I'm just beginning to learn C++ through my university course.
I'm revising for a C++ exam and I was hoping I could just ask some simple questions about things I'm not so sure on.
First of all, if I have this:
int i = 4;
What is the difference between these two lines:
int* p = &i;
int& r = i;
By my understanding, at this point both p and r are like conduits to i; changing their value will change the value of i...
Secondly, about assembly, I have a question in a past exam that is fooling me:
mov eax, dword ptr[b]
push eax
mov eax, dword ptr[a]
push eax
call printCode (411186h)
add esp,8
Q: Which of the following prototypes
when dissassembled, best match the
assembly code?
A: int printCode(int a, int b)
My question is: If the assembly is moving ptr[a] and ptr[b]... Why is the prototype not int* a, int* b?
Thank you.
The difference is that changing p's value won't change i, it will change what p points to:
p = &j;
With p, you can access, and change, i via indirection:
*p = 5;
Whereas with r, the indirection is not required, since r is an alias for i. On the flip side, this means that there is no way to change r to reference a different variable. i.e., there is no equivalent to p = &j;.
You should split the second question into a second question (if you get my drift).
int* p = &i;
p is a pointer pointing to the address of i. p will be a variable stored in your memory that contains the (virtual) memory address of i.
int& r = i;
r is a reference to i. You should think of references as aliases: you could think that r is an alias for the name i
I'd suggest you to read C++Faq chapter about references.
int* p = &i;
int& r = i;
The first statement means assigning the address of variable i to a pointer p. Here later you can change the value of the pointer p.
The second statement is "create a reference or alias r for the variable i". It is just another name for the same variable. You cannot change the reference later to point to a different variable in c++.
For the assembly snippet. I sense the snippet points to calling code. the call might have been like
printCode(*a,*b); // Here a and b are pointer to integers
So the prototype is
int printCode(int a,int b)
is right.
Ferose Khan J
What is the difference between these two lines:
int* p = &i;
int& r = i;
By my understanding, at this point both p and r are like conduits to i; changing their value will change the value of i...
The sentence is not true if taken literally. Changing p value will only make it point to a different int, changing the memory pointed to will change i:
int j = 0;
*p = 5; // equivalent to r = 5, or i = 5 (semantically, can be different in assembler)
p = &j; // changes p to point to j instead of i
You should think of pointers as proxy objects that refer you to a real instance, but that can be modified to lead to a different place. References are best understood as aliases to the object on the right hand side of the definition. You never operate on the reference, you operate on the referred object, they are the same.
References are more or less only syntactic sugars, but there are few differences between them and pointers which are described very well here.
For disassembly question.
When you say:
mov eax, dword ptr[b]
You are asking CPU to move the value from a specified address (b). So it depends what is b. If b is a int* then the answer is correct.
Related
Can someone explain to me what's going on in this code block? Specifically on line 3. I have a hunch the * before ptr is significant. And (uint8_t *) looks like a cast to a byte... But what's up with the *? It also looks like r, g, and b will all evaluate to the same value.
case TRUECOLOR: { // 24-bit ('truecolor') image (no palette)
uint8_t pixelNum, r, g, b,
*ptr = (uint8_t *)&imagePixels[imageLine * NUM_LEDS * 3];
for(pixelNum = 0; pixelNum < NUM_LEDS; pixelNum++) {
r = *ptr++;
g = *ptr++;
b = *ptr++;
strip.setPixelColor(pixelNum, r, g, b);
}
I work primarily in C#.
The second and third line can be expressed more cleanly:
uint8_t pixelNum;
uint8_t r;
uint8_t g;
uint8_t b;
uint8_t *ptr = (uint8_t *)&imagePixels[imageLine * NUM_LEDS * 3];
The first four variable declarations should be fairly simple, the fifth one is something C# does not have. It declares ptr as a pointer to a uint8_t. This pointer is set to the address of the value which is the imageLine * NUM_LEDS * 3th element in the imagePixels array. As this might be a different type (maybe a pointer to a char, who knows), this value is cast to a pointer to an uint8_t.
The next occurence of the asterisk (*) is in the for-loop body, where it is used as the dereference operator, which basically resolves a pointer to get the actual value.
Pointers 101
A pointer is like the street address of a house. It shows you where the house is so you can find it, but when you pass it around, you don't pass around the whole house. You can dereference it, meaning you can actually visit the house.
