How Would I Call Increments To Increase The Count?

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1.00/5 (1 vote)

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My Script:

C++

int increment(int x)
{
  return x + 1;
}

void increment2(int *x)
{
  *x = *x + 1; 
}

int main(void)
{
  int count = 0;
  // call increment...
  // call increment2...

Posted 10-Dec-15 13:18pm

Member 12197225

Updated 10-Dec-15 13:39pm

User 11060979

v3

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Sergey Alexandrovich Kryukov 10-Dec-15 22:24pm

Where do you see a script? C++ is not a scripting language.
—SA

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barneyman · Answer 1 · 2015-12-10T13:40:00

Solution 1

Increment takes an int as a param and returns an int, so count would go in, and be assigned to whatever came out

Increment2 is slightly different, it directly increments the int at the address it was passed (it should check for a NULL pointer), so you'd have to pass the address of count in

I can't make the explanation simpler than that without actually doing your homework for you :)

Posted 10-Dec-15 13:40pm

barneyman

Comments

Member 12197225 10-Dec-15 19:43pm

Thanks but i know what each one does as i wrote it. But I'm unsure on how to call them to increment the count without a loop....

barneyman 10-Dec-15 19:48pm

the word 'loop' wasn't mentioned in your original question

you just call them ? they're functions .... it's a procedural language ...

Perhaps you've not outlined your *actual* requirement fully?

Member 12197225 10-Dec-15 19:49pm

thanks. But I'm confused to how u call them

barneyman 10-Dec-15 20:25pm

see my other comment down the thread ... google

i could give you a salmon, but you learning to fish is best

[no name] 10-Dec-15 19:43pm

A 5. Maybe a hint to a Parameter ....(int&) would complete it ;)

Member 12197225 10-Dec-15 19:44pm

ok. So just write int& ??

barneyman 10-Dec-15 19:58pm

no - 426 was being obtuse :)

the function 'Increment' is simple - google 'c++ how to call a function' will help you

Increment2 is more complex because of the pointer (*) qualifier - you need to get the addressof count, which is where the ampersand comes in .. again google 'c++ pass pointer to function'