Weird memory overwrite issues causing integers to become hex numbers

Question

I am in the midst of debugging someone's code, and I have code like

int i = their_val;
std::cout << "output: " << i << std::endl;

When I look at the log output I see lines like

output: a

Should this happen? Is something changing the cout formatting or could it be something odder?

"causing integers to become hex numbers" is a phrase to think about... those poor integers. — Kerrek SB, Oct 31 '11 at 17:25
As the answers below have noted, a preceding `std::hex` is causing the issue. See this previous discussion: http://stackoverflow.com/questions/1532640/which-iomanip-manipulators-are-sticky — Gnawme, Oct 31 '11 at 18:20

score 3 · Answer 1 · answered Oct 31 '11 at 17:15

3

Did someone cout << std::hex prior to that output? It would cause it to print in hexadecimal.

answered Oct 31 '11 at 17:15

Michael Kristofik

I think it's probably that (upvoted) but it's a huge enough code base that I can't immediately find it. I realised this was probably the case just after submitting... – invisiblerhino Oct 31 '11 at 17:27

score 3 · Answer 2 · answered Oct 31 '11 at 17:16

3

Check and see if std::hex gets passed into std::cout anywhere. That would result in the behavior you're seeing.

You can force things to be in decimal using:

std::cout << "output: " << std::dec << i << std::endl;

answered Oct 31 '11 at 17:16

Billy ONeal

score 0 · Answer 3 · answered Oct 31 '11 at 17:16

0

You probably did std::cout << std::hex somewhere earlier. You can undo this with std::cout << std::dec.

std::cout << "output: " << std::dec << i << std::endl;

answered Oct 31 '11 at 17:16

David Heffernan

3 Answers3