I've been thinking about this in terms of incrementing a pointer, but i guess in general now I don't know the semantic difference between these two operations/ operators. For example, my professor said that if you have int a[10] you can't say a++ to point at the next element, but I know from experience that a+1 does work. I asked why and he said something like "a++ is an action and a+1 is an expression". What did he mean by it's an "action"? If anyone could tell me more about this and the inherent difference between the two operations I'd greatly appreciate it. Thank you.
x++ and ++x
The increment operator x++ will modify and usually returns a copy of the old x. On a side note the prefixed ++x will still modify x but will returns the new x.
In fact x++ can be seen as a sort of:
{
int temp = x;
x = x + 1;
return temp;
}
while ++x will be more like:
{
x = x + 1;
return x;
}
x + 1
The x+1 operation will just return the value of the expression and will not modify x. And it can be seen as:
{
return (x + 1);
}
x++ is an action in the sense that it changes x
x+1 does not change x
x++ is a const expression that modifies the value of x (It increases it by 1). If you reference x++, the expression will return the value of x before it is incremented.
The expression ++x will return the value of x after it is incremented.
x + 1 however, is an expression that represents the value of x + 1. It does not modify the value of x.
a++ will translate to a=a+1 which is an action (due to the contained assignment operation)
a+1 is just an expression which refers to a+1 (either in pointer terms or in terms of a number depending upon a's type)
x++ is equivalent to x = x + 1. It is an action in that it is actually changing the value of x.
Every expression returns a result (unless it's void).
x + 1 returns the value of x + 1.
x++ returns the value of x, and as a side effect the value of x is incremented at some point, not necessarily immediately.
This means you can have:
x = x + 1;
but this is illegal:
x = x++;
Related
I have this question for a practice test in C that I don't understand.
Show the output of the following program
x = y = 3;
y = x++, x++;
printf("x = %d, y = %d\n", x, y);
Answer:
1 x = 5, y = 3
I dont understand what
x++, x++
Does x++ mean to implement the value of x into y then add one to it, but why is there a comma ? Would it just first add the value of x in y, and do x=x+1 twice?
I tried putting it in a compiler, found some struggles.
In this statement:
y = x++, x++;
It contains both the assignment operator and the comma operator as well as the postfix increment operator. Of these the postfix increment operator has the highest precedence followed by the assignment operator, then the comma operator. So the expression parses as this:
(y = (x++)), (x++);
The comma operator first evaluates its left operand for any side effects and discards the value of that operand. There is a sequence point here before it then evaluates its right operand, which means evaluating left side and all side effect are guaranteed to occur before those on the right side.
So let's first look at the left operand of the comma operator:
y = (x++)
The subexpression x++ evaluates to the current value of x and increments x as a side effect. At this point x has the value 3, so y gets assigned the value 3 and x is incremented to 4.
Now for the right side:
x++
Since x currently has the value 4, it is incremented to now contain the value 5.
So at the end x is 5 and y is 3.
There is a wikipedia page explaining the comma operator in C and C++: https://en.wikipedia.org/wiki/Comma_operator
It basically evaluates the two expressions left-to-right and returns the value of the second one.
Here you can see the precedence of operators in c++:
https://en.cppreference.com/w/cpp/language/operator_precedence
The comma operator has the lowest precedence.
In your example.
The first line sets both x and y to 3.
The second line increments x twice: x++, x++.
However, since the assignment operator has higher precedence than the comma operator, y gets the value returned by the first increment.
The statement is evaluated as (y = x++), x++.
The first x++ is executed and x gets the value of 4 and returns 3.
This value is assigned to y: y = x++.
Then, the last x++ is evaluated and x gets the value 5.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 6 years ago.
Please explain me the outcome of this code.
//code a
when I run this code on my laptop, value of y is 4. And I think, logically value of y should be 5 because by doing x++ it should return 2 without incrementing as it is post increment and then when we add x which now contains an incremented value ie 3. So 2+3 is 5, according to me. But according to my turbo c++ 3.0 answer is 4.
#include <stdio.h>
void main() {
int x = 2,y;
**int y = x++ + x;** // ans 4
printf("y is :%d", y);
}
// code B
When I run this code, the answer is 6. Here turbo c++ 3.0 in ++x is picking up an incremented value of x++, which is the opposite of above mention code. Logically here answer is correct but in the above code it's not.
