In the first code sample, a is a main method local variable. Method local variables need to be initialized before using them. In the second code sample, a is class member variable, hence it will be initialized to the default value.
I just tested exec and it works in Python 2.6.2 >>> def test(): … exec “a = 5” … print a … >>> test() 5 If you are using Python 3.x, it does not work anymore because locals are optimized as an array at runtime, instead of using a dictionary. When Python detects the “exec … Read more
Either allocate the string on the stack on the caller side and pass it to your function: void getStr(char *wordd, int length) { … } int main(void) { char wordd[10 + 1]; getStr(wordd, sizeof(wordd) – 1); … } Or make the string static in getStr: char *getStr(void) { static char wordd[10 + 1]; … return … Read more
Yes, lifetime of a local variable is within the scope({,}) in which it is created. Local variables have automatic or local storage. Automatic because they are automatically destroyed once the scope within which they are created ends. However, What you have here is an string literal, which is allocated in an implementation defined read-only memory. … Read more
Python variables are scoped to the innermost function, class, or module in which they’re assigned. Control blocks like if and while blocks don’t count, so a variable assigned inside an if is still scoped to a function, class, or module. (Implicit functions defined by a generator expression or list/set/dict comprehension do count, as do lambda … Read more
You could do something along these lines (which worked in both Python v2.7.17 and v3.8.1 when I tested it/them): def hi(): # other code… hi.bye = 42 # Create function attribute. sigh = 10 hi() print(hi.bye) # -> 42 Functions are objects in Python and can have arbitrary attributes assigned to them. If you’re going … Read more
It’s an extreme optimization Doug Lea, the author of the class, likes to use. Here’s a post on a recent thread on the core-libs-dev mailing list about this exact subject which answers your question pretty well. from the post: …copying to locals produces the smallest bytecode, and for low-level code it’s nice to write code … Read more
The issue here is that when you create the local variable it is allocated on the stack and is therefore unavailable once the function finishes execution (implementation varies here). The preferable way would be to use malloc() to reserve non-local memory. the danger here is that you have to deallocate (free()) everything you allocated using … Read more
myArray is a local variable and as thus the pointer is only valid until the end of its scope (which is in this case the containing function getArray) is left. If you access it later you get undefined behavior. In practice what happens is that the call to printf overwrites the part of the stack … Read more
How can it be? Isn’t the memory of a local variable inaccessible outside its function? You rent a hotel room. You put a book in the top drawer of the bedside table and go to sleep. You check out the next morning, but “forget” to give back your key. You steal the key! A week … Read more