I tried these lines of code and found out shocking output. I am expecting some reason related to initialisation either in general or in for loop.
1.)
int i = 0;
for(i++; i++; i++){
if(i>10) break;
}
printf("%d",i);
Output - 12
2.)
int i;
for(i++; i++; i++){
if(i>10) break;
}
printf("%d",i);
Output - 1
I expected the statements "int i = 0" and "int i" to be the same.What is the difference between them?
I expected the statements "int i = 0" and "int i" to be the same.
No, that was a wrong expectation on your part. If a variable is declared outside of a function (as a "global" variable), or if it is declared with the static keyword, it's guaranteed to be initialized to 0 even if you don't write = 0. But variables defined inside functions (ordinary "local" variables without static) do not have this guaranteed initialization. If you don't explicitly initialize them, they start out containing indeterminate values.
(Note, though, that in this context "indeterminate" does not mean "random". If you write a program that uses or prints an uninitialized variable, often you'll find that it starts out containing the same value every time you run your program. By chance, it might even be 0. On most machines, what happens is that the variable takes on whatever value was left "on the stack" by the previous function that was called.)
See also these related questions:
Non-static variable initialization
Static variable initialization?
See also section 4.2 and section 4.3 in these class notes.
See also question 1.30 in the C FAQ list.
Addendum: Based on your comments, it sounds like when you fail to initialize i, the indeterminate value it happens to start out with is 0, so your question is now:
"Given the program
#include <stdio.h>
int main()
{
int i; // note uninitialized
printf("%d\n", i); // prints 0
for(i++; i++; i++){
if(i>10) break;
}
printf("%d\n", i); // prints 1
}
what possible sequence of operations could the compiler be emitting that would cause it to compute a final value of 1?"
This can be a difficult question to answer. Several people have tried to answer it, in this question's other answer and in the comments, but for some reason you haven't accepted that answer.
That answer again is, "An uninitialized local variable leads to undefined behavior. Undefined behavior means anything can happen."
The important thing about this answer is that it says that "anything can happen", and "anything" means absolutely anything. It absolutely does not have to make sense.
The second question, as I have phrased it, does not really even make sense, because it contains an inherent contradiction, because it asks, "what possible sequence of operations could the compiler be emitting", but since the program contains Undefined behavior, the compiler isn't even obliged to emit a sensible sequence of operations at all.
If you really want to know what sequence of operations your compiler is emitting, you'll have to ask it. Under Unix/Linux, compile with the -S flag. Under other compilers, I don't know how to view the assembly-language output. But please don't expect the output to make any sense, and please don't ask me to explain it to you (because I already know it won't make any sense).
Because the compiler is allowed to do anything, it might be emitting code as if your program had been written, for example, as
#include <stdio.h>
int main()
{
int i; // note uninitialized
printf("%d\n", i); // prints 0
i++;
printf("%d\n", i); // prints 1
}
"But that doesn't make any sense!", you say. "How could the compiler turn "for(i++; i++; i++) ..." into just "i++"? And the answer -- you've heard it, but maybe you still didn't quite believe it -- is that when a program contains undefined behavior, the compiler is allowed to do anything.
The difference is what you already observed. The first code initializes i the other does not. Using an unitialized value is undefined behaviour (UB) in c++. The compiler assumes UB does not happen in a correct program, and hence is allowed to emit code that does whatever.
Simpler example is:
int i;
i++;
Compiler knows that i++ cannot happen in a correct program, and the compiler does not bother to emit correct output for wrong input, hece when you run this code anything could happen.
For further reading see here: https://en.cppreference.com/w/cpp/language/ub
The is a rule of thumb that (among other things) helps to avoid uninitialized variables. It is called Almost-Always-Auto, and it suggests to use auto almost always. If you write
auto i = 0;
You cannot forget to initialize i, because auto requires an initialzer to be able to deduce the type.
PS: C and C++ are two different languages with different rules. Your second code is UB in C++, but I cannot answer your question for C.
Related
In the following code:
#include<iostream>
using namespace std;
int main()
{
int A[5] = {10,20,30,40,50};
// Let us try to print A[5] which does NOT exist but still
cout <<"First A[5] = "<< A[5] << endl<<endl;
//Now let us print A[5] inside the for loop
for(int i=0; i<=5; i++)
{
cout<<"Second A["<<i<<"]"<<" = "<<A[i]<<endl;
}
}
Output:
The first A[5] is giving different output (is it called garbage value?) and the second A[5] which is inside the for loop is giving different output (in this case, A[i] is giving the output as i). Can anyone explain me why?