#include <stdio.h>
void main() {
int x = 2,y;
**int y = x++ + ++x;** //and 6
printf("y is :%d", y);
}
Firstly, assignment operator that is = works from right to left, which means if you write x = 2 + 4 + 1; your compiler starts reading it from rightmost digit that is 1 then it add 4 to it and so on and then it assigns that value to x.
So, in your case statement y = x++ + x; compiler starts seeing it from right that is it first sees x i.e. 2 and then sees x++ i.e. also 2 as it is post increment operator finally it adds them and assigns y as 2 + 2 that is 4.
In the second case, that is y = x++ + ++x;, compiler first looks at ++x and as it is pre increment operator it increases x with one, i.e. x is now 3. After that x++ is seen and as stated above because it is post operator it would be treated as x in this operation and that value is 3 (remember we incremented x by one earlier) and hence, compiler assigns 3 + 3 i.e. 6 to y.
First program where the result is 4
#include <stdio.h>
void main() {
int x = 2,y;
**int y = x++ + x; // x=2, so y=2+2=4
printf("y is :%d", y); // after post increment x=3
}
Since a variable only gets it post incremented value only after the execution of the statement is completed so y=2+2
Second program where the result is 6.
Associativity of ++ operator is "right to left"
void main() {
int x = 2,y;
**int y = x++ + ++x; //Since associativity is left to right so y=3+3=6
printf("y is :%d", y);// after the post increment x=4
}
Here the pre-increment is performed first since because of the associativity rule so y=3+3
What is the difference between a++ and a=a+1 in C if a is a variable.
Tell me the difference between the following assignments;
x=a+1;
x=a++;
I am not able to find a convincing explanation on Google.Please explain the difference clearly and step by step.
a = a+1; calculates the result of a+1 and assigns it to a.
a++; increments a by 1, but evaluates to the previous value of a.
++a; increments a by 1, and evaluates to the new value of a.
a += 1 is identical to ++a;.
So for example:
x = a++; → a will be incremented by 1, and the previous value of a will be assigned to x.
x = ++a; → a will be incremented by 1, and the new value of a will be assigned to x.
x=a+1 just sets the value of x.
x=a++ first sets x to the value of a, and then increments a. a++ is an operation in itself, and you could call it as a single statement to increment a, without assigning it to x whatsoever.
FYI, there is also ++a, which could be used like this:
x=++a. In this case, a is incremented first, and then the incremented value is assigned to x. In either case, the ++ operator modifies a, which is not the case for a+1.
From the C11 Standard (draft):
6.5.2.4 Postfix increment [...] operator[...]
[...]
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
int a = 0;
int x = a++;
results in a being 1 and x being 0.
x = a++ returns the value of a to the variable x first and later it is incremented.
x = a+1 Here a+1 is evaluated and the result is stored in x
3 differences
Return value and side effect
// a is incremented, but its former value is assigned to x
x=a++; //
// a is unchanged. The sum of `a+1`is assigned to x
x=a+1;
Operator precedence. post-fix ++ (aka Suffix increment) is higher precedence than simple addition.
++ only works with integer and non-void* pointer types. +1 works with all scalar type including double.
int a;
x=a++; // valid
x=a+1; // valid
double a;
x=a++; // invalid
x=a+1; // valid
What is the difference between increment operator ++ and an addition + operator? Why can't we can use + instead of ++?
What are the advantages of ++/-- operators over +/-? Where exactly are they applicable?
x++;
v.s.
x = x + 1;
The main advantage comes from pre-increment v.s. post increment:
e.g.
x = 1;
y = 1;
a = x + 1; // a is 2, x is 1 - e.g. does not modify x
a = ++x; // a is 1, x is 2
b = y++; // b is 2, y is 2
The major downside is that stuff like
a = ++x + x--;
is undefined behavior. Completely compiler dependent and WILL make life hell for anyone trying to figure out the "bug".
The only difference that is given by the C standard is the number of evaluations of x. For normal variables the difference usually doesn't matter. If the compiler can prove that in x = x + 1 the two evaluations of x should give the same value it might optimize this out.
If x is e.g declared volatile or involves the evaluation of a function, the evaluation must be done twice. Example:
unsigned* f(void);
then
*f() = *f() + 1;
is quite different from
++(*f());
The unary operators (++, --) are mainly there for convenience - it's easier to write x++ than it is to write x = x + 1 for example.
