I am trying to understand how this code works,
int main () {
int m, k;
m = (k=5)+(k=8)+(k=9)+(k=7);
printf("m=%d\n",m);
printf("k=%d\n",k);
}
The output:
m=32
k=7
I have no idea how is the value of m become 32.
I hope someone can help me to understand how this code works and how the outputs end up like this
Simplified explanation:
When you use k=... multiple times in the same expression, all assignments to that same variable are so-called "unsequenced side-effects". Simply put, it means that C doesn't specify which operand of + to evaluate/execute first nor does it specify the order in which the assignments will get carried out.
So the compiler has no way of knowing which k to evaluate/assign to first and therefore gets all confused. This is so-called "undefined behavior", anything can happen.
You have to solve this by splitting the expression up in several, each separated by a semicolon, which acts as a "sequence point", meaning all prior evaluations need to be done at the point where the ; is encounterd. Example:
k=5;
k+=8;
k+=9;
m = k + 7;
Detailed explanation with standard references here: Why can't we mix increment operators like i++ with other operators?
This is undefined behavior. Your compiler warn about this
warning: multiple unsequenced modifications to 'k' [-Wunsequenced]
You can learn more about this here:
A Guide to Undefined Behavior
What Every C Programmer Should Know About Undefined Behavior
What are “sequence points” and how do they affect undefined behavior?
The behaviour of the program is undefined.
There are multiple unsequenced writes on k in the expression
(k = 5) + (k = 8) + (k = 9) + (k = 7)
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 1 year ago.
Why i can't do the following in Objective-C?
a = (a < 10) ? (a++) : a;
or
a = (a++)
The ++ in a++ is a post-increment. In short, a = a++ does the following:
int tmp = a; // get the value of a
a = a + 1 //post-increment a
a = tmp // do the assignment
As noted by others, in C this is actually undefined behavior and different compilers can order the operations differently.
Try to avoid using ++ and -- operators when you can. The operators introduce side effects which are hard to read. Why not simply write:
if (a < 10) {
a += 1;
}
To extend Sulthan's answer, there are several problem with your expressions, at least the simple assignment (case 2).
A. There is no sense in doing so. Even a++ has a value (is a non-void expression) that can be assigned, it automatically assigns the result to a itself. So the best you can expect is equivalent to
a++;
The assignment cannot improve the assignment at all. But this is not the error message.
B. Sulthans replacement of the statement is a better case. It is even worse: The ++ operator has the value of a (at the beginning of the expression) and the effect to increment a at some point in future: The increment can be delayed up to the next sequence point.
The side effect of updating the stored value of the operand shall occur between the previous and the next sequence point.
(ISO/IEC 9899:TC3, 6.5.2.4, 2)
But the assignment operator = is not a sequence point (Annex C).
The following are the sequence points described in 5.1.2.3:
[Neither assignment operator nor ) is included in the list]
Therefore the expression can be replaced with what Sulthan said:
int tmp = a; // get the value of a
a = a + 1 //post-increment a
a = tmp // do the assignment
with the result, that a still contains the old value.
Or the expression can be replaced with this code …:
int tmp = a; // get the value of a
a = tmp // assignemnt
a = a + 1 // increment
… with a different result (a is incremented). This is what the error message says: There is no defined sequence (order the operations has to be applied.)
You can insert a sequence point using the comma operator , (what is the primary use of it), …
a++, a=a; // first increment, then assign
… but this shows the whole leak of meaning of what you want to do.
It is the same with your first example. Though ? is a sequence point …:
The following are the sequence points described in 5.1.2.3:
… The end of the first operand of the following operators: […] conditional ? (6.5.15);[…].
… both the increment (a++) and the assignment (a=) are after the ? operator is evaluated and therefore unsequenced ("in random order") again.
To make the comment "Be careful" more concrete: Don't use an incremented object in an expression twice. (Unless there is a clear sequence point).
int a = 1;
… = a++ * a;
… evaluates to what? 2? 1? Undefined, because the increment can take place after reading a "the second time".
BTW: The Q is not related to Objective-C, but to pure C. There is no Objective-C influence to C in that point. I changed the tagging.
This question already has answers here:
Order of operations for pre-increment and post-increment in a function argument? [duplicate]
(4 answers)
Closed 7 years ago.
i am a beginner in c, and i am finding it difficult to understand the post and pre increment i have given my code below,i already compiled it in a turbo c++ compiler and i got output as
a = 6 and b = 10 but since the post increment operator is used the output should be a = 6 and b = 11 ,why is it not happening?could someone explain it..
