Simple Code failing

[LTXN8vqX]

I have been window gui's with winapi and game hacks, but i suddenly realized that i don't fully understand everything. So i quit hacking and crap like that and now i am trying to restudy c++. So i wanted to make a program that would print every other letter of a string, but it ended up printing random characters. Here is my code:

#include <iostream>
using namespace std;
int main()
{
	char szbuffer[10]="hello how";
	for(int i = 0; i = sizeof(szbuffer); i+=2){ 
	cout<<szbuffer[i];
	if(i = 10){break;}
	
	}
	cin.get();
	return 0 ;
}

[Disch]

Line 6... your condition is wrong:

i = sizeof(szbuffer);

You probably meant this:

i < sizeof(szbuffer);


You also have the same problem on line 8:

if(i = 10){break;} // you probalby meant i == 10

Remember that = is assignment. In both cases you are reassigning the value of i so that it is something else. You don't want to do that in either or those situations.

[matsom]

The format is: for (initialization; condition; increase) statement;
ref: http://www.cplusplus.com/doc/tutorial/control/

...printing random characters.

you first declare i = 0, then you change it to i = sizeof(szbuffer), so your cout will start printing from szbuffer[sizeof(szbuffer)] which gives you garbage.
you're lucky because you make another mistake in if(i = 10) break; which cause your loop to exit ;)

[Branflakes91093]

Your problem is the for loop.

Look here: http://cplusplus.com/doc/tutorial/control/

Also, the if statement will not do what you expect it to.

//Right way:
int a = 0; // one '=' for assignment/definition
if(a == 1) { } // two '='s for comparison

// Wrong way:
int a = 0;
if(a = 2) { } // sets a equal to 2 and returns true (the code within the if statement will always be run) 

That if statement will be redundant when you fix the for loop anyway.