++ can also be used to do a 'pre-increment' or a 'post-increment'. If you write x++ then the value of x is increased and the original value of x is returned. For example:
int a = 0;
int x = 0;
a = x++; // x is now equal to 1, but a is equal to 0.
If you write ++x, x is still incremented, but the new value is returned:
int a = 0;
int x = 0;
a = ++x; // Both a and x now equal 1.
There is also usually a minor difference in the compiler's implementation as well. Post-increment (x++) will do something like this:
Create a temporary variable
Copy x to the temporary variable
Increment x
Return the temporary variable
Whereas pre-increment (++x) will do something like this:
Increment x
Return x
So using pre-increment requires less operations than post-increment, but in modern day systems this usually makes no worthwile difference to be a decent way of optimising code.
You could in fact use addition:
a = a + 1
But most people prefer the shorter version. In some languages it actually avoids the need to copy the value to a new location, but as nneonneo has helpfully pointed out, the C compiler is likely to optimise this for you.
"++" means "plus one"
eg
int x = 5;
x++; // the same as x = x + 1
cout << x; // will print 6
"+" is the known plus operator
++ is a convenience syntax. It does not really add capability to the language, but it adds a way of writing some common operations more concisely.
As a standalone statement a++; is identical to a+=1; is identical to a=a+1;
a++ can be useful in some situations that would otherwise need two statements:
while (a < N) doSomethingWith(a++);
is just a shorter form of
while (a<N)
{
doSomethingWith(a);
a=a+1;
}
I don't think there is anything you can write with an a++ that you couldn't also write with an a=a+1, but you can't just do a 1 for 1 substitution. Sometimes the 2nd form will require more code to be equivalent, since the 1st performs two things: produce the value of a, and then increment a. The a=a+1 form produces the value of a after the increment, so if you need the original value, you need a separate statement to process that first.
The difference between using the increment operator(ie. value++) vs using the addition operator(ie. value + 1) is that the first one sometimes can cause mutation especially if we are accessing a global variable inside a function;
Using increment operator.
// The global variable
var fixedValue = 4;
function incrementer () {
// Only change code below this line
return fixedValue ++;
// Only change code above this line
}
var newValue = incrementer(); // Should equal 5
console.log(fixedValue); // Should print 5
Using addition operator.
// The global variable
var fixedValue = 4;
function incrementer () {
// Only change code below this line
return fixedValue + 1;
// Only change code above this line
}
var newValue = incrementer(); // Should equal 5
console.log(fixedValue); // Should print 4
increment doing on register but addition do by ALU we can use + instead of increment but increment is faster
Is there a difference between the two expressions:
(*x)++
and
++(*x)
I can see both of these statements replace the contents of (*x+1) in *x. But are there any differences between them?
(*x)++ evaluates to the value of*x; as a side effect, the value in *x will be incremented by 1.
++(*x) evaluates to the value of *x + 1; as a side effect, the value in *x will be incremented by 1.
Assuming the code:
int a = 5;
int *x = &a;
(*x)++ will evaluate to 5, while ++(*x) will evaluate to 6.
Note that the side effect does not have to be applied immediately after the expression has been evaluated; it only needs to be applied before the next sequence point.
Also note that the parentheses matter for (*x)++; postfix ++ has higher precedence than unary *, so *x++ would be parsed as *(x++); it still evaluates to the value of *x, but as a side effect advances the pointer, rather than increment the thing being pointed to. Prefix ++ and unary * have the same precedence, so they're applied left-to-right; thus ++(*x) and ++*x would have the same result (*x gets incremented, not x).
Let us say the value pointed by x is 10 i:e (*x is 10)
y = (*x)++;
the above statement will be executed as
1. y = *x
2. *x = *x + 1
so y = 10 & *x is 11
y = ++(*x);
the above statement will be executed as
1. *x = *x + 1
2. y = *x
so y = 11 & *x is 11
One increments the values stored in x before it is used (pre), while the other does it after it is used (post).
Also note that ++(*x) is not the same as (*x + 1). The first increments the value and stores it back, the other just increments the value.
One is "pre" and the other is "post". That's the difference. One is evaluated before the the increment (the first option), the other is evaluated after the increment (the second option).