#include<stdio.h>
int main()
{
int a=5,b;
b = a++ + a;
printf("\na = %d and b = %d",a,b);
return 0;
}
The behaviour of a++ + a; is undefined in C. This is because the + is not a sequencing point and you're essentially attempting to increment and read a in the same expression.
So you can't guarantee a particular answer.
In order to understand prefix and postfix increments, use statements like b = a++; and b = ++a;
What happens in the following?
b = a++ + a;
1) Is a incremented and its original value is then added to the original value of a?
2) Is a incremented and its original value is then added to the new value of a?
3) Is a on the right side fetched first and then added to the original value of an incremented a?
C allows any of theses approaches (and likely others) as this line of code lacks a sequence point which would define evaluation order. This lack of restriction allows compilers often to make optimized code. It comes at a cost as the approaches do not generate the same result when accessing a in the various ways above.
Therefore it is implementation defined behavior. Instead:
b = a++;
b = b + a;
or
b = a;
b = b + a++;
After int a = 5; the value of a is 5
b = a; // b is 5;
After int a = 5; the value of a++ is 5
b = a++; // b is 5
but the side effect of a++ is to increase the value of a. That increase can happen anytime between the last and next sequence points (basically the last and next semicolon).
So
/* ... */;
b = a++ + a;
#if 0
/* side-effect */ 5 + 6
5 /* side-effect */ + 6
5 + /* side effect mixed with reading the value originating a strange value */ BOOM
5 + 5 /* side effect */
#endif
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.
This question already has answers here:
Is a += b more efficient than a = a + b in C?
(7 answers)
Closed 9 years ago.
what is the difference between i = i + j; and i += j; in C?
Are they equivalent? Is there any side effect of i?
I was trying to check the assignment mechanism in C using the GCC compiler.
They're almost the same. The only difference is that i is only evaluated once in the += case versus twice in the other case.
There is almost no difference, but if i is a complex expression, it is only computed once. Suppose you had:
int ia[] = {1, 2, 3, 4, 5};
int *pi = &(ia[0]); // Yes, I know. I could just have written pi = ia;
*pi++ += 10;
// ia now is {11, 2, 3, 4, 5}.
// pi now points to ia[1].
// Note this would be undefined behavior:
*pi++ = *pi++ + 10;
i = i + j is equivalent to i += j but not same.
In some cases(rare) i += j differs from i = i + j because i itself has a side effect.
Also one more problem is operator precedence i.e
i = i * j + k;
is not same as
i *= j + k;
The two statements i = i + j and i += j, are functionally same, in first case you are using the general assignment operation, while the second one uses the combinatorial assignment operator. += is additive assignment operator (addition followed by assignment).
The use of combinatorial assignment operators generates smaller source code that is less susceptible to maintenance errors and also possibly a smaller object code where it would also run faster. Compilation is also likely to be a little faster.
Syntactic sugar baby.
Any differences are just going to come down to compiler implementation.
http://en.wikipedia.org/wiki/Syntactic_sugar
In both cases i (the variable or expression being assigned) must be an lvalue. In most simple cases this will yield code that is identical in both cases so long as i is not declared volatile.
However there are a few cases where a lvalue can be an expression involving operators, and this may cause evaluation of i twice. The most plausible example of an lvalue expression that might be used in that way is perhaps simple dereferencing of a pointer (*p):
*p = *p + j ;
*p += j ;
may generate different code, but it is trivially optimised so I would expect not even without optimisation enabled. Again p cannot be volatile, otherwise the expressions are semantically different.
A less plausible scenario is to use a conditional operator expression as an lvalue. For example the following adds j to b or c depending on a:
(a ? b : c) += j ;
(a ? b : c) = (a ? b : c) + j ;
These might generate different code - the compiler might reasonably not spot that idiom and apply an optimisation. If the expression a has side effects - for example were the expression getchar() == '\n' or a is volatile (regardless of b or c), then they are not equivalent since the second would evaluate to:
c = b + j for the input "Y\n",
b = b + j for input "\n\n",
c = c + j for input "YN".
These points are of course mostly irrelevant - if you write code like that and it does things you did not expect, sympathy may be in short supply!