Confused about the output for the second code snippet. Why is the output different than the first program?
#include <iostream>
using namespace std;
int main() {
int s[5] = {1, 2 , 3, 4, 5};
int *p = s;
int first = *(p++);
int second = *++p;
int third = ++*p;
int fourth = *p++;
cout << "*p++ is " << first << endl
<< "*++p is " << second << endl
<< "++*p is " << third << endl
<< "*p++ is " << fourth << endl;
return 0;
}
output:
*p++ is 1
*++p is 3
++*p is 4
*p++ is 4
https://ideone.com/Qu2uIJ
I expected the output would be the same in the code below:
#include <iostream>
using namespace std;
int main() {
int s[5] = {1, 2 , 3, 4, 5};
int *p = s;
cout << "*p++ is " << *p++ << endl
<< "*++p is " << *++p << endl
<< "++*p is " << ++*p << endl
<< "*p++ is " << *p++ << endl;
return 0;
}
output:
*p++ is 3
*++p is 3
++*p is 3
*p++ is 1
https://ideone.com/nwd7xR
What's going on?
Your statement cout << "*p++ is " << *p++ << endl << ...; is treated as one expression, and C++ is almost free in the order of evaluating the arguments used in expressions. SO it is undefined (behaviour actually) in which order the p++ and ++-statements are evaluated.
In the first approach, evaluation order is according to the variables to which you assign. In the second, C++ is free (and treats it as UB if there is no sequence point in the expression; in your's, there isn't a sequence point).
As per operator precedence, operator << is left to right which occurs in cout object in sequence one after the other and sequence of evaluation of arguments to operator << is unspecified.
Between consecutive "sequence points" an object's value can be
modified only once by an expression.
https://msdn.microsoft.com/en-us/library/azk8zbxd.aspx
The second code snippet attempts to modify the pointer's value multiple times within one sequence.
List of C Sequence Points:
Left operand of the logical-AND operator (&&).
Left operand of the logical-OR operator (||).
Left operand of the comma operator (,)
Function-call operator ()
First operand of the conditional operator aka ternary operator ( ? : )
The end of a full initialization expression (that is, an expression
that is not part of another expression such as the end of an
initialization in a declaration statement).
The expression in an expression statement. Expression statements
consist of an optional expression followed by a semicolon (;). The
expression is evaluated for its side effects and there is a sequence
point following this evaluation.
The controlling expression in a selection (if or switch) statement.
The controlling expression of a while or do statement.
Each of the three expressions of a for statement.
The expression in a return statement.
Related
The << and >> operators have two meanings in C++, bit-shifting and stream operations. How does the compiler resolve this ambiguity when the meaning isn't obvious from context? Take this line for example:
std::cout << 1 << 2 << std::endl;
Would the output be 12, as if the second << were treated as a stream insertion, or 4, as if the second << were treated as a bit shift?
operator >> and operator << have left to right associativity. That means that with the addition of some parentheses, the actual expression is
((std::cout << 1) << 2) << std::endl;
and here you can see that with each call, the stream object, or the return of the stream expression, is used as the left hand side of each expression. That means all of the values will inserted into the stream.
There is no ambiguity, because compilers interprets the expression from left to right, so this:
std::cout << 1 << 2 << std::endl;
is equivalent to:
((std::cout << 1) << 2) << std::endl;
Consider that << has left-to-right associativity (see here) and that
std::cout << 1 << 2 << std::endl;
can be thought of as the short way of writing:
std::cout.operator<<(1).operator<<(2).operator<<(std::endl);
In other words: There is no ambiguity.
PS: Also consider that the "problem" you see is not only about two meanings of <<. The operator can be overloaded to have any meaning for a custom type. Nevertheless std::cout << custom_object_1 << custom_object_2; calls std::ostream& operator<<(std::ostream&,const custom_type&). For example: https://godbolt.org/z/fn3PTz.
Why does this giving compilation problem at bold line?
#include<iostream>
static int i = 10;
int main() {
**(i) ? (std::cout << "First i = " << i << std::endl) : ( i = 10);**
std::cout << "Second i = " << i << std::endl;
}
Compilation message:
test.c:8: error: invalid conversion from ‘void*’ to ‘int’
Your usage of the ternary operator is a bit weird: based on the value of i, you either print something to std::cout or assign a new value to it. Those actions don't share a connection through the return value of an expression, so don't do it like this. When using the ternary operator, it's best to stay closer to its intended purpose: a short notation for two possible expressions with a dispatch based on a simple predicate. Example:
const int n = i == 0 ? 42 : 43;
Your code should look like this:
if (i == 0)
i = 10;
else
std::cout << "First i = " << i << "\n";
The reason the original snippet did not compile is that there is no common return type of the ternary operator. "Common" means that both expressions can be converted to the return type. E.g., in const int n = i == 0 ? 42 : 43; the return type is int.
The problem comes from the fact that the return values of the expressions in your conditional operator (ternary operator) (std::ofstream in the case of the std::cout ..., and int for i = 10) are incompatible and therefore the conditional operator is ill-formed. Please check the rules for return type of the conditional operator.
In this case, just use a normal conditional:
if (i)
std::cout << "First i = " << i << std::endl;
else
i = 10;
I am quite new to C++, I know that shift operator in C++ is overloaded. But as how we can do shift operation within printf statement in C can we do similar shift operation in cout statement.
Well, just try it...
