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In some project, I must access the machine code instructions of a function for debugging reasons.
My first approach (I decided to do it differently) was to convert a function pointer to the function to a data pointer using a union like this:
void exampleCode(void)
{
volatile union {
uint8_t ** pptr;
void (* pFunc)(void);
} u;
uint8_t ** copyPptr;
uint8_t * resultPtr;
#if 1
u.pptr = (uint8_t **)0x10000000;
#endif
u.pFunc = &myFunction;
copyPptr = u.pptr;
/* The problem is here: */
resultPtr = copyPptr[109];
*resultPtr = 0xA5;
}
If I remove the line u.pptr = (uint32_t **)0x10000000, the compiler assumes that copyPptr two lines below has an undefined value (according to the C standard, converting function pointers to data pointers is not allowed).
For this reason, it "optimizes" the last two lines out (which is of course not wanted)!
However, the volatile keyword means that the C compiler must not optimize away the read operation of u.pptr in the line copyPptr = u.pptr.
From my understanding, this also means that the C compiler must assume that it is at least possible that a "valid" result is read in this read operation, so copyPptr contains some "valid" pointer.
(At least this is how I understand the answers to this question.)
Therefore, the compiler should not optimize away the last two lines.
Is this a compiler bug (GCC for ARM) or is this behavior allowed according to the C standard?
EDIT
As I have already written, I already have another solution for my problem.
I asked this question because I'm trying to understand why the union solution did not work to avoid similar problems in the future.
In the comments, you asked me for my exact use case:
For debugging reasons, I need to access some static variable in some library that comes as object code.
The disassembly looks like this:
.text
.global myFunction
myFunction:
...
.L1:
.word .theVariableOfInterest
.data:
# unfortunately not .global!
.theVariableOfInterest:
.word 0
What I'm doing is this:
I set copyPtr to myFunction. If K is (.L1-myFunction)/4, then copyPtr+K points to .L1 and copyPtr[K] points to .theVariableOfInterest.
So I can access the variable using *(copyPtr[K]).
From the disassembly of the object file, I can see that K is 109.
All this works fine if copyPtr contains the correct pointer.
EDIT 2
It's much simpler to reproduce the problem with gcc -O3:
char a, c;
char * b = &a;
void testCode(void)
{
char * volatile x;
c = 'A';
*x = 'X';
}
It's true that uninitialized variables cause undefined behavior.
However, if I understand the description of volatile in the draft "ISO/IEC 9899:TC3" of the C99 standard correctly, a volatile variable would not be "uninitialized" if I stop the program in the debugger (using a breakpoint) at the line c = 'A' and copy the value of b to x in the debugger.
On the other hand, some other statement was saying that using volatile for local variables in current C++ (not C) versions already causes undefined behavior ...
Now I'm wondering what is true for modern C (not C++) versions.
I'd personally try compiling with the no optimization flag -O0 (capital o, zero) to avoid the compiler "optimizing" out your code and seeing if it works then. If that's still acting problematic, consider adding the assembly -S flag so you can see what it's actually doing translating those instructions as and retranslating them yourself using the inline assembly __asm__ function.
#include <stdio.h>
int main()
{
const int a = 12;
int *p;
p = &a;
*p = 70;
}
Will it work?
It's "undefined behavior," meaning that based on the standard you can't predict what will happen when you try this. It may do different things depending on the particular machine, compiler, and state of the program.
In this case, what will most often happen is that the answer will be "yes." A variable, const or not, is just a location in memory, and you can break the rules of constness and simply overwrite it. (Of course this will cause a severe bug if some other part of the program is depending on its const data being constant!)
However in some cases -- most typically for const static data -- the compiler may put such variables in a read-only region of memory. MSVC, for example, usually puts const static ints in .text segment of the executable, which means that the operating system will throw a protection fault if you try to write to it, and the program will crash.
In some other combination of compiler and machine, something entirely different may happen. The one thing you can predict for sure is that this pattern will annoy whoever has to read your code.