Also inside the for loop, if I declare a random variable like int sax = 100; then A[5] will take the value 100 and I don't have the slightest of clue why is this happening.
I am on Windows, CodeBlocks, GNUGCC Compiler
Well you invoke Undefined Behaviour, so behaviour is err... undefined and anything can happen including what your show here.
In common implementations, data past the end of array could be used by a different element, and only implementation details in the compiler could tell which one.
Here your implementation has placed the next variable (i) just after the array, so A[5] is an (invalid) accessor for i.
But please do not rely on that. Different compilers or different compilation options could give a different result. And as a compiler is free to assume that you code shall not invoke UB an optimizing compiler could just optimize out all of your code and only you would be to blame.
TL/DR: Never, ever try to experiment UB: anything can happen from a consistent behaviour to an immediate crash passing by various inconsistent outputs. And what you see will not be reproduced in a different context (context here can even be just a different run of same code)
In your Program, I think "there is no any syntax issue" because when I execute this same code in my compiler. Then there is no any issue likes you.
It gives same garbage value at direct as well as in loop.
enter image description here
The problem is that when you wrote:
cout <<"First A[5] = "<< A[5] << endl<<endl;//this is Undefined behavior
In the above statement you're going out of bounds. This is because array index starts from 0 and not 1.
Since your array size is 5. This means you can safely access A[0],A[1],A[2],A[3] and A[4].
On the other hand you cannot access A[5]. If you try to do so, you will get undefined behavior.
Undefined behavior means anything1 can happen including but not limited to the program giving your expected output. But never rely(or make conclusions based) on the output of a program that has undefined behavior.
So the output that you're seeing is a result of undefined behavior. And as i said don't rely on the output of a program that has UB.
So the first step to make the program correct would be to remove UB. Then and only then you can start reasoning about the output of the program.
For the same reason, in your for loop you should replace i<=5 with i<5.
1For a more technically accurate definition of undefined behavior see this where it is mentioned that: there are no restrictions on the behavior of the program.
During a discussion I had with a couple of colleagues the other day I threw together a piece of code in C++ to illustrate a memory access violation.
I am currently in the process of slowly returning to C++ after a long spell of almost exclusively using languages with garbage collection and, I guess, my loss of touch shows, since I've been quite puzzled by the behaviour my short program exhibited.
The code in question is as such:
#include <iostream>
using std::cout;
using std::endl;
struct A
{
int value;
};
void f()
{
A* pa; // Uninitialized pointer
cout<< pa << endl;
pa->value = 42; // Writing via an uninitialized pointer
}
int main(int argc, char** argv)
{
f();
cout<< "Returned to main()" << endl;
return 0;
}
I compiled it with GCC 4.9.2 on Ubuntu 15.04 with -O2 compiler flag set. My expectations when running it were that it would crash when the line, denoted by my comment as "writing via an uninitialized pointer", got executed.
Contrary to my expectations, however, the program ran successfully to the end, producing the following output:
0
Returned to main()
I recompiled the code with a -O0 flag (to disable all optimizations) and ran the program again. This time, the behaviour was as I expected:
0
Segmentation fault
(Well, almost: I didn't expect a pointer to be initialized to 0.) Based on this observation, I presume that when compiling with -O2 set, the fatal instruction got optimized away. This makes sense, since no further code accesses the pa->value after it's set by the offending line, so, presumably, the compiler determined that its removal would not modify the observable behaviour of the program.
I reproduced this several times and every time the program would crash when compiled without optimization and miraculously work, when compiled with -O2.
My hypothesis was further confirmed when I added a line, which outputs the pa->value, to the end of f()'s body:
cout<< pa->value << endl;
Just as expected, with this line in place, the program consistently crashes, regardless of the optimization level, with which it was compiled.
This all makes sense, if my assumptions so far are correct.
However, where my understanding breaks somewhat is in case where I move the code from the body of f() directly to main(), like so:
int main(int argc, char** argv)
{
A* pa;
cout<< pa << endl;
pa->value = 42;
cout<< pa->value << endl;
return 0;
}
With optimizations disabled, this program crashes, just as expected. With -O2, however, the program successfully runs to the end and produces the following output:
0
42
And this makes no sense to me.
