 Hex-Rays (hexrays) <info hex-rays com> |
Tuesday, March 14 2006 09:29.46 CST |
Quite busy week, sorry for being silent.
I wanted to talk about an annoyance I discovered with all my C/C++ compilers.
Here is quite interesting presentation from Halvar Flake:
Attacks on uninitialized local variables
After reading it I wanted to verify my compilers and created a small C file. I wanted to check if the compilers would warn me of a potential uninitialized variable. The source code was pretty simple:
void const_ptr_acceptor(const int *);
int control_func(void)
{
int x;
return x + 1; // compiler emits a warning
}
int check_func(void)
{
int x;
const_ptr_acceptor(&x;); // we do not modify x here!
return x + 1; // compiler does not emit a warning
}
We have two functions, they both use an uninitialized variable. The only difference is the call to const_ptr_acceptor() which promises not to modify x. I compiled this source code with all warnings turned on. I was expecting two warnings from the compiler: the first warning about control_func and the second warning about check_func. However, there was only one warning:
E:hexconst_ptr>cl /Wall /c const_ptr.cpp
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 14.00.50215.44 for 80x86
Copyright (C) Microsoft Corporation. All rights reserved.
const_ptr.cpp
const_ptr.cpp
const_ptr.cpp
e:hexconst_ptrconst_ptr.cpp(6) : warning C4700: uninitialized local variable x used
I tried with all available compilers, but they were unanimous in their behavior: as soon as we pass a pointer to a variable, the compiler thinks that it is initialized. We explicitly specify with the const specifier that the function does not modify the variable, but the compilers seems to ignore it.
I compiled the code with Microsoft Visual Studio, Borland BCB6, GNU C, Intel compilers.
Still have no explanation why all compilers behave this way.
| Posted: Wednesday, December 31 1969 18:00.00 CST | |