The two operators used in conjunction with pointers are the asterisk (*) and the ampersand (&). The asterisk is used in declarations of pointers and to dereference a pointer, the ampersand is used to get the address of something.
Take a look at the following example:
int x = 12;
int *y = &x;
std::cout << "X is " << *y; // Will print "X is 12"
We declare x as an int holding the value 12. Now we declare y as a pointer to an int, and set it to point at x by storing x's address. By using *y, we access the actual value of x, the int that y points at.
Since a pointer is a type of reference, modifying the value via the pointer changes the actual value of the thing pointed at.
int x = 12;
int *y = &x;
*y = 10;
std::cout << "X is " << x; // Will print "X is 10"
Pointers 102
Pointers are a large topic, and I suggest you take your time to read about them from different sources if necessary.
Used in a variable definition, the * means ptr is a pointer. The value it stores is an address in memory for another variable or a part of another variable. In this case ptr is a pointer to a block of memory inside imagePixels and from the names of the variables involved it's a line in an image. Since the type is uint8_t, this is taking whatever imagePixels is and using it as a block of individual bytes.
Used outside a varable definition, the * takes on a different meaning: dereference the pointer. Go to the location in memory stored in the pointer and get the value.
And yeah, * can also be used for multiplication, upping the code-reading fun level.
Incrementing (++) a pointer moves the address to the next address. If you had a uint32_t * the address would advance by 4 to point at the next uint32_t. In this case we have uint8_t, so the address is advanced one byte. So
r = *ptr++;
A) Get value at pointer.
After A) Advance the pointer.
After A) Assign value to r.
Exactly where the "advance the pointer" stage goes is tricky. It is after step A. In C++17 or greater it is before "Assign the value" because there is now a separation between the stuff on the right and the stuff on the left of an equals sign. But before C++17 all we can say is it's after step A. Search keyterm: "Sequence Points".
g = *ptr++;
b = *ptr++;
Do it again, get and assign the current value at ptr, advance the pointer.
strip.setPixelColor(pixelNum, r, g, b);
From the naming I presume this sets a given pixel to the colours read above.
You can't just
strip.setPixelColor(pixelNum, *ptr++, *ptr++, *ptr++);
Because of sequencing again. There are no guarantees of the order in which the parameters will be computed. This is to allow compiler developers to make optimizations for speed and size that they cannot if the ordering is specified, but it's a kick in the teeth to those expecting left-to-right resolution. My understanding is this still holds true in the C++17 standard.
OK. So what is this doing?
There is a big block of memory from which you want one and only one line.
*ptr = (uint8_t *)&imagePixels[imageLine * NUM_LEDS * 3];
pinpoints the beginning of that line and sets it up to be treated like a dumb array of bytes.
for(pixelNum = 0; pixelNum < NUM_LEDS; pixelNum++) {
Generic for loop. For all the pixels on the line of LEDs.
r = *ptr++;
g = *ptr++;
b = *ptr++;
Get the colour of one pixel on the line in the standard 8 bit RGB format and point at the next pixel
strip.setPixelColor(pixelNum, r, g, b);
writes the read colour to one pixel.
The for loop will then loop around and start working on the next pixel until there are no more pixels on the line.
The asterisk(*) is the symbol for a pointer. So the (uint8_t *) is a cast to a pointer that is pointing to a uint8_t. Then within the loop, where the asterisk is prefixed to a symbol (ie *ptr) that is dereferencing that pointer. Dereferencing the pointer returns the data that the pointer is pointing to.
I suggest reading a bit about pointers as they are critical to understanding C/C++. Here is the C++ Docs on Pointers
MildlyInformed, I would need more code to run through it to explain it. One tool I found really, really useful though is the C visualizer. It's an online debug tool that helps you figure out what's happening in code by running you through step-by-step, line by line. It can be found at: http://www.pythontutor.com/visualize.html#mode=edit
Even though the URL talks about python, it can do C and a bunch of languages. I would have commented instead of posting an answer, but my rep isn't high enough. I hope this helps!
(I'm not affiliated with the above website, other than to use it occasionally when I'm baffled)
This question already has answers here:
Two different values at the same memory address
(7 answers)
Closed 5 years ago.
#include <iostream>
using namespace std;
int main() {
const int a = 10;
auto *b = const_cast<int *>(&a);
*b = 20;
cout << a << " " << *b;
cout << endl << &a << " " << b;
}
The output looks like:
10 20
0x7ffeeb1d396c 0x7ffeeb1d396c
The a and *b are at the same address, why do they have different value?