It's undefined behaviour. Proof:
/* program.c */
int main()
{
const int a = 12;
int* p;
p = &a;
*p = 70;
printf("%d\n", a);
return 0;
}
gcc program.c
and run it. Output will be 70 (gcc 4.3)
Then compile it like this:
gcc -O2 program.c
and run it. The output will be 12. When it does optimisation, the compiler presumably loads 12 into a register and doesn't bother to load it again when it needs to access a for the printf because it "knows" that a can't change.
Modifying a const qualified object through a pointer invokes undefined behaviour, and such is the result. It may be something you'd expect from a particular implementation, e.g. the previous value unchanged, if it has been placed in .text, etc.
It does indeed work with gcc. It didn't like it though:
test.c:6: warning: assignment discards qualifiers from pointer target type
But the value did change when executed. I won't point out the obvious no-no...
yes, you can make it done by using such code. but the code do not apply when when a is global (a gcc-compiled program gave me segmentation fault.)
generally speaking, in beloved C, you can almost always find someway to hack things that are not supposed to be changed or exposed. const here being a example.
But thinking about the poor guy(maybe myself after 6 months) maintains our code, I often choose not do so.
Here the type of pointer p is int*, which is being assigned the value of type const int* (&a => address of a const int variable).
Implicit cast eliminates the constness, though gcc throws a warning (please note this largely depends on the implementation).
Since the pointer is not declared as a const, value can be changed using such pointer.
if the pointer would be declared as const int* p = &a, you won't be able to do *p = 70.
This code contains a constraint violation:
const int a = 12;
int *p;
p = &a;
The constraint violated is C11 6.5.16.1/1 "Simple assignment"; if both operands are pointers then the type pointed to by the left must have all the qualifiers of the type pointed to by the right. (And the types, sans qualifiers, must be compatible).
So the constraint is violated because &a has type const int *, which has const as a qualifier; but that qualifier does not appear in the type of p which is int *.
The compiler must emit a diagnostic and might not generate an executable. The behaviour of any executable would be completely undefined, since the program does not comply with the rules of the language.
You cannot change the value of a constant variable by using a pointer pointing to it. This type of pointer is called as Pointer to a constant.
There is also another concept called Constant Pointer. It means that once a pointer points to a memory location you cannot make it point to the another location.
Bad, BAD idea.
Also, the behavior is platform- and implementation-specific. If you're running on a platform where the constant is stored in non-writable memory, this obviously won't work.
And, why on earth would you want to? Either update the constant in your source, or make it a variable.
The problem with changing the value of const variable is that the compiler will not expect that to happen. Consider this code:
const int a = 12;
int * p = &a;
*p = 70;
printf("%d\n", a);
Why would the compiler read a in the last statement? The compiler knows that a is 12 and since it is const, it will never change. So the optimizer may transform the code above into this:
const int a = 12;
int * p = &a;
*p = 70;
printf("%d\n", 12);
This can lead to strange issues. E.g. the code might work as desired in debug builds without optimization but it will fail in release builds with optimization.
Actually a good optimizer might transform the entire code to this:
printf("%d\n", 12);
As all other code before has no effect in the eye of the compiler. Leaving out code that has no effect will also have no effect on the overall program.
On the other hand, a decent compiler will recognize, that your code is faulty and warn you, since
int * p = &a;
is actually wrong. Correct would be:
const int * p = &a;
as p is not a pointer to int, it is a pointer to const int and when declared like that, the next line will cause a hard compile error.
To get rid of the warning, you have to cast:
int * p = (int *)&a;
And an even better compiler will recognize that this cast breaks the const promise and instruct the optimizer to not treat a as const.
As you can see, the quality, capabilities and settings of the compilerwill decide in the end what behavior you can expect. This implies that the same code may show different behavior on different platforms or when using different compilers on the same platform.
If the C standard had defined a behavior for that case, all compilers would have to implement it and no matter what the standard had defined, it would have been hard to implement, putting a huge burden on everyone who wants to write a compiler. Even if the standard had just said "This is forbidden", all compilers would have to perform complex data flow analysis to enforce this rule. So the standard just doesn't define it. It defines that const values cannot be changed and if you find a way to change them anyway, there is no behavior you can rely on.
Yes, you can change the value of a constant variable.
