I am trying to use this example:
std::size_t s2 = boost::asio::buffer_size(buffer);
const void* p2 = boost::asio::buffer_cast<const void*>(buffer);
And I am getting a vaild size s2 and some seemingly valid address p2.
Now, how could I create a cout or printf loop or phrase a debug-statement, to see the content of p2?
I bet this is quite basic, but currently I can't see what I am missing.
Tried this:
std::cout << "TEST: " << boost::asio::buffer_cast<const void*>(buffer) << std::endl;
but it only prints out the address, not the content
and this:
for(int i =0; i!=s2; i++){
std::cout << "TEST: " << p2[i];
}
std::cout << std::endl;
but I am ending up with compile errors, like C0253 - unknown size.
So, how can I print out the content of p2?
You can print the bytes pointed to by buffer as:
std::size_t s2 = boost::asio::buffer_size(buffer);
const void* p2 = boost::asio::buffer_cast<const void*>(buffer);
unsigned char const* bytes = static_cast<unsigned char const*>(p2);
for(std::size_t i = 0 ; i < s2 ; ++i)
std::cout << static_cast<unsigned int>(bytes[i]);
Of course, you have to interpret the output.
You can print the hexadecimal values instead which is easier to interpret:
std::cout << std::hex << static_cast<unsigned int>(bytes[i]);
// ^^^^^^^^ note this
I think you have to #include<iomanip> for this.
The content of p2 is a memory address...since p2 is a pointer.
A void* means that the language + libraries are not aware of the type of data being referenced.
You can't print the value being pointed to if you don't know of what type it is.
Related
I want to copy a struct content in memory via char* pc the print it back but here I have an exception (reading violation)
struct af {
bool a;
uint8_t b;
uint16_t c;
};
int main() {
af t;
t.a = true;
t.b = 3;
t.c = 20;
char* pc = nullptr;
memcpy(&pc, &t, sizeof(t));
std::cout << "msg is " << pc << std::endl; // here the exception
return 0;
}
then I want to recover data from memory to another structure of same type.
I did af* tt = (af*)(pc); then tried to access to tt->a but always an exception.
You need to allocate memory before you can copy something into it. Also, pc is already the pointer, you need not take the address of it again. Moreover, the byte representation is very likely to contain non-printable characters. To see the actual effect the following copies from the buffer back to an af and prints its members (note that a cast is needed to prevent std::cout to interpret the uint8_t as a character):
#include <iostream>
#include <cstring>
struct af {
bool a;
uint8_t b;
uint16_t c;
};
int main() {
af t;
t.a = true;
t.b = 3;
t.c = 20;
char pc[sizeof(af)];
std::memcpy(pc, &t, sizeof(t)); // array pc decays to pointer to first element
for (int i=0;i<sizeof(af); ++i){
std::cout << i << " " << pc[i] << "\n";
}
af t2;
std::memcpy(&t2, pc,sizeof(t));
std::cout << t2.a << " " << static_cast<unsigned>(t2.b) << " " << t2.c;
}
Output:
0
1
2
3
1 3 20
Note that I replaced the output of pc with a loop that prints individual characters, because the binary representation might contain null terminators and pc is not a null terminated string. If you want it to be a null-terminated string, it must be of size sizeof(af) +1 and have a terminating '\0'.
After a pointer is initialized, do you have to use the * dereference operator to call the pointer in a condition?
Example:
int main()
{
int var = 10;
int *ptr = &var;
if(ptr) // does this need to be if(*ptr) ???
{.......}
}
And can I have a short explanation as to why?
Thank you.
if (ptr)
check if the pointer is not Null but
if (*ptr)
check if the value it points to is not zero (in this example is 10)
So for checking the value you shoud add *.
It depends on what you want to do.
if(ptr) checks if the pointer value is nullptr or not. Note that this is shorthand for if(ptr != nullptr).
if(*ptr) checks if what the pointer points to is nullptr (or 0) - and in that case, since you dereference (follow) the pointer to answer the question, the pointer itself had better not be nullptr in that case.