Most probably this is caused by optimization.
As molbdnilo already said in his comment: "Compilers trust you blindly. If you lie to them, strange things can happen."
So when optimization is enabled, the compiler finds the declaration
const int a = 10;
and thinks "ah, this will never change, so we don't need a "real" variable and can replace all appearances of a in the code simply with 10". This behavior is called constant folding.
Now in the next line you "cheat" the compiler:
auto *b = const_cast<int *>(&a);
*b = 20;
and change a, although you have promised not to do so.
The result is confusion.
Like a lot of people have already mentioned and Christian Hackl has thoroughly analyzed in his excellent in-depth answer, it's undefined behavior. The compiler is generally allowed to apply constant folding on constants, which are explicitly declared const.
Your question is a very good example (I don't know how anyone can vote it down!) why const_cast is very dangerous (and even more when combined with raw pointers), should be only used if absolutely necessary, and should be at least thoroughly commented why it was used if it was unavoidable. Comments are important in that case because it's not only the compiler you are "cheating":
Also your co-workers will rely on the information const and rely on the information that it won't change. If it DOES change though, and you didn't inform them, and did not comment, and did not exactly know what you were doing, you'll have a bad day at work :)
Try it out: Perhaps your program will even behave different in debug build (not optimized) and release build (optimized), and those bugs are usually pretty annoying to fix.
I think it helps to view const like this:
If you declare something as const it is actually not the compiler that ensures that you dont change it, but rather the opposite: you make a promise to the compiler that you wont change it.
The language helps you a bit to keep your promise, eg you cannot do:
const int a = 5;
a = 6; // error
but as you discovered, you indeed can attempt to modify something that you declared const. However, then you broke your promise and as always in c++, once you break the rules you are on your own (aka undefined behaviour).
The a and *b are at the same address, why do they have different
value?
They don't, not according to C++. Neither do they, according to C++.
How can that be? The answer is that the C++ language does not define the behaviour of your program because you modify a const object. That's not allowed, and there are no rules for what will happen in such a case.
You are only allowed to use const_cast in order to modify something which wasn't const in the first place:
int a = 123;
int const* ptr1 = &a;
// *ptr1 = 456; // compilation error
int* ptr2 = const_cast<int*>(ptr1);
*ptr2 = 456; // OK
So you've ended up with a program whose behaviour is not defined by C++. The behaviour is instead defined by compiler writers, compiler settings and pure chance.
That's also what's wrong with your question. It's impossible to give you a correct answer without knowing exactly how you compiled this and on which system it runs, and perhaps even then there are random factors involved.
You can inspect your binary to find out what the compiler thought it was doing; there are even online tools like https://gcc.godbolt.org/ for that. But try to use printf instead of std::cout for such an analysis; it will produce easier-to-read assembly code. You may also use easier-to-spot numbers, like 123 and 456.
For example, assuming that you even use GCC, compare these two programs; the only difference is the const modifier for a:
#include <stdio.h>
int main() {
int const a = 123;
auto *b = const_cast<int *>(&a);
*b = 456;
printf("%d", a);
printf("%d", *b);
}
#include <stdio.h>
int main() {
int a = 123;
auto *b = const_cast<int *>(&a);
*b = 456;
printf("%d", a);
printf("%d", *b);
}
Look at what the printf("%d", a); call becomes. In the const version, it becomes:
mov esi, 123
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
In the non-const version:
mov eax, DWORD PTR [rbp-12]
mov esi, eax
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
Without going too much into the details of assembly code, one can see that the compiler optimised the const variable such that the value 123 is pushed directly to printf. It doesn't really treat a as a variable for the printf call, but it does do so for the pointer initialisation.
Such chaos is the nature of undefined behaviour. The moral of the story is that you should always avoid undefined behaviour, and thinking twice before casting is an important part of allowing the compiler to help you in that endeavor.
Compiling your code once with const and once without const using gcc, shows that there either no space is allocated for a, or it's content is being ignored in const version. (even with 0 optimization) The only difference in their assembly result is the line which you print the a. Compiler simply prints 10 instead of refering to memory and fetching contents of a: (left side is const int a and right side is int a version)
movl $10, %esi | movl -36(%rbp), %eax
> movl %eax, %esi
As you can see there is an extra memory access for non-const version, which we know is expensive in terms of time. So the compiler decided to trust your code, and concluded that a const is never to be changed, and replaced its value wherever the variable is referenced.
