Well, I'm going to have to do a bit more searching then. I found an article that makes ReadProcessMemory feasible, but like you said, it doesn't do any good if I can't find the pointer.
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Show posts MenuQuote from: Adron on April 07, 2006, 04:33 AMYeah, I realized that after awhile. I guess I'm going to have to find where in the code the memory address is determined.
Though at this time it might be appropriate to point out that values in registers do not quite stay the same throughout the execution of a program.
MOV BYTE PTR DS:[EDX],AL
$a = pack('V', 0x00);
$b = pack('v', 0x00);
Quote from: quasi-modo on January 10, 2006, 12:31 PMso you are saying we need to throw old people and babies into fires because they wont run away?Jeez, what's wrong with you? You sell the babies on the black market! Why waste perfectly good produce? As for the old people, I just said that because a lot of them (not all) tend to be rather feeble and they wouldn't be able to get out.
Quote from: Yegg on January 09, 2006, 02:23 PMWhat is the value of endl? Is it simply '\n'?I found this with my Googling expertise:
QuoteMany programmers append the endl manipulator after each cout expression, as in:cout << "user login" << endl;
cout << "enter your name: " << endl;
cin >> name;
In the first line, the endl manipulator is added to force a line break. However, it's cheaper in terms of performance to use the '\n' character instead of endl because endl also flushes the stream rather than just adding a line break. Flushing a stream can be a costly operation. The second cout expression too has a redundant endl manipulator because the display must appear on the screen before the cin expression is executed, as cin and cout are tied.
The bottom line: avoid using the endl manipulator if you can because it slows down your program.
Quote from: Shout on January 05, 2006, 09:47 PM
I graphed it on my TI87 and used ZBox around the intersection. It is D.
So how is your answer wrong?
y = 1 - x3
x = 1 - y3
1 - x = y3
3√(1 - x) = y
Page created in 0.133 seconds with 16 queries.