First of all, a pointer is only a variable. However, there are different contexts in which you can use it.
As any other variable you can access the pointers content (which is the adress of the underlying memory) as follows:
int i = 1;
int * p = &i;
std::cout << p << std::endl
this would output the adress of i since this is what is stored in p
If you however want to access the content of the underlying memory (the value of i), you need to dereference the pointer first by using the * operator:
std::cout << *p << std::endl;
This would print the value of iso 1.
of course you can also access the pointer's adress (since the adress of i is a numeric value as well and needs to be stored somewhere too):
std::cout << &p << std::endl;
That would output the adress of p so the adress where the adress of i is stored :)
As a little example try to run this code:
#include <iostream>
int main() {
int i = 1;
int * p = &i;
std::cout << "Value of i(i): " << i << std::endl
<< "Adress of i(&i): " << &i << std::endl
<< "Value of p(p): " << p << std::endl
<< "Dereferenced p(*p): " << *p << std::endl
<< "Adress of p(&p): " << &p << std::endl
<< "Dereferenced adress of p(*(&p)): " << *(&p) << std::endl;
}
I have a vector of unsigned char where I copy bytes in C++. I convert all primitive types to bytes and copy to this vector of char (which is interpreted as bytes in C++). Now I am copying also strings. But I am not sure if I am converting strings to bytes. If you take a look at my output when I am printing the vector of unsigned char I am printing bytes from double int float but I am printing the real string of my variable testString. So I suppose that I am not inserting bytes of this testString on my vector of unsigned char. How should I do that?
Thanks
const std::string lat = "lat->", alt = "alt->", lon = "lon->", testString = "TEST-STRING";
double latitude = 10.123456;
double longitude = 50.123456;
double altitude = 1.123456;
std::vector<unsigned char> result(
sizeof(latitude) + sizeof(longitude) + sizeof(altitude) + testString.length());
std::cout << "copying to the vector" << std::endl;
memcpy(result.data(), &longitude, sizeof(longitude));
memcpy(result.data() + sizeof(longitude), &latitude, sizeof(latitude));
memcpy(result.data() + sizeof(longitude) + sizeof(latitude), &altitude, sizeof(altitude));
memcpy(result.data() + sizeof(longitude) + sizeof(latitude) + sizeof(altitude), testString.c_str(),
testString.length() + 1);
std::cout << "copied to the vector\n" << std::endl;
std::cout << "printing the vector" << std::endl;
for (unsigned int j = 0; j < result.size(); j++) {
std::cout << result[j];
}
std::cout << std::endl;
std::cout << "printed the vector\n" << std::endl;
// testing converting back ...................
std::cout << "printing back the original value" << std::endl;
double dLat, dLon, dAlt;
std::string value;
memcpy(&dLon, result.data(), sizeof(longitude));
memcpy(&dLat, result.data() + sizeof(longitude), sizeof(latitude));
memcpy(&dAlt, result.data() + sizeof(longitude) + sizeof(latitude), sizeof(altitude));
value.resize(testString.length());
memcpy(&value[0], result.data() + sizeof(longitude) + sizeof(latitude) + sizeof(altitude),
sizeof(value.data()) + testString.size());
std::cout << alt << dAlt;
std::cout << lat << dLat;
std::cout << lon << dLon;
std::cout << " " << value << std::endl;
std::cout << "printed back the original value\n" << std::endl;
output:
copying to the vector
copied to the vector
printing the vector
[?�gI#m���5?$#l������?TEST-STRING
printed the vector
printing back the original value
alt->1.12346lat->10.1235lon->50.1235 TEST-STRING
printed back the original value
There's no problem with your code! You're printing the actual bytes of your variables. The bytes in a double can't really be interpreted as a text string (at least, it doesn't make sense if you do) but the bytes in a text string can, producing what you see.