Try this code:
#include <stdio.h>
int main()
{
const int x=10;
int *p;
p=(int*)&x;
*p=12;
printf("%d",x);
}
#include <stdio.h>
int main()
{
const int a = 12;
int *p;
p = &a;
*p = 70;
}
Will it work?
It's "undefined behavior," meaning that based on the standard you can't predict what will happen when you try this. It may do different things depending on the particular machine, compiler, and state of the program.
In this case, what will most often happen is that the answer will be "yes." A variable, const or not, is just a location in memory, and you can break the rules of constness and simply overwrite it. (Of course this will cause a severe bug if some other part of the program is depending on its const data being constant!)
However in some cases -- most typically for const static data -- the compiler may put such variables in a read-only region of memory. MSVC, for example, usually puts const static ints in .text segment of the executable, which means that the operating system will throw a protection fault if you try to write to it, and the program will crash.
In some other combination of compiler and machine, something entirely different may happen. The one thing you can predict for sure is that this pattern will annoy whoever has to read your code.
It's undefined behaviour. Proof:
/* program.c */
int main()
{
const int a = 12;
int* p;
p = &a;
*p = 70;
printf("%d\n", a);
return 0;
}
gcc program.c
and run it. Output will be 70 (gcc 4.3)
Then compile it like this:
gcc -O2 program.c
and run it. The output will be 12. When it does optimisation, the compiler presumably loads 12 into a register and doesn't bother to load it again when it needs to access a for the printf because it "knows" that a can't change.
Modifying a const qualified object through a pointer invokes undefined behaviour, and such is the result. It may be something you'd expect from a particular implementation, e.g. the previous value unchanged, if it has been placed in .text, etc.
It does indeed work with gcc. It didn't like it though:
test.c:6: warning: assignment discards qualifiers from pointer target type
But the value did change when executed. I won't point out the obvious no-no...
yes, you can make it done by using such code. but the code do not apply when when a is global (a gcc-compiled program gave me segmentation fault.)
generally speaking, in beloved C, you can almost always find someway to hack things that are not supposed to be changed or exposed. const here being a example.
But thinking about the poor guy(maybe myself after 6 months) maintains our code, I often choose not do so.
Here the type of pointer p is int*, which is being assigned the value of type const int* (&a => address of a const int variable).
Implicit cast eliminates the constness, though gcc throws a warning (please note this largely depends on the implementation).
Since the pointer is not declared as a const, value can be changed using such pointer.
if the pointer would be declared as const int* p = &a, you won't be able to do *p = 70.
This code contains a constraint violation:
const int a = 12;
int *p;
p = &a;
The constraint violated is C11 6.5.16.1/1 "Simple assignment"; if both operands are pointers then the type pointed to by the left must have all the qualifiers of the type pointed to by the right. (And the types, sans qualifiers, must be compatible).
So the constraint is violated because &a has type const int *, which has const as a qualifier; but that qualifier does not appear in the type of p which is int *.
The compiler must emit a diagnostic and might not generate an executable. The behaviour of any executable would be completely undefined, since the program does not comply with the rules of the language.
You cannot change the value of a constant variable by using a pointer pointing to it. This type of pointer is called as Pointer to a constant.
There is also another concept called Constant Pointer. It means that once a pointer points to a memory location you cannot make it point to the another location.
Bad, BAD idea.
Also, the behavior is platform- and implementation-specific. If you're running on a platform where the constant is stored in non-writable memory, this obviously won't work.
And, why on earth would you want to? Either update the constant in your source, or make it a variable.
The problem with changing the value of const variable is that the compiler will not expect that to happen. Consider this code:
const int a = 12;
int * p = &a;
*p = 70;
printf("%d\n", a);
Why would the compiler read a in the last statement? The compiler knows that a is 12 and since it is const, it will never change. So the optimizer may transform the code above into this:
const int a = 12;
int * p = &a;
*p = 70;
printf("%d\n", 12);
This can lead to strange issues. E.g. the code might work as desired in debug builds without optimization but it will fail in release builds with optimization.
Actually a good optimizer might transform the entire code to this:
printf("%d\n", 12);
As all other code before has no effect in the eye of the compiler. Leaving out code that has no effect will also have no effect on the overall program.