#include <iostream>
int main() {
int k = 1;
std::cout << (k << 1) << std::endl; // Correct shifting - notice the parentheses
std::cout << k << 1 << std::endl; // Wrong
return 0;
}
Output:
2
11
What matters here is the type of the variables used for the << operator.
The parentheses causes it to be int << int which is the bitwise shifting. Without the parentheses it will be ostream << int which will write the int to the stream.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Undefined behavior and sequence points
(5 answers)
Closed 9 years ago.
Why the output of next code is 2 1 2?
#include "iostream"
int main(int argc, const char *argv[])
{
int i = 0;
std::cout << i << std::endl << i++ << std::endl << ++i << std::endl;
return 0;
}
Because first i is equal 2 but not zero, it means that the whole like of cout is evaluated first
and then printed (not part by part). If so, then first value should be 1, but not 2, because i++ should increment i after printing. Could you clarify?
EDIT:
The output of next code is 2 2 0.
#include "iostream"
int main(int argc, const char *argv[])
{
int i = 0;
std::cout << i << std::endl << ++i << std::endl << i++ << std::endl;
return 0;
}
why?
There is no sense reasoning in the output of your code because as it stands your program exhibits Undefined Behavior.
Per paragraph 1.9/15 of the C++11 Standard:
The value computations of the operands of an operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.
Because there is no sequence point separating both mutations of i, Undefined Behavior ensues. Your compiler might not output anything, and the program might output differently on different compilers. But arguing about the output is unnecessary in this context.
If you separate the statements, the result will then come out as expected:
std::cout << i << std::endl; // 0
std::cout << i++ << std::endl; // 0
std::cout << ++i << std::endl; // 2
the evalution goes from right to left.
i = 0
++i -> i = 1
i++ -> i = 1, post incrementation, a copy occurs. then i = 2
i -> i = 2
As all this occurs before being send to cout, the value of i is 2, and the middle one have been copied and its value is 1.
Please tell me if I don't understand your question clearly:
cout << i++;
is the equivalent of
cout << i;
i+=1;
while cout << ++i
is the equivalent of
i += 1;
cout << i;
in otherwords any time you use i++, post-increment it returns the current value then changes while ++i means increment first then return the new value. It has nothing to do with cout
I use the stream operator << and the bit shifting operator << in one line.
I am a bit confused, why does code A) not produce the same output than code B)?
A)
int i = 4;
std::cout << i << " " << (i << 1) << std::endl; //4 8
B)
myint m = 4;
std::cout << m << " " << (m << 1) << std::endl; //8 8
class myint:
class myint {
int i;
public:
myint(int ii) {
i = ii;
}
inline myint operator <<(int n){
i = i << n;
return *this;
}
inline operator int(){
return i;
}
};
thanks in advance
Oops
Your second example is undefined behavior.
You have defined the << operator on your myint class as if it were actually <<=. When you execute i << 1, the value in i is not modified, but when you execute m << 1, the value in m is modified.
In C++, it is undefined behavior to both read and write (or write more than once) to a variable without an intervening sequence point, which function calls and operators are not, with respect to their arguments. It is nondeterministic whether the code
std::cout << m << " " << (m << 1) << std::endl;
will output the first m before or after m is updated by m << 1. In fact, your code may do something totally bizarre, or crash. Undefined behavior can lead to literally anything, so avoid it.
One of the proper ways to define the << operator for myint is:
myint operator<< (int n) const
{
return myint(this->i << n);
}
(the this-> is not strictly necessary, just my style when I overload operators)
Because int << X returns a new int. myint << X modifies the current myint. Your myint << operator should be fixed to do the former.
The reason you're getting 8 for the first is that apparently m << 1 is called first in your implementation. The implementation is free to do them in any order.
Your << operator is in fact a <<= operator. If you replace the line with
std::cout << i << " " << (i <<= 1) << std::endl; //8 8
you should get 8 8.
since m is a myInt your second example could be rewritten as:
std::cout << m << " " << (m.operator<<( 1)) << std::endl;
The order of evaluation for the subexpressions m and (m.operator<<( 1)) is unspecified, so there's no saying which "m" you'll get for the 1'st expression that m is used in (which is a simple m expression). So you might get a result of "4 8" or you might get "8 8".
Note that the statement doesn't result in undefined behavior because there are sequence points (at least one function call) between when m is modified and when it's 'read'. But the order of evaluation of the subexpressions is unspecified, so while the compiler has to produce a result (it can't crash - at least not legitimately), there's no saying which of the two possible results it should produce.
So the statement is about as useful as one that has undefined behavior, which is to say it's not very useful.
Well (m << 1) is evaluated before m and therefore m holds 8 already, as in your operator<< you overwrite your own value.
This is wrong behaviour on your side, the operator<< should be const and not change your object.
Because the << operator of myint modifies its lhs. So after evaluating m << 1, m will actually have the value 8 (while i << 1 only returns 8, but does not make i equal to 8). Since it is not specified whether m<<1 executes before cout << m (because it's not specified in which order the arguments of a function or operator are evaluated), it is not specified whether the output will be 8 8 or 4 8.
The C++ language does not define the order of evaluation of operators. It only defines their associativity.
Since your results depend on when the operator<< function is evaluated within the expression, the results are undefined.
Algebraic operator $ functions should always be const and return a new object:
inline myint operator <<(int n) const { // ensure that "this" doesn't change
return i << n; // implicit conversion: call myint::myint(int)
}