Thread: What does ++ mean?

the ++ increases the variables value by 1. It is just like var = var+1; just made short

c++
c = c + 1

c += 1

I was going to reply that. But I decided to test, because I'm a little unclear on the subject in reference to pointers. I really need to read a reference.

Anyway, I'm not sure how its happening. But it appears *z++ will change y to point to the next (or previous, can't tell) value in memory, and then you would be assigning the value pointed to by z to w. The thing is that the assignment doesn't carry through when its in one statement for some reason. However if you split the statement up it will work. Which is almost exactly what the OP asked if it was doing.

Pointers just became fun.

Code:

#include <iostream>

int main()
{
    float a = 10.0;
    float b = 5.0;
    float c = 2.5;

    float* y = &c;

    std::cout << *y << std::endl; //Output: 2.5

    *y--; //no longer points to c, and now points to undefined
    std::cout << *y << std::endl; //Output: something x10^-39
    
    *y++; //no longer points to undefined, and now points back to c
    std::cout << *y << std::endl; //Output: 2.5
    
    *y++; //no longer points to c, and now points to b
    std::cout << *y << std::endl; //Output: 5.0
    
    *y++; //no longer points to b, and now points to a
    std::cout << *y << std::endl; //Output: 10.0

    ++*y; //increments the value of pointed to by the pointer
    std::cout << *y << std::endl; //Output: 11.0

    ++*y; //increments the value of pointed to by the pointer
    std::cout << *y << std::endl; //Output: 12.0

    --*y; //decrements the value of pointed to by the pointer
    std::cout << *y << std::endl; //Output: 11.0

    *y--; //no longer points to a, and now points to b
    *y = c; //change the value pointed to by the pointer (b) to c
    std::cout << b << std::endl; //Output: 2.5

    *y++ = a; //no longer points to b, and now points back to a, and attempt to change the value pointed to by the pointer (b) to c
    std::cout << b << std::endl; //Output: 5.0 (should output 2.5 because its being assigned c's value, but outputs a's value of 11.0)

    std::cin.get();
}

Edit: Looks like 7stud and Daved explained it. But still, code to match here.

Edit: I understand now. I forgot. The reason *y++ = a; doesn't work is because it increments the pointer to the next place after the assignment.

Originally Posted by rodrigorules

c++ adds 1 to c

c =+ 1 means c= c + 1

well thats the same, but i think it shouldn't be =+, it should +=;
And what does
++c;
means?
I know it's possible.

Originally Posted by ElastoManiac

well thats the same, but i think it shouldn't be =+, it should +=;
And what does
++c;
means?
I know it's possible.

It meants c is incremented before it returns its value.

int c = 5, a = 0;

a = ++c; //would be 6
a = c++; //would be 5

and just to make this thread even more confusing (*x)++ will incremenet the value pointer to by x rather than x pointer itself.

Originally Posted by Ancient Dragon

and just to make this thread even more confusing (*x)++ will incremenet the value pointer to by x rather than x pointer itself.

interesting

Originally Posted by ElastoManiac

well thats the same, but i think it shouldn't be =+, it should +=;
And what does
++c;
means?
I know it's possible.

think of it like this
var++
is equivalent to the following function

Code:

int Postfix(int& var)
{
    int temp = var;
    var = var + 1;
    return temp;
}

whereas
++var
is equivalent to the following function

Code:

int Prefix(int& var)
{
    var = var + 1;
    return var;
}