Let's say you've got the following code (which is really just disguised C):
#include <cstdio>
int main(int argc, char *argv[]) {
struct {
double latitude;
double longitude;
char name[30];
} structure = {
53.6344,
126.5223167,
"Keyboard Mash"
};
printf("%f %f %s\n", structure.latitude, structure.longitude, structure.name);
for (size_t i = 0; i < sizeof(structure); i += 1) {
printf("%c", ((char*)&structure)[i]);
}
printf("\n");
}
This code would (probably) print:
53.6344 126.5223167 Keyboard Mash
����������������Keyboard Mash�����������������
The first 16 bytes are from the doubles, and the next 30 are from the char[]. That's just how char[]s are stored! Your code is doing what you'd expect it to.
Of course, you can't rely on it doing this in exactly this way; that's undefined behaviour.
I feel like you were expecting something like: 128565TESTSTRING where 12, 85 and 65 are values of longitude, latitude and altitude. Well, that's not going to happen be cause you wrote 12 in the data, not "12"; therefore, it will return you the character whose ASCII code is 12. Maybe you could use something like sprintf() instead.
I want to compare the memory address and pointer value of p, p + 1, q , and q + 1.
I want to understand, what the following values actually mean. I can't quite wrap my head around whats going on.
When I run the code:
I get an answer of 00EFF680 for everytime I compare the adresss p with another pointer.
I get an answer of 00EFF670 for everytime I compare the address of q with another pointer.
I get an answer of 15726208 when I look at the pointer value of p.
And I get an answer of 15726212 When I look at the pointer value of p + 1.
I get an answer of 15726192 when I look at the pointer value of q
And I get an answer of 15726200 Wehn I look at the pointer value of q + 1.
Code
#include <iostream>
#include <string>
using namespace std;
int main()
{
int val = 20;
double valD = 20;
int *p = &val;
double *q;
q = &valD;
cout << "Memory Address" << endl;
cout << p == p + 1;
cout << endl;
cout << q == q + 1;
cout << endl;
cout << p == q;
cout << endl;
cout << q == p;
cout << endl;
cout << p == q + 1;
cout << endl;
cout << q == p + 1;
cout << endl;
cout << "Now Compare Pointer Value" << endl;
cout << (unsigned long)(p) << endl;
cout << (unsigned long) (p + 1) << endl;
cout << (unsigned long)(q) << endl;
cout << (unsigned long) (q + 1) << endl;
cout <<"--------" << endl;
return 0;
}
There are a few warnings and/or errors.
The first is that overloaded operator << has higher precedence than the comparison operator (on clang++ -Woverloaded-shift-op-parentheses is the flag).
The second is that there is a comparison of distinct pointer types ('int *' and 'double *').
For the former, parentheses must be placed around the comparison to allow for the comparison to take precedence. For the latter, the pointers should be cast to a type that allows for safe comparison (e.g., size_t).
For instance on line 20, the following would work nicely.
cout << ((size_t) p == (size_t) (q + 1));
As for lines 25-28, this is standard pointer arithmetic. See the explanation here.
As to your question:
I want to compare p, p +1 , q , and q + 1. And Understand what the results mean.
If p is at address 0x80000000 then p+1 is at address 0x80000000 + sizeof(*p). If *p is int then this is 0x80000000 + 0x8 = 0x80000008. And the same reasoning applies for q.
So if you do p == p + 1 then compiler will first do the additon: p+1 then comparison, so you will have 0x80000000 == 0x80000008 which results in false.
Now to your code:
cout << p == p + 1;
is actually equivalent to:
(cout << p) == p + 1;
and that is because << has higher precedence than ==. Actually you should get a compilation error for this.
Another thing is comparision of pointers of non related types like double* with int*, without cast it should not compile.
In C and C++ pointer arithmetic is very closely tied with array manipulation. The goal is that
int array[3] = { 1, 10, 100 };
int *ptr = { 1, 10, 100 };
std::cout << array[2] << '\n';
std::cout << *(ptr + 2) << '\n';
outputs two 100s. This allows the language to treat arrays and pointers as equivalent - that's not the same thing as "the same" or "equal", see the C FAQ for clarification.