On the other hand, a decent compiler will recognize, that your code is faulty and warn you, since
int * p = &a;
is actually wrong. Correct would be:
const int * p = &a;
as p is not a pointer to int, it is a pointer to const int and when declared like that, the next line will cause a hard compile error.
To get rid of the warning, you have to cast:
int * p = (int *)&a;
And an even better compiler will recognize that this cast breaks the const promise and instruct the optimizer to not treat a as const.
As you can see, the quality, capabilities and settings of the compilerwill decide in the end what behavior you can expect. This implies that the same code may show different behavior on different platforms or when using different compilers on the same platform.
If the C standard had defined a behavior for that case, all compilers would have to implement it and no matter what the standard had defined, it would have been hard to implement, putting a huge burden on everyone who wants to write a compiler. Even if the standard had just said "This is forbidden", all compilers would have to perform complex data flow analysis to enforce this rule. So the standard just doesn't define it. It defines that const values cannot be changed and if you find a way to change them anyway, there is no behavior you can rely on.
Yes, you can change the value of a constant variable.
Try this code:
#include <stdio.h>
int main()
{
const int x=10;
int *p;
p=(int*)&x;
*p=12;
printf("%d",x);
}
This question already has answers here:
Can we change the value of an object defined with const through pointers?
(11 answers)
Closed 5 years ago.
I am able to change the value of const modified variable in gcc but not in other compilers.
I have tried this code on gcc, which updates the value of i and j (11). With an online compiler, I get different values.
#include<stdio.h>
void main() {
const int i=10;
int *j;
j = &i;
(*j)++;
printf("address of j is %p address of i is %p\n",j,&i);
printf("i is %d and j is %d\n",i,*j);
}
Yes, you can do it with a little hack.
#include <stdio.h>
int main(){
const int a = 0;
*(int *)&a = 39;
printf("%d", a);
}
In the above code, a is a const int. With the little hack, you can change the constant value.
Update: Explanation
In the above code, a is defined as a const. For example a has a memory addr 0x01 and therefore &a returns the same. When it is casted with (int *) it becomes another variable referred as a pointer to the const. When it is accessed with * again, the another variable can be accessed without violation of the const policy because it is not the original variable, but the changes are reflected because it is referred as address to the pointer.
This will work on older versions like Borland C++ or Turbo C++, however no one is using it now a days.
It's "undefined behaviour", meaning that based on the standard you can't predict what will happen when you try this. It may do different things depending on the particular machine, compiler, and state of the program.
In this case, what will most often happen is that the answer will be "yes". A variable, const or not, is just a location in memory, and you can break the rules of const and simply overwrite it. (Of course this will cause a severe bug if some other part of the program is depending on its const data being constant!)
However in some cases -- most typically for const static data -- the compiler may put such variables in a read-only region of memory. MSVC, for example, usually puts const static ints in .text segment of the executable, which means that the operating system will throw a protection fault if you try to write to it, and the program will crash.
In some other combination of compiler and machine, something entirely different may happen. The one thing you can predict for sure is that this pattern will annoy whoever has to read your code.
Try this and let me know.
How to modify value of const variable?
No! You shouldn't modify a const variable.
The whole point of having a const variable is to be not able to modify it. If you want a variable which you should be able to modify, simply don't add a const qualifier on it.
Any code which modify's a const forcibly through (pointer)hackery invokes Undefined Behavior.
An Undefined Behavior means that the code is non conforming to the standard specifications laid out by the C standard and hence not a valid code. Such a code can show any behavior and it is allowed to do so.
By defining i as const, you promised not to modify it. The compiler can rely on that promise and assume that it's not modified. When you print the value of i, the compiler can just print 10 rather than loading whatever value is currently stored in i.
Or it can choose to load the value. Or it can cause your program to crash when you try to modify i. The behavior is undefined.
You'll likely see different behavior with gcc depending on the optimization options (-O1, -O3).
Oh, and void main() is incorrect; it should be int main(void). If your textbook tells you to use void main(), you should get a better book.
Take a look at Can we change the value of an object defined with const through pointers?
Long story short, it's an undefined behaviour. It can result dependently on compiler/machine.
