This question already has answers here:
Closed 11 years ago.
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Undefined Behavior and Sequence Points
The variable i is changed twice, but is the next example going to cause an undefined behaviour?
#include <iostream>
int main()
{
int i = 5;
std::cout << "before i=" << i << std::endl;
++ i %= 4;
std::cout << "after i=" << i << std::endl;
}
The output I get is :
before i=5
after i=2
Yes, it's undefined. There is no sequence point on assignment, % or ++ And you cannot change a variable more than once within a sequence point.
A compiler could evaluate this as:
++i;
i = i % 4;
or
i = i % 4;
++i;
(or something else)
Related
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 3 years ago.
I understand the basic differences between prefix/postfix notation for decrement/increment operators in C++. However, there is something going on in the next example that stumps me.
The code I shared below prints the following.
5*4**3***4****2*****1
But I would have guessed it would print this.
5*4**4***3****2*****1
How is this happening? Is something going on with pushing/popping to/from the stack?
int var = 5;
cout << var-- << '*'; //prints 5 and then decrements to 4.
cout << var << "**"; //The value of var (now 4)
//is printed to the console.
//The next line prints 3***4****.
//I would have guessed it prints 4***3****.
cout << var-- << "***" << var-- << "****";
//No matter what the above line prints,
//the value of var after the above lines is 2, so...
cout << var-- << "*****" << var << endl; //...Print 2, decrement to 1
//and then 1 is finally printed.
Welcome to the strange world of undefined behaviour. Calling an increment or decrement operator twice on the same variable, in the same statement, is undefined behaviour, so don't do it :)
#include <iostream>
int main()
{
int i = 1;
// should this print 9, 10, or 12? Compilers will vary...
std::cout << (++i + ++i + ++i) << std::endl;
return 0;
}
The problem in this line:
cout << var-- << "***" << var-- << "****";
is undefined behaviour because you using post-decrement twice in a single statement.
This question already has answers here:
Unexpected order of evaluation (compiler bug?) [duplicate]
(3 answers)
Undefined behavior and sequence points
(5 answers)
Closed 6 years ago.
I came across this piece of code:
#include <iostream>
int main()
{
int m = 44;
std::cout << "m = " << m << ", m++ = " << m++ << ", ++m = " << ++m <<
std::endl;
return 0;
}
My first thought was, this is undefined behavior, but it turns out this code was part of an exam in my university, so I'm not so sure now.
The question on the exam is: what is the output of this program?
I also heard that C++11 or C++14 changed the rules about UB in some way, so it might be defined now.
Anyway, the output is (on gcc in windows)
m = 46, m++ = 45, ++m = 46
Is that output correct?
This question already has answers here:
Has C++ standard changed with respect to the use of indeterminate values and undefined behavior in C++14?
(1 answer)
Does initialization entail lvalue-to-rvalue conversion? Is `int x = x;` UB?
(4 answers)
Closed 6 years ago.
I am confused about the following code:
#include <iostream>
int i = 1;
int main()
{
int i = i;
std::cout << "i: " << i << "\n";
return 0;
}
Output:
i: 0
I had expected running the above code would print 1. Can someone please explain the reason for this strange behavior?
You are initializing i with itself. The both i's in int i = i; are the inner one not the outer one. This is undefined behavior and you may get 0 or anything may happen.
This is the right way if you want to assign the outer i to the inner i.
#include <iostream>
int i = 1;
int main()
{
int i = ::i;
std::cout << "i: " << i << "\n";
return 0;
}
Live Demo
BTW, You should carefully read all the compiler warnings. If you did you could see the problem yourself:
warning 'i' is used uninitialized in this function
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
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is the correct answer for cout << c++ << c;?
I just ouputted text, when I suddenly noticed.
#include <iostream>
int main()
{
int array[] = {1,2,3,4};
int *p = array;
std::cout << *p << "___" << *(p++) << "\n";
// output is 1__1. Strange, but I used brackets! it should be at
// first incremented, not clear.
p = array;
std::cout << *p << "___" << *(++p) << "\n";
// output is 2_2 fine, why first number was affected? I didn't intend
// to increment it, but it was incremented
p=array;
std::cout << *p << "___" << *(p + 1) << "\n";
// output is 1_2 - as it was expected
p = array;
return 0;
}
Such behaviour is strange for me, why is it so?
You are causing undefined behaviour, so anything can happen and there's no point in speculating about why.
The expression
std::cout<<*p<<"___"<<*(p++)<<"\n"
Is one example: the order of evaluation of all the things between << is unspecified, so *p and *(p++) are unsequenced with respect to each other (i.e. the compiler is not required do do either one first). You are not allowed to modify a variable and then use it without the modification and usage being sequenced, and so this causes undefined behaviour.
The same thing applies to all the other places in that program where a variable is modified and used unsequenced separately in the same expression.