This answer mentions "dereferencing a pointer that has not yet been definitely initialized", which is exactly what I'm doing, as one of the sources of undefined behaviour in C++.
So, is this difference in the way optimization affects the code in main(), compared to the code in f(), entirely explained by the fact that my program contains UB, and thus compiler is technically free to "go nuts", or is there some fundamental difference, which I don't know of, between the way code in main() is optimized, compared to code in other routines?
Your program has undefined behaviour. This means that anything may happen. The program is not covered at all by the C++ Standard. You should not go in with any expectations.
It's often said that undefined behaviour may "launch missiles" or "cause demons to fly out of your nose", to reinforce that point. The latter is more far-fetched but the former is feasible, imagine your code is on a nuclear launch site and the wild pointer happens to write a piece of memory that starts global thermouclear war..
Writing unknown pointers has always been something which could have unknown consequences. What's nastier is a currently-fashionable philosophy which suggests that compilers should assume that programs will never receive inputs that cause UB, and should thus optimize out any code which would test for such inputs if such tests would not prevent UB from occurring.
Thus, for example, given:
uint32_t hey(uint16_t x, uint16_t y)
{
if (x < 60000)
launch_missiles();
else
return x*y;
}
void wow(uint16_t x)
{
return hey(x,40000);
}
a 32-bit compiler could legitimately replace wow with an unconditional call to
launch_missiles without regard for the value of x, since x "can't possibly" be greater than 53687 (any value beyond that would cause the calculation of x*y to overflow. Even though the authors of C89 noted that the majority of compilers of that era would calculate the correct result in a situation like the above, since the Standard doesn't impose any requirements on compilers, hyper-modern philosophy regards it as "more efficient" for compilers to assume programs will never receive inputs that would necessitate reliance upon such things.
I would like to save typing in some loop, creating reference to an array element, which might not exist. Is it legal to do so? A short example:
#include<vector>
#include<iostream>
#include<initializer_list>
using namespace std;
int main(void){
vector<int> nn={0,1,2,3,4};
for(size_t i=0; i<10; i++){
int& n(nn[i]); // this is just to save typing, and is not used if invalid
if(i<nn.size()) cout<<n<<endl;
}
};
https://ideone.com/nJGKdW compiles and runs the code just fine (I tried locally with both g++ and clang++), but I am not sure if I can count on that.
PS: Neither gcc not clang complain, even when compiled+run with -Wall and -g.
EDIT 2: The discussion focuses on array indexing. The real code actually uses std::list and a fragment would look like this:
std::list<int> l;
// the list contains something or not, don't know yet
const int& i(*l.begin());
if(!l.empty()) /* use i here */ ;
EDIT 3: Legal solution to what I was doing is to use iterator:
std::list<int> l;
const std::list<int>::iterator I(l.begin()); // if empty, I==l.end()
if(!l.empty()) /* use (*I) here */ ;
No it's not legal. You are reading data out of bounds from the vector in the declaration of n and therefore your program have undefined behavior.
No, for two reasons:
The standard states (8.3.2):
A reference shall be initialized to refer to a valid object or function
std::vector::operator[] guarantees that even if N exceeds the container size, the function never throws exceptions (no-throw guarantee, no bounds checking other than at()). However, in that case, the behavior is undefined.
Therefore, your program is not well-formed (bullet point 1) and invoke undefined behaviour (bullet point 2).
I'd be surprised if this is "allowed" by the specification. However, what it does is store the address of an element that is outside the range of its allocation, which shouldn't in itself cause a problem in most cases - in extreme cases, it may overflow the pointer type, which could cause problems, I suppose.
In other words, if i is WAY outside the size of nn, it could be a problem, not necessarily saying i has to be enormous - if each element in the vector is several megabytes (or gigabytes in a 64-bit machine), you can quite quickly run into problems with address range.
But don't ask me to quote the specification - someone else will probably do that.
Edit: As per comment, since you are requesting the address of a value outside of the valid size, at least in debug builds, this may well cause the vector implementation to assert or otherwise "warn you that this is wrong".
Why this piece of code compiles?
#include <iostream>
int foo(int x)
{
if(x == 10)
return x*10;
}
int main()
{
int a;
std::cin>>a;
std::cout<<foo(a)<<'\n';
}
The compiler shouldn't give me an error like "not all code paths returns a value"? What happens/returns my function when x isn't equal to ten?
