If it is a global variable that is not declared static or inline, then they have external linkage by default. The problem with your extern declaration is that it is also a definition because you provided an initialisation value. Now since this definition is in a header file, this definition is going to appear across multiple source files (translation units) that include this header file. Defining the same variable more than once in multiple translation units (barring a few exceptions) violates what is known as One Definition Rule (ODR) and will be flagged as an error when it can be detected. There are ways to get around this. 1. You can not initialize the extern variable in the header and instead initialize it in one and only one source file that includes the header file. 2. You can remove extern and declare the variable as either static (not recommended) or inline 3. You can remove extern and provide the definition within an anonymous namespace in the header file.

Does anyone knows how to use loop and nested loop on c programming?

Seriously?

How to differentiate loop & nested loop from how they used?

How to differentiate loop & nested loop from how they used?

generally nested loops are not so encouraged.As Tazz explained, unless you want to change value and require to keep track with.

Not sure what you mean by "not encouraged", they increase complexity but some algorithms require them

generally nested loops are not so encouraged.As Tazz explained, unless you want to change value and require to keep track with.

Define a function to input variable length string and store it in an array without memory wastage.

"input variable length string" probably means that a user should type a string and your program should read it. Without memory wastage means that you need to use no more memory than required to store the string (length + 1). So you can't just allocate a buffer of 10000 chars and store the string in that buffer.

7 3 1 2 3 4 5 6 7

read by line or read by char and skip space and \n

int x,y,a[10],i; scanf("%d",&x); scanf("%d",&y); for(i=0;i<10;i++) { scanf("%d",&a[i])!=EOF; } can i write like this

how about if（1 ！= scanf（））break；

Guys can you see whats wrong with my code🥲

First of all, it doesn't compile, crashes, or produces wrong result?

just not output correct answer

Ok, what is data that you provide to it, expected answer, and what does it output instead?

https://onlinegdb.com/eBPdFwrHj

Can't edit it.

Clicking on fork this in the top left corner might help?

https://onlinegdb.com/eBPdFwrHj

You set m[2] = 29; if at least one of the years from 2000 to the given year passed the condition for the leap year? Is that correct? In this condition you divide month index by 400 if (((year % 4 == 0 && year % 100 != 0) || (i % 400 == 0)) then you account for the leap year one more time the next line these three things don't feel correct

You set m[2] = 29; if at least one of the years from 2000 to the given year passed the condition for the leap year? Is that correct? In this condition you divide month index by 400 if (((year % 4 == 0 && year % 100 != 0) || (i % 400 == 0)) then you account for the leap year one more time the next line these three things don't feel correct

If you're doing in cmd line then knowing any TUI library would help in creating an UI with cooler interfaces.

Print the contents of the iostream library using pointers ?

Whot

Print the contents of the iostream library using pointers ?

u mean stringstream？

Huh ?

u wish to get pointer that wrapped by iostream and manipulation the raw data？

I wanna contents the (iostream) Library using pointers !

I wanna contents the (iostream) Library using pointers !

How this code does not produce UB? Value is destructed before Ref which holds a reference to it But it still prints in Ref destructor struct Value{ int value; Value(int value) : value{value}{} ~Value(){ std::cout << "Value: " << value << " Out\n"; } }; struct Ref{ Ref(Value& r) : my_ref{r}{} Value& my_ref; ~Ref(){ std::cout << "Ref: " << my_ref.value << " Out\n"; } }; struct S{ Ref ref{value}; Value value{10}; }; int main(){ S s; return 0; }

How this code does not produce UB? Value is destructed before Ref which holds a reference to it But it still prints in Ref destructor struct Value{ int value; Value(int value) : value{value}{} ~Value(){ std::cout << "Value: " << value << " Out\n"; } }; struct Ref{ Ref(Value& r) : my_ref{r}{} Value& my_ref; ~Ref(){ std::cout << "Ref: " << my_ref.value << " Out\n"; } }; struct S{ Ref ref{value}; Value value{10}; }; int main(){ S s; return 0; }

Undefined behavior means it is not defined it the standard. It doesnt mean it wont compile or run

when it runs，the result is undefined

int computeMax(int * array, int arrayLength) { // todo int i, maxInt; maxInt=array[0]; for (i = 0; i < arrayLength - 1; i++) { if(array[i] > maxInt) { maxInt= array[i]; } } return maxInt; }

This is Undefined Behavior. Why do you think it is not? Use a pastebin next time to post code.

Undefined behavior means it is not defined it the standard. It doesnt mean it wont compile or run

I know But it prints the value in Ref destructor which is 10

The value that ref refers to will still exist in the stack since main is the only function executing. The stack is not clobbered. So the reference to this value may still print 10 as the value 10 is still in the stack. It is Undefined Behavior nevertheless.

or deterministic is the wrong word I think... to put it simply: you can know what the result will be on an actual machine

nt checkIncreasing(int * array, int arrayLength) { // todo int i; for(i = 0; i < arrayLength - 1; i++){ if(array[i - 1] >= array[i + 1]){ return 1; } else if(array[i - 1] < array[i + 1]) return 0; }

do you see anything wrong here? and why isn't i starting by 0?

do you see anything wrong here? and why isn't i starting by 0?

when i =0，array[i-1] out of bound

struct Word *words[size]; addStruct(){ words[counter++] = createWord(isims...); } emptyStruct{ //In this method, I want to empty this struct(words). } In C, How can I get memory space for this Struct(words) in emptyStruct method.

cast pointer from word to emptystruct

yes but i is not equal to 0 because it worked like that. and that's the question, why isn't it 0 and out of bound?

When the first enter, i = 0 if segment will be executed

How?

sorry i'm wrong. in empty method，u can use counter to make a for loop，each time u can use word[index] to delete data

No. That is not what Undefined Behavior means. There is no way for you to predict the output of a program that exhibits Undefined Behavior. It may exhibit the same behavior sometimes and could behave totally unpredictably at other times.