You can not modify the value of const variable if you compile this code in gcc it will show
error: invalid conversion from ‘const int*’ to ‘int*’ [-fpermissive]
and show undefined behave
The maiin thing is that we can only modify the variable when we can access the address without address we can't do anything. That's the same thing with Register storage class.
We are currently developing an application for a msp430 MCU, and are running into some weird problems. We discovered that declaring arrays withing a scope after declaration of "normal" variables, sometimes causes what seems to be undefined behavior. Like this:
foo(int a, int *b);
int main(void)
{
int x = 2;
int arr[5];
foo(x, arr);
return 0;
}
foo is passed a pointer as the second variable, that sometimes does not point to the arr array. We verify this by single stepping through the program, and see that the value of the arr array-as-a-pointer variable in the main scope is not the same as the value of the b pointer variable in the foo scope. And no, this is not really reproduceable, we have just observed this behavior once in a while.
This is observable even before a single line of the foo function is executed, the passed pointer parameter (b) is simply not pointing to the address that arr is.
Changing the example seems to solve the problem, like this:
foo(int a, int *b);
int main(void)
{
int arr[5];
int x = 2;
foo(x, arr);
return 0;
}
Does anybody have any input or hints as to why we experience this behavior? Or similar experiences? The MSP430 programming guide specifies that code should conform to the ANSI C89 spec. and so I was wondering if it says that arrays has to be declared before non-array variables?
Any input on this would be appreciated.
Update
#Adam Shiemke and tomlogic:
I'm wondering what C89 specifies about different ways of initializing values within declarations. Are you allowed to write something like:
int bar(void)
{
int x = 2;
int y;
foo(x);
}
And if so, what about:
int bar(int z)
{
int x = z;
int y;
foo(x);
}
Is that allowed? I assume the following must be illegal C89:
int bar(void)
{
int x = baz();
int y;
foo(x);
}
Thanks in advance.
Update 2
Problem solved. Basically we where disabling interrupts before calling the function (foo) and after declarations of variables. We where able to reproduce the problem in a simple example, and the solution seems to be to add a _NOP() statement after the disable interrupt call.
If anybody is interested I can post the complete example reproducing the problem, and the fix?
Thanks for all the input on this.
That looks like a compiler bug.
If you use your first example (the problematic one) and write your function call as foo(x, &arr[0]);, do you see the same results? What about if you initialize the array like int arr[5] = {0};? Neither of these should change anything, but if they do it would hint at a compiler bug.
In your updated question:
Basically we where disabling interrupts before calling the function (foo) and after declarations of variables. We where able to reproduce the problem in a simple example, and the solution seems to be to add a _NOP() statement after the disable interrupt call.
It sounds as if the interrupt disabling intrinsic/function/macro (or however interrupts are disabled) might be causing an instruction to be 'skipped' or something. I'd investigate whether it is coded/working correctly.
You should be able to determine if it is a compiler bug based on the assembly code that is produced. Is the assembly different when you change the order of the variable declarations? If your debugger allows you, try single stepping through the assembly.
If you do find a compiler bug, also, check your optimization. I have seen bugs like this introduced by the optimizer.
Both examples look to be conforming C89 to me. There should be no observable difference in behaviour assuming that foo isn't accessing beyond the bounds of the array.
For C89, the variables need to be declared in a list at the start of the scope prior to any assignment. C99 allows you to mix assignment an declaration. So:
{
int x;
int arr[5];
x=5;
...
is legal c89 style. I'm surprised your compiler didn't throw some sort of error on that if it doesn't support c99.
Assuming the real code is much more complex, heres some things i would check, keep in mind they are guesses:
Could you be overflowing the stack on occasion? If so could this be some artifact of "stack defense" by the compiler/uC? Does the incorrect value of &foo fall inside a predictable memory range? if so does that range have any significance (inside the stack, etc)?
Does the mcu430 have different ranges for ram and rom addressing? That is, is the address space for ram 16bit while the program address space 24bit? PIC's have such an architecture for example. If so it would be feasible that arr is getting allocated as rom (24bit) and the function expects a pointer to ram (16bit) the code would work when the arr was allocated in the first 16bit's of address space but brick if its above that range.
Maybe you have at some place in your program in illegal memory write which corrupts your stack.
Did you have a look at the disassembly?