This means that the language allows:
int array[3] = { 1, 10, 100 };
int *ptr = { 1, 10, 100 };
And then
std::cout << (void*)array << ", " << (void*)&array[0] << '\n';
outputs the address of the first element twice, the first array behaves like a pointer.
std::cout << (void*)(array + 1) << ", " << (void*)&array[1] << '\n';
prints the address of the second element of array, again array behaving like a pointer in the first case.
std::cout << ptr[2] << ", " << *(ptr + 2) << '\n';
prints element #3 of ptr (100) twice, here ptr is behaving like an array in the first use,
std::cout << (void*)ptr << ", " << (void*)&ptr[0] << '\n';
prints the value of ptr twice, again ptr behaving like an array in the second use,
But this can catch people unaware.
const char* h = "hello"; // h points to the character 'h'.
std::cout << (void*)h << ", " << (void*)(h+1);
This prints the value of h and then a value one higher. But this is purely because the type of h is a pointer to a one-byte-sized data type.
h + 1;
is
h + (sizeof(*h)*1);
If we write:
const char* hp = "hello";
short int* sip = { 1 };
int* ip = { 1 };
std::cout << (void*)hp << ", " << (void*)(hp + 1) << "\n";
std::cout << (void*)sip << ", " << (void*)(sip + 1) << "\n";
std::cout << (void*)ip << ", " << (void*)(ip + 1) << "\n";
The first line of output will show two values 1 byte (sizeof char) apart, the second two values will be 2 bytes (sizeof short int) apart and the last will be four bytes (sizeof int) apart.
The << operator invokes
template<typename T>
std::ostream& operator << (std::ostream& stream, const T& instance);
The operator itself has very high precedence, higher than == so what you are actually writing is:
(std::cout << p) == p + 1
what you need to write is
std::cout << (p == p + 1)
this is going to print 0 (the result of int(false)) if the values are different and 1 (the result of int(true)) if the values are the same.
Perhaps a picture will help (For a 64bit machine)
p is a 64bit pointer to a 32bit (4byte) int. The green pointer p takes up 8 bytes. The data pointed to by p, the yellow int val takes up 4 bytes. Adding 1 to p goes to the address just after the 4th byte of val.
Similar for pointer q, which points to a 64bit (8byte) double. Adding 1 to q goes to the address just after the 8th byte of valD.
If you want to print the value of a pointer, you can cast it to void *, for example:
cout << static_cast<void*>(p) << endl;
A void* is a pointer of indefinite type. C code uses it often to point to arbitrary data whose type isn’t known at compile time; C++ normally uses a class hierarchy for that. Here, though, it means: treat this pointer as nothing but a memory location.
Adding an integer to a pointer gets you another pointer, so you want to use the same technique there:
cout << static_cast<void*>(p+1) << endl;
However, the difference between two pointers is a signed whole number (the precise type, if you ever need it, is defined as ptrdiff_t in <cstddef>, but fortunately you don’t need to worry about that with cout), so you just want to use that directly:
cout << (p+1) - p << endl;
cout << reinterpret_cast<char*>(p+1) - reinterpret_cast<char*>(p) << endl;
cout << (q - p) << endl;
That second line casts to char* because the size of a char is always 1. That’s a big hint what’s going on.
As for what’s going on under the hood: compare the numbers you get to sizeof(*p) and sizeof(*q), which are the sizes of the objects p and q point to.
The pointer values that are printed are likely to change on every execution (see why the addresses of local variables can be different every time and Address Space Layout Randomization)
I get an answer of 00EFF680 for everytime I compare the adresss p with another pointer.
int val = 20;
double valD = 20;
int *p = &val;
cout << p == p + 1;
It is translated into (cout << p) == p + 1; due to the higher precedence of operator << on operator ==.
It print the hexadecimal value of &val, first address on the stack frame of the main function.
Note that in the stack, address are decreasing (see why does the stack address grow towards decreasing memory addresses).