The result is undefined, so the compiler is free to choose -- you probably get what happens to sit at the appropriate stack address where the caller expects the result. Activate compiler warnings, and your compiler will inform you about your omission.
The compiler is not required to give you an error in this circumstance. Many will, some will only issue warnings. Some apparently won't notice.
This is because it's possible that your code ensures outside of this function that the condition will always be true. Therefore, it isn't necessarily bad (though it almost always is, which is why most compilers will issue at least a warning).
The specification will state that the result of exiting a function that should return a value but doesn't is undefined behavior. A value may be returned. Or the program might crash. Or anything might happen. It's undefined.
I have written 2 programs. Please go through both the programs and help me in understanding why variable 'i' and '*ptr' giving different values.
//Program I:
//Assumption: Address of i = 100, address of ptr = 500
int i = 5;
int *ptr = (int *) &i;
*ptr = 99;
cout<<i; // 99
cout<<&i;// 100
cout<<ptr; // 100
cout<<*ptr; // 99
cout<<&ptr; // 500
//END_Program_I===============
//Program II:
//Assumption: Address of i = 100, address of ptr = 500
const int i = 5;
int *ptr = (int *) &i;
*ptr = 99;
cout<<i; // 5
cout<<&i;// 100
cout<<ptr; // 100
cout<<*ptr; // 99
cout<<&ptr; // 500
//END_PROGRAM_II===============
The confusion is: Why variable i still coming as 5, even though *ptr ==99?
In the following three lines, you are modifying a constant:
const int i = 5;
int *ptr = (int *) &i;
*ptr = 99;
This is undefined behavior. Anything can happen. So don't do it.
As for what's happening underneath in this particular case:
Since i is const, the compiler assumes it will not change. Therefore, it simply inlines the 5 to each place where it is used. That's why printing out i shows the original value of 5.
All answer will probably talk about "undefined behavior", since you are attempting the logical nonsense of modifying a constant.
Although this is technically perfect, let me give you some hints about why this happens (about "how", see Mysticial answer).
It happens because C++ is by design an "imperfectly specified language". The "imperfection" consist in a number of "undefined behaviors" that pervade the language specification.
In fact, language designers deliberately choose that -in some circumstances- instead of say "if you do this, will gave you that", (that may be: you got this code, or you got this error) thay prefer to say "we don't define what will happen".
This lets the compiler manufacturers free to decide what to do. And since there are many compiler working on many platforms, may be the optimal solution for one in not necessarily the optimal solution for another (that may have rely to a machine with a different instruction set) and hence you (as a programmer) are left in the dramatic situation that you'll never know what to expect, and even if you test it, you cannot trust the result of the test, since in another situation (compiling the same code with a different compiler or just a different version of it, or for a different platform) it will be different.
The "bad" thing, here, is that a compiler should warn when an undefined behavior is hit (forcing a const should be warned as a potential bug, especially if the compiler does const-inlining otimizations, since it is a nonsense if a const is allowed to be changed), as mot likely it does, if you specify the proper flag (may be -W4 or -wall or -pedantic or similar, depending of the compiler you have).
In particular the line
int *ptr = (int *) &i;
should issue a warning like:
warning: removing cv-qualifier from &i.
So that, if you correct your program as
const int *ptr = (const int *) &i;
to satisfy the waarning, you wil get an error at
*ptr = 99;
as
error: *ptr is const
thus making the problem evident.
Moral of the story:
From a legal point of view, you wrote bad code since it is -by language definition- relying on undefined behavior.
From a moral point of view: the compiler kept an unfair behavior: performing const-inlining (replacing cout << i with cout << 5) after accepting (int*)&i is a self-contradition, and incoherent behavior should at least be warned.
If it wants to do one thing must not accept the other, or vice-versa.
So check if there is a flag you can set to be warned, and if not, report to the compiler manufacturer its unfairness: it didn't warn about its own contradiction.
const int i = 5;
Implies that the variable i is a const and it cannot/should not be changed, it is Imuttable and changing it through a pointer results in Undefined Behavior.
An Undefined Behavior means that the program is ill-formed and any behavior is possible. Your program might seem to work as desired, or not or it might even crash. All safe bets are off.
Remember the Rule:
It is Undefined Behavior to modify an const variable. Don't ever do it.
You're attempting to modify a constant through a pointer, which is undefined. This means anything unexpected can happen, from the correct output, to the wrong output, to the program crashing.