Compiler options: --std=c++11 -O0 -S
How are pointers made to increment by their type.
For example if we have
int *ptr;
ptr++; //would point to the next integer i.e. it would increment ptr by 4bytes in 32 bit system
I wanted to know that how is this done internally.
The compiler compiling the code knows the base type of the pointer, and it puts the code to increment pointer (offset) appropriately. For example:
int* p = new int[10];
*(p+2) = 100;
The second line will be like:
p + sizeof(int) * 2 ... // Not C/C++
And, similary:
p++;
Would mean:
p = p + sizeof(int); // Not C/C++
If type of p is something else (like float or some structure), the calculations would be performed appropriately. There is no magic. Types are compile time defined - a type of variable don't change at runtime, hence the calculations.
If I declare
int x = 5 ;
int* p = &x;
unsigned int y = 10 ;
cout << p+y ;
Is this a valid thing to do in C++, and if not, why?
It has no practical use, but is it possible?
The math is valid; the resulting pointer isn't.
When you say ptr + i (where ptr is an int*), that evaluates to the address of an int that's i * sizeof(int) bytes past ptr. In this case, since your pointer points to a single int rather than an array of them, you have no idea (and C++ doesn't say) what's at p+10.
If, however, you had something like
int ii[20] = { 0 };
int *p = ii;
unsigned int y = 10;
cout << p + y;
Then you'd have a pointer you could actually use, because it still points to some location within the array it originally pointed into.
What you are doing in your code snippet is not converting unsigned int to pointer. Instead you are incrementing a pointer by an integer offset, which is a perfectly valid thing to do. When you access the index of an array, you basically take the pointer to the first element and increase it by the integer index value. The result of this operation is another pointer.
If p is a pointer/array, the following two lines are equivalent and valid (supposing the pointed-to-array is large enough)
p[5] = 1;
*(p + 5) = 1;
To convert unsigned int to pointer, you must use a cast
unsigned int i = 5;
char *p = reinterpret_cast<char *>(i);
However this is dangerous. How do you know 5 is a valid address?
A pointer is represented in memory as an unsigned integer type, the address. You CAN store a pointer in an integer. However you must be careful that the integer data type is large enough to hold all the bits in a pointer. If unsigned int is 32-bits and pointers are 64-bits, some of the address information will be lost.
C++11 introduces a new type uintptr_t which is guaranteed to be big enough to hold a pointer. Thus it is safe to cast a pointer to uintptr_t and back again.
It is very rare (should be never in run-of-the-mill programming) that you need to store pointers in integers.
However, modifying pointers by integer offsets is totally valid and common.
Is this a valid thing to do in c++, and if not why?
Yes. cout << p+y; is valid as you can see trying to compile it. Actually p+y is so valid that *(p+y) can be translated to p[y] which is used in C-style arrays (not that I'm suggesting its use in C++).
Valid doesn't mean it actually make sense or that the resulting pointer is valid. Since p points to an int the resulting pointer will be an offset of sizeof(int) * 10 from the location of x. And you are not certain about what's in there.
A variable of type int is a variable capable of containing an integer value. A variable of type int* is a pointer to a variable copable of containing an integer value.
Every pointer type has the same size and contains the same stuff: A memory address, which the size is 4 bytes for 32-bit arquitectures and 8 bytes for 64-bit arquitectures. What distinguish them is the type of the variable they are poiting to.
Pointers are useful to address buffers and structures allocated dynamically at run time or any sort of variable that is to be used but is stored somewhere else and you have to tell where.
Arithmetic operations with pointers are possible, but they won't do what you think. For instance, summing + 1 to a pointer of type int will increase its value by sizeof(int), not by literally 1, because its a pointer, and the logic here is that you want the next object of this array.
For instance:
int a[] = { 10, 20, 30, 40 };
int *b = a;
printf("%d\n", *b);
b = b + 1;
printf("%d\n", *b);
It will output:
10
20
Because b is pointing to the integer value 10, and when you sum 1 to it, or any variable containing an integer, its then poiting to the next value, 20.
If you want to perform operations with the variable stored at b, you can use:
*b = *b + 3;
Now b is the same pointer, the address has not changed. But the array 10, 20, 30, 40 now contains the values 13, 20, 30, 40, because you increased the element b was poiting to by 3.
