To make it short:
b++ is incremented after the statement is done
But even after that, the result of b++ is put to a.
Because of that parentheses do not change the value here.
The following snippet of C code (where a and b are both type double) is what my question is about:
if(1.0-a < b && b <= 1.0)
Based on the order of operations shown in Wikipedia I understand this as evaluating the same as the following code snippet with parentheses:
if( ( (1.0-a) < b ) && ( b <= 1.0) )
which is what I want. I just want to double check my understanding that the two code snippets are indeed equivalent by the order of operations in C.
Note: obviously I could just use the second code snippet and make explicit what I want if() to evaluate; I ask because I've used the first snippet in my code for a while and I want to make sure my previous results from the code are okay.
Quick answer: yes, it is equivalent.
This means that the result of both code snippets is the same; the meaning is the same, but be careful when you talk about order of operations. It looks to me like your question here is about precedence and associativity. The latter tells you what an expression means, not the order of evaluation of its operands. To learn about order of evaluation, read about sequence points: Undefined behavior and sequence points
You ask about "order of operations", but I don't think that's what you really want to know.
The phrase "order of operations" refers to the time order in which operations are performed. In most cases, the order in which operations are performed within an expression is unspecified. The && operator is one of the few exceptions to this; it guarantees that its left operand is evaluated before its right operand (and the right operand might not be evaluated at all).
The parentheses you added can affect which operands are associated with which operators -- and yes, the two expressions
1.0-a < b && b <= 1.0
and
( (1.0-a) < b ) && ( b <= 1.0)
are equivalent.
Parentheses can be used to override operator precedence. They do not generally affect the order in which the operators are evaluated.
An example: this:
x + y * z
is equivalent to this:
x + (y * z)
because multiplication has a higher precedence than addition. But the three operands x, y, and z may be evaluated in any of the 6 possible orders:
x, y, z
x, z, y
y, x, z
y, z, x
z, x, y
z, y, x
The order makes no difference in this case (unless some of them are volatile), but it can matter if they're subexpressions with side effects.
Consider the code:
if(x = a/b, a%b)
printf("do this");
else
printf("do that");
Now if I replace if statement by if(x = a/b && a % b). Then it works also. So I want to know that replacing comma by && operator and vice-versa works always or not in if-else and loops
They are quite different!
In the first example, , is the comma operator, which 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).
So,
if(x = a/b, a%b)
is equivalent to (thanks to #jaket)
x = a/b;
if(a%b)
a%b will always be evaluated.
For the second one:
if(x = a/b && a % b)
is equivalent to
x = (a/b) && (a%b);
if( x )
a%b will only be evaluated if a/b (i.e. a/b is not zero) is true.
if you replace the comma operator by &&, you're changing the logic slightly. What the comma operator does is it will execute x = a/b and "discard the result", i.e. it's not taken into consideration for the if's condition, only a%b is taken into account.
if you do
if(x = a/b && a % b)
then "do this" will be printed if and only if a/b is non zero and a % b is non zero. Whereas with the comma operator, "do this" is printed only if a % b is non zero.
if(x = a/b, a%b)
is an obfuscated, unreadable way of writing
x = a/b;
if(a%b)
But
x = a/b && a % b
is equivalent to
x = ( (a/b) && (a%b) );
To sum it up, don't mix a whole lot of different operators on the same line. The only thing you achieved with it was to confuse yourself, while creating bugs.
The && operator forces the the result of a/b as well into consideration in determining the expression x = a/b && a%b, In case of ,, the result of a/b is ignored. So replacing , with && will not work always. And not the right thing to do,
int x = 0;
x^=x || x++ || ++x;
and the answer for x at last is 3.
How to analysis this expression?
little confused about this.
Thanks a lot.
This is undefined behaviour. The result could be anything. This is because there is no sequence point between the ++x and the x ^=, so there is no guarantee which will be "done" first.
It's undefined behaviour - so you can get any answer you'd like.
As others have already noted, this is undefined behavior. But why?
When programming in C, there is an inherent difference between a statement and an expression. Expression evaluation should give you the same observable results in any case (e.g., (x + 5) + 2 is the same as x + (5 + 2)). Statements, on the other hand, are used for sequencing of side-effects, that is to say, will generally result in, say, writing to some memory location.
Considering the above, expressions are safe to "nest" into statements, whereas nesting statements into expressions isn't. By "safe" I mean "no surprising results".
In your example, we have
x^=x || x++ || ++x;
Which order should the evaluation go about? Since || operates on expressions, it shouldn't matter whether we go (x || x++) || ++x or x || (x++ || ++x) or even ++x || (x || x++). However, since x++ and ++x are statements (even though C allows them to be used as expressions), we cannot proceed by algebraic reasoning. So, you will need to express the order of operations explicitly, by writing multiple statements.
XOR 0 with 0 is 0. Then ++ twice is equal to 2. Nevertheless, as pointed in other answers, there's no sequence point. So the output could be anything.