I get an answer of 00EFF670 for everytime I compare the address of q with another pointer.
double *q = &valD;
cout << q == q + 1;
It is translated into (cout << q) == q + 1; due to the precedence of operator << on operator ==.
It prints the hexadecimal value of &valD, second address on the stack frame of the main function.
Note that &valD <= &val - sizeof(decltype(valD) = double) == &val - 8 since val is just after valD on the stack. It is a compiler choice that respects some alignment constraints.
I get an answer of 15726208 when I look at the pointer value of p.
cout << (unsigned long)(p) << endl;
It just prints the decimal value of &val
And I get an answer of 15726212 When I look at the pointer value of p + 1.
int *p = &val;
cout << (unsigned long) (p + 1) << endl;
It prints the decimal value of &val + sizeof(*decltype(p)) = &val + sizeof(int) = &val + 4 since on your machine int = 32 bits
Note that if p is a pointer to type t, p+1 is p + sizeof(t) to avoid memory overlapping in array indexing.
Note that if p is a pointer to void, p+1 should be undefined (see void pointer arithmetic)
I get an answer of 15726192 when I look at the pointer value of q
cout << (unsigned long)(q) << endl;
It prints the decimal value of &valD
And I get an answer of 15726200 Wehn I look at the pointer value of q + 1.
cout << (unsigned long) (q + 1) << endl;
It prints the decimal value of &val + sizeof(*decltype(p)) = &valD + sizeof(double) = &valD + 8
in a function, that gets unsigned char && unsigned char length,
void pcap_callback(u_char *args, const struct pcap_pkthdr* pkthdr, const u_char* packet)
{
std::vector<unsigned char> vec(packet, packet+pkthdr->len); // optimized from foo.
std::stringstream scp;
for (int i=0;i<pkthdr->len;i++) {
scp<<vec[i];
}
std::string mystr = std::string(scp.rdbuf()->str());
std::cout << "WAS: " << packet << std::endl;
std::cout << "GOOD: " << scp.str() << std::endl;
std::cout << "BAD: " << scp.str().c_str() << std::endl;
std::cout << "TEST: " << mystr.size() << std::endl;
assert(mystr.size() == pkthdr->len);
}
Results:
WAS: prints nothing (guess there is a pointer to const.. case)
GOOD: prints data
BAD: prints nothing
TEST, assert: prints that mystr.size() is equal to passed unsigned char size.
I tried:
string.assign(scp.rdbuf());
memcpy(char, scp.str(), 10);
different methods of creating/allocating temporary chars, strings
No help.. it is wanted to get a std::cout'able std::string that contains data, (which was picked from foo, which was unsigned char, which was packet data).
Guessing either the original foo may not be null-terminated, or the problem is something like this - simple, but can't get in.. what are the things to look for here?
(this code is another attempt to use libpcap, just to print packets in C++ way, without using known C++ magic wrappers like libpcapp).
For a quick test, throw in a check for scp.str().size() == strlen(scp.str().c_str()) to see if there are embedded '\0' characters in the string, which is what I suspect is happening.
I think you're going about this the wrong way. It looks like you're dealing with binary data here, in which case you can't expect to meaningfully output it to the screen as text. What you really need is a hex dump.
const unsigned char* ucopy = packet;
std::ios_base::fmtflags old_flags = std::cout.flags();
std::cout.setf(std::ios::hex, std::ios::basefield);
for (const unsigned char* p = ucopy, *e = p + pkthdr->len; p != e; ++p) {
std::cout << std::setw(2) << std::setfill('0') << static_cast<unsigned>(*p) << " ";
}
std::cout.flags(old_flags);
This will output the data byte-by-byte, and let you examine the individual hex values of the binary data. A null byte will simply be output as 00.
Check std::cout.good() after the failed output attempt. My guess is that there's some failure on output (i.e. trying to write a nonprintable character to the console), which is setting failbit on cout.
Also check to ensure the string does not start with a NULL, which would cause empty output to be the expected behavior :)
(Side note, please use reinterpret_cast for unsigned char *ucopy = (unsigned char*)packet; if you're in C++ ;) )