What happens to the values or text without assignment. How does compiler handle them? Do they impact performance of the code (What does the CPU/processor do with them)?
Here is the example code: I just put 3, 5 and "abcd", code doesn't throw any error.
int main()
{
// Here I declared the variables
int a, b;
// Here there are some numbers
3;
5;
"abcd";
return 0;
}
The compiler will tell you:
<source>:9:5: warning: expression result unused [-Wunused-value]
If the expression result is not used then the compiler will not generate any resulting code. It is very likely that the string literal will be not present in the object and executable files
As #EricPostpischil stated if the expression fas side effect it will be evaluated.
examples:
p++;
1 + foo(p);
1 + p++;
foo(p);
also if you declare objects as side effects prone this kind of expressions will be evaluated.
example
volatile int a, b;
a; //this will generate the code
b; //this will generate the code
https://godbolt.org/z/fa4sa51nq
C 2018 6.8.3 says:
The expression in an expression statement is evaluated as a void expression for its side effects.
Thus no use is made of the value. Only side effects are useful. In a statement such as printf("Hello, world.\n");, a so-called side effect is to send the string to standard output.
Statements such as 3; and 5; have no side effects, and the main value is ignored, so they have no effects.
All expression statements you think of as “doing something”, such as:
printf("x = %d.\n", x);
b = 4;
FindMatchingThings(a, b, c);
++y;
actually do those things by way of their side effects: Sending data to a stream, updating the value of an object, or calling a function which has its own side effects.
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.
I have read that comma operator is used to assign expression, and the right expression is supplied to lvalue.
But why does this program assign left expression to lvalue when not using parenthesis? I am using turbo c compiler.
int b=2;
int a;
a=(b+2,b*5); // prints 10 as expected
a=b+2,b*5; // prints 4 when not using parenthesis
Also the following works:
int a =(b+2,b*5);
But this generates an error:
int a =b+2,b*5; // Error
I can't understand why.
Because precedence of , operator is lower than of = one, this...
a=b+2,b*5;
... will actually be evaluated as...
a = b + 2;
b * 5;
With int i = b + 2, b * 5; is a bit different, because comma has different meaning in declaration statements, separating different declarations from each other. Consider this:
int a = 3, b = 4;
You still have comma here, but now it separates two variable assignment-on-declarations. And that's how the compiler attempts to treat that line from your example - but fails to get any meaning from b * 5 line (it's neither assignment nor declaration).
Now, int a = (b + 2, b * 5) is different: you assign a value of b + 2, b * 5 expression to a variable a of type int. The first sub-expression is discarded, leaving you just with b * 5.
This question already has answers here:
lvalue required as increment operand error
(4 answers)
Closed 8 years ago.
Why lvalue required as increment operand Error In a=b+(++c++); ?
Just Wanted to assign 'b+(c+1)' to 'a' and Increment 'C' by 2 at the same time.
I'M A Beginner Just Wanted A Clarification About What "LVALUE ERROR" Actually Is?
main()
{
int a=1,b=5,c=3;
a=b+(++c++);
printf("a=%d b= %d c= %d \n",a,b,c);
}
Postfix increment binds tighter than prefix increment so what you would want would be something like:
a = b + (++c)++;
This is not legal C, though, as the the result of prefix increment (like the result of postfix increment in your example) is not an lvalue. This means that it's just a value; it doesn't refer to a particular object like 'c' any more so trying to change it makes no sense. It would have no visible effect as no object would be updated.
Personally I think that doing it in two statements is clearer in any case.
a = b + c + 1;
c += 2;
LVALUE means, that there isn't a variable the operation is supposed to be performed on.
C files are basically nothing but text files, which require a particular formatting, so the compiler can understand it.
Writing something like ++Variable++ is complete nonsense for the compiler.
You can basically imagine ++c as:
Var += 1;
return Var;
while c++ is:
int Buf = Var;
Var += 1;
return Buf;
To 'repair' your code:
void main() {
int a=1,b=5,c=3;
a = b + (++c); //Equals 5 + 4
printf("a=%d b= %d c= %d \n",a,b, ++c); //a = 9, b = 5, c = 5
}
This way, you'll get the result you wanted, without the compiler complaining.
Please remember, that when using ++c or c++ in a combined operation, the order DOES matter.
When using ++c, the higher value will be used in the operation, when using c++, it will operate with the old value.
That means:
int a, c = 5;
a = 5 + ++c; //a = 11
while
int a, c = 5;
a = 5 + c++; //a = 10
Because in the latter case, c will only be '6' AFTER it is added to 5 and stored in a.
When the evaluation of l-value precedes the evaluation of r-value and the assignment also returns a value, which of the following is evaluated first?
int i = 2;
int x[] = {1, 2, 3};
int y[] = {4, 5, 6};
int z[] = {7, 8, 9};
x[--i] = y[++i] = z[i++]; // Out of bound exception or not?
NOTE: generic C-like language with l-value evaluation coming first. From my textbook:
In some languages, for example C,
assignment is considered to be an
operator whose evaluation, in addition
to producing a side effect, also
returns the r-value thus computed.
Thus, if we write in C:
x = 2;
the evaluation of such a command, in
addition to assigning the value 2 to x,
returns the value 2. Therefore, in C,
we can also write:
y = x = 2;
which should be interpreted as:
(y = (x = 2));
I'm quite certain that the behaviour in this case is undefined, because you are modifying and reading the value of the variable i multiple times between consecutive sequence points.
Also, in C, arrays are declared by placing the [] after the variable name, not after the type:
int x[] = {1, 2, 3};
Edit:
Remove the arrays from your example, because they are [for the most part] irrelevant. Consider now the following code:
int main(void)
{
int i = 2;
int x = --i + ++i + i++;
return x;
}
This code demonstrates the operations that are performed on the variable i in your original code but without the arrays. You can see more clearly that the variable i is being modified more than once in this statement. When you rely on the state of a variable that is modified between consecutive sequence points, the behaviour is undefined. Different compilers will (and do, GCC returns 6, Clang returns 5) give different results, and the same compiler can give different results with different optimization options, or for no apparent reason at all.
If this statement has no defined behaviour because i is modified several times between comsecutive sequence points, then the same can be said for your original code. The assignment operator does not introduce a new sequence point.
General
In C, the order of any operation between two sequence points should not be dependent on. I do not remember the exact wording from the standard, but it is for this reason
i = i++;
is undefined behaviour. The standard defines a list of things that makes up sequence points, from memory this is
the semicolon after a statement
the comma operator
evaluation of all function arguments before the call to the function
the && and || operand
Looking up the page on wikipedia, the lists is more complete and describes more in detail. Sequence points is an extremely important concept in C and if you do not already know what it means, do learn it immediately.
Specific
No matter how well defined the order of evaluation and assignment of the x, y and z variables are, for
x[--i] = y[++i] = z[i++];
this statement cannot be anything but undefined behaviour because of the i--, i++ and i++.
On the other hand
x[i] = y[i] = z[i];
is well defined, but I am not sure what the status for the order of evaluation for this. If this is important however I would rather prefer this to be split into two statements along with a comment "It is important that ... is assigned/initialized before ... because ...".
i think its the same as
x[3] = y[4] = z[2];
i = 3;