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What happens when I print an uninitialized variable in C++? [duplicate]
(4 answers)
Closed 7 years ago.
Why uninitialized variable print a strange negative value ?
int x;
cout << x << endl;
What you're doing (reading the value of an uninitialised variable) is undefined behaviour; anything can happen, from it appearing to work, to printing random values, to crashing, to buying pizza with your credit card.
An uninitialized variable is a variable that is declared but is not set to a definite known value before it is used. It will have some value, but not a predictable one.
When a variable is not initialized , it shows you "Garbage Value".
What that mean is it can be any arbitrary number from anywhere, may be from another running application or random number from big pool of memory.
Related
This question already has answers here:
C++: Uninitialized variables garbage
(5 answers)
Closed 4 years ago.
A programmer told me that if we don't assign a value to any variable e.g like max then it consider as null value.
{
int max=0,x[5];
for(int a=0;a<5;a++)
{
cout<<"Enter no "<<a+1<<" : ";
cin>>x[a];
if(max<x[a])
{
max=x[a];
}
}
cout<<endl<<max;
}
output:
Enter no 1 : 1
Enter no 2 : 5
Enter no 3 : 8
Enter no 4 : 7
Enter no 5 : 5
8
and when type max=0 means at the declaration assign 0 to max
it gives me the same result that's mean null value is equal to 0.
if yes then what is difference between null value and 0
It actually depends on what programming language are you writing your code in.
For instance, if you access in read an uninitialized variable in c/c++ you cannot know which value it will give you.
Memory is not zero-ed when the variable is allocated, so whatever bits were there, they will be read as a value in the type of that variable.
Game Maker's own language, GML, allows you to decide at compile time if uninitialized variables have to be set to 0 by default or not; but any decent programmer would set that option off and not rely on it; i'm not aware of other languages that automatically initialize variable values.
I'd say relying on such a feature is bad practice.
About the difference between null and 0, it actually depends on the base type of your variable.
An integer can only store integers, so null will still be readable as an integer.
If you default initialize a fundamental-type object (such as int) in automatic storage, the value will be indeterminate. The behaviour of reading an indeterminate value is undefined.
Undefined behaviour means that the standard doensn't constrain the behaviour of the program in any way. As far as the C++ standard is concerned, possible behaviours include, none which are guaranteed:
Output that you expect.
Output that you don't expect.
Same output as some program which doesn't have UB.
Different output than some program which doesn't have UB.
Any output.
No output whatsoever.
Side-effects that you expect.
Side-effects that you don't expect.
Any Side-effects.
No Side-effects whatsoever.
Possible side-effects include:
Corruption of data.
Security vulnerabilities.
Anything within the capability of the process, hopefully limited by the OS.
Inconsistent behaviour on other systems.
Inconsistent behaviour even on same system if you re-compile using another compiler.
Inconsistent behaviour even if you re-compile using same compiler.
Inconsistent behaviour even without recompilation during another execution:
Possibly only when you're on vacation.
Possibly only when you're demonstrating your program to your employer or important client.
Consistent behaviour in all of the above cases.
A programmer told me that if we don't assign a value to any variable e.g like max then it consider as null value.
There is no such thing as null value integer in C++. There is such thing as a null pointer, as well as null character, neither of which are directly related to each other, and neither have anything to do with an uninitialized variable.
Your programmer is using confusing terminology, but if by null value, they meant indeterminate value, then they are correct. And in that case your question becomes what is difference between indeterminate value and 0, and the answer is that reading an indeterminate value is undefined behaviour (see above).
when type max=0 means at the declaration assign 0 to max
To be pedantic, you have int max=0 which is not an assignment, but initialization.
There's no 'null' value in c++. Local primitives are uninitialized (you'll get garbage that just happens to be in memory at this time), while static and global variables are zero-initialized. So by saying int max; You'll either get random garbage or zero (depending on where you declared it).
This question already has answers here:
How are variable names stored in memory in C?
(5 answers)
Closed 5 years ago.
It is always advised to use bigger and clearer variable names in programming suppose I define :
int captain=10 ;
I know that 10 is stored but where is the "captain" or the variable name stored and who allocates memory for it?
In the end they aren't stored anywhere unless you have linked in additional debug information.
They're translated into memory (be it RAM or ROM) addesses and address offsets by your toolchains linker.
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When and why will a compiler initialise memory to 0xCD, 0xDD, etc. on malloc/free/new/delete?
(9 answers)
Closed 7 years ago.
So after allocating char array in debugging windows I can see my array is filled with 0xCC. What does it mean? (-52 = 0xCC)
Uninitialized built-in types have an in-determined value, trying to read it is undefined behavior.
The actual values you can see depend on the compiler: You might for example see garbage, zeroes or (what seems to be the case in your example) some special value indicating "data uninitialized".
It's there as a sentinel value so that you know that the memory is uninitialized.
See the /GZ compiler switch.
This question already has answers here:
Accessing an array out of bounds gives no error, why?
(18 answers)
Closed 9 years ago.
Why does this work without any errors?
int array[2];
array[5] = 21;
cout << array[5];
It printed out 21 just fine. But check this out! I changed 5 to 46 and it still worked. But when I put 47, it didn't print anything. And showed no errors anywhere. What's up with that!?!?
Because it's simply undefined behaviour (there is no checks for bounds of arrays in C++). Anything can happen.
Simply array[5] is equivalent to *(&array[0] + 5), you are trying to write/read to memory, that you are not allocate.
In the C and C++ language there are very few runtime errors.
When you make a mistake what happens instead is that you get "undefined behavior" and that means that anything may happen. You can get a crash (i.e. the OS will stop the process because it's doing something nasty) or you can just corrupt memory in your program and things seems to work anyway until someone needs to use that memory. Unfortunately the second case is by far the most common so when a program writes outside an array normally the program crashes only one million instructions executed later, in a perfectly innocent and correct part.
The main philosophical assumption of C and C++ is that a programmer never makes mistakes like accessing an array with an out-of-bounds index, deallocating twice the same pointer, generating a signed integer overflow during a computation, dereferencing a null pointer and so on.
This is also the reason for which trying to learn C/C++ just using a compiler and experimenting by writing code is a terrible idea because you will not be notified of this pretty common kind of error.
The array has 2 elements, but you are assigning array[5] = 21; which means 21 is in memory outside the array. Depends on your system and environment array[46] is a valid memory to hold a number but array[47] isn't.
You should do this
int array[47];
array[47] = 21;
cout << array[47];
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
C and C++ : Partial initialization of automatic structure
I'm looking for a fast way to init local array with zeros. (By "Fast", I mean "Fast to type.") When I do the following:
HANDLE hHandles[32] = {0};
does it zero out the first element or all 32?
It initializes all the 32 elements to zero.
See this surprisingly popular answer for details/alternatives. The difference between C and C++ seems to be that in C++ {} will do zero-initialization as well.