This question already has answers here:
Weird Behaviour with const_cast [duplicate]
(2 answers)
Closed 6 years ago.
I am trying to change the value of a variable which is defined as int const as below.
const int w = 10;
int* wp = const_cast <int*> (&w);
*wp = 20;
The value of w didn't change and was 10 even after the assignment, though it shows as if both w and wp are pointing to the same memory location. But I am able to the change the value of w, if defined as below while declaring
int i = 10;
const int w = i;
If I change the declaration of i to make it const like in
const int i = 10;
The value of w doesn't change.
In the first case, how come the value of w didn't change, even though w and wp point to the same memory location [ that was my impression I get when I print their addresses ]
What difference it's to the compiler that it treats both the cases differently?
Is there a way to make sure that w doesn't lose constness, irrespective of the way it is defined?
This is one of the cases where a const cast is undefined, since the code was probably optimized such that w isn't really a variable and does not really exist in the compiled code.
Try the following:
const volatile int w = 10;
int &wr = const_cast <int &> (w);
wr = 20;
std::cout << w << std::endl;
Anyhow, I would not advise abusing const_cast like that.
The code in the above example translates into the following assembler:
movl $10, 28(%esp) //const int i = 10;
leal 28(%esp), %eax //int* wp = const_cast <int*>(&i);
movl %eax, 24(%esp) //store the pointer on the stack
movl 24(%esp), %eax //place the value of wp in eax
movl $20, (%eax) //*wp = 20; - so all good until here
movl $10, 4(%esp) //place constant value 10 onto the the stack for use in printf
movl $.LC0, (%esp) // load string
call printf //call printf
Because the original int i was declared constant, the compiler reserves the right to use the literal value instead of the value stored on the stack. This means that the value does not get changed and you are stuck with the original 10.
The moral of the story is compile time constants should remain constant because that is what you are telling the compiler. The moral of the story is that casting away constness in order to change a constant can lead to bad things.
const_cast doesn't take away the const-ness of a variable as defined. If you were to pass a non-const variable by reference in to a method taking a const reference like void foo(const int& x) then you could use const_cast to modify the value of x within foo, but only if the variable you actually passed in was not const in the first place.
Why can't you just re-bind the constant? So instead of
const int w = 10;
int* wp = const_cast <int*> (&w);
*wp = 20;
// some code
just introduce the different constant with the same name
const int w = 10;
{
const int w = 20;
// the same code
}
If the "new" constant should depend on its own value, you should introduce another constant (const int _w = w; const int w = _w * 2;). The unnecessary assignments will be optimized out by compiler--because we've seen it has done such optimisation, as it's the reason why you asked your question.
You should not being changing the const value. There is a reason it is const and trying to change it will most likely just result in errors. If the const is stored in a read only memory section then you will get access violations.
Good question. I think the confusion comes from the fact that C++ uses the keyword ‘const’ for two different concepts depending on the context. These concepts are constant and read-only variables.
When a value of a ‘const’ variable can be calculated during the compilation, it creates a true constant. References to such constant are replaced with its value whenever it is used. That’s why there is no location in the memory that can be changed to affect all places where it is used. It is like using #define.
When value of a ‘const’ variable cannot be calculated during the compilation, it creates a read-only variable. It has a location in the memory that contains a value but compiler enforces a read-only behavior.
Here's a refresher, it should be noted that this is in C. This is a deceptively tricky underpinnings of the usage of a variable or pointer using the const keyword. This highlights the difference between the pointer variable foo and how the meaning of it can change by using the said keyword.
char const *foo;
char * const foo;
const char *foo;
The first and last declarations, makes the data pointed to by ‘foo’ read-only, but, you can change the address pointed to by ‘foo’ e.g.
const *char foo; /* OR char const *foo */
char str[] = "Hello";
foo = &str[0]; /* OK! */
foo[1] = 'h'; /* BZZZZTTT! Compile Fails! */
The middle declaration in the above, makes the pointer read-only, i.e. you cannot change the address of the data pointed to by ‘foo’
char * const foo;
char str[] = "Hello";
foo = &str[0]; /* BZZZZTTT! Compile Fails! */
My guess would be that declaring w const allows the compiler to perform more aggressive optimizations such as inlining w's value and reordering instructions. Wether w appears to change or not depends on which optimizations were applied in the precise case and is not under your control.
You can't force w to be totally const. The cons_cast should be a hint to the programmer that they might be doing something fishy.