Comparing ip addresses in bpftrace? - networking

I am writing some bpftrace code in which I would like to compare an IP address (stored as a 32 bit integer) against the string representation of an address. That is, I want to do something like this:
kprobe:netif_receive_skb {
$skb = (struct sk_buff *)arg0;
$dev = $skb->dev;
$name = $dev->name;
$ipheader = ((struct iphdr *) ($skb->head + $skb->network_header));
// This pseudocode conditional is what I am trying to figure out
if ($ipheader->daddr == "8.8.8.8") {
printf("%s %s %s -> %s\n",
func, $name, ntop($ipheader->saddr), ntop($ipheader->daddr))
}
}
In the above, $ipheader is a struct iphdr *, and struct iphdr looks like this:
struct iphdr {
...
__struct_group(/* no tag */, addrs, /* no attrs */,
__be32 saddr;
__be32 daddr;
);
};
So both saddr and daddr are unsigned 32-bit integers.
I thought perhaps I could use the pton() function, like this:
if ($ipheader->daddr == pton("8.8.8.8")) {
But unfortunately pton() returns an int8[4] array instead of a single 32 bit integer, so this fails with:
traceit.bpf:8:5-41: ERROR: Type mismatch for '==': comparing 'unsigned int32' with 'unsigned int8[4]'
if ($ipheader->daddr == pton("8.8.8.8")) {
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I next tried to see if I could normalize both values into int8[4] arrays, like this:
$target = pton("8.8.8.8");
$daddr = pton(ntop($ipheader->daddr));
if ($target == $daddr) {
But this fails because ntop() doesn't actually return a string, so we get:
ERROR: Expected string literal, got inet
traceit.bpf:9:10-38: ERROR: pton() expects an string argument of an IPv4/IPv6 address, got
$daddr = pton(ntop($ipheader->daddr));
I realize I could just specify the target as an integer myself:
if ($ipheader->daddr == 0x08080808) {
But while that's easy to interpret for a contrived address like 8.8.8.8, it's opaque when the hex form of the address is something like 0xacd9004e.
Is it possible to compare a 32-bit integer representation of an address with the string representation of an address using bpftrace (and if so, what's the canonical way of doing so)?

Related

char array variable name want to use in variable in Arduino IDE

i created char array
char Jan1[] = "1,2,3,4";
char Jan2[] = "5,7,3,4";
char Jan3[] = "10,9,3,4";`
the above char arrays i want to use it as shown below in for loop each time it will iterate and print, print is an example i am processing this string further in code, but it is giving error. if instead of yy i use Jan1 then it is printing properly. what is the other way i can use yy as to get print char array as string.
for(int i=1;i<=2;i++)
{
char yy[4];
sprintf(yy,"Jan%d",i);
String presentMonth = String(yy);
Serial.print(presntMonth);
}
So by the comments it seems that what you want is to print the contents of the JanX variable based on an index variable.
Missunderstanding
First of all you need to understand the difference between strings as datatypes and variable names. The former are ways to represent a sequence of characters (used mainly, but not only, to display output messages to any sort of output like a file, a console etc..). The latter are names that you use while coding and they can be whatever you want but they will never appear in the final program.
This for instance creates a String (datatype) named hello which contains the sequence of characters h e l l o.
String hello = "Hello";
Nothing prevents me to assign goodbye to it:
hello = "Goodbye"
Serial.print(hello); // This will print "Goodbye"
In general (apart from very hacky ways) you can't retrieve the name of the variable from your program and have them ready in your executable.
Issue
char Jan1[] = {'1','2','3'};
String yy = "Jan1"
Serial.print(yy); // This will print "Jan1"
To print the items in Jan1 you need to iterate through the values.
void printItems(char* s,int N){
for ( i = 0; i<N; i++ )
{ Serial.print(s[i]); }
}
Since Arduino provides the String class, however, it would be better to do this:
String Jan1 = "1,2,3,4";
Serial.print(Jan1); // This iterates under the hood, takes care of the length, and all the good stuff.
Solution
You want to do something a little more advanced, you want to point to a particular string based on a variable, then print the content of the retrieved string.
I can think of two ways for doing so: by using an list of strings or via a hashmap.
List of strings
String list[] = {"1,2,3,4" , "4,5,6,7", ... };
for(int i = 0; i < sizeof(list)/sizeof(String) ; i++ ){
Serial.print(list[i])
}
Hashmap
The reason I am thinking about this is because you want a string as the "index" that let's you lookup the string, so you can "search" by name. The easiest and quickest method I can think of is to declare an array of structs string_name;string_content and use strcmp to iterate through the array of structs until the needed one is found.
typedef struct{ String name; String content;} element_t;
element_t dict[] = { {"Jan1","1,2,3,4"} , {"Jan2","2,3,4,5"} ... }'
// Note this is not even close to perfect (for instance lacks check if key does not exists)
String lookup(element_t DICT, int DICT_SIZE, String key){
// Iterate through the elements, use strcmp to retrieve it
for(int i = 0; i < DICT_SIZE ; i++ ){
if(strcmp(DICT[i].name,key) {
return DICT[i].content;
}
}
}
// Now create the key, as in your code, and then lookup.
for(int i=1;i<=2;i++)
{
char yy[4];
sprintf(yy,"Jan%d",i);
String presentMonth = lookup(dict,dict_size,yy);
Serial.print(presntMonth);
}

Different result between Serial.print and Serial.printf in ESP32

I want to print my string variable using printf method:
id = 6415F1BF713C
Serial.printf("id: %s\n\n", id);
Serial.print(id);
The result that I got was:
id: ⸮⸮⸮?
6415F1BF713C
is there any thing that's wrong?
Thanks.
Update :
//get device id
String getDeviceID() {
uint64_t chipid = ESP.getEfuseMac(); // The chip ID is essentially its MAC address(length: 6 bytes).
uint16_t chip = (uint16_t)(chipid >> 32);
char devID[40];
snprintf(devID, 40, "%04X%08X", chip, (uint32_t)chipid);
return devID;
}
String id = getDeviceID();
Serial.printf("id: %s\n\n", id);
Serial.print(id);
You didn't offer enough code to properly debug this, but I'm guessing what you mean is
String id = "6415F1BF713C";
Serial.printf("id: %s\n\n", id);
Serial.print(id);
The %s format in the printf() method expects to take a C/C++ char *, not a String. When you passed it a String, it printed the memory address of the String object - four characters which would appear as garbage, as you saw.
C and C++ use char * (pointers to characters) and char [] (arrays of characters) to represent strings. These are different from the Arduino String class, and people often confuse them.
To use the printf() method with a String you need to get its C string pointer, like this:
Serial.printf("id: %s\n\n", id.c_str());
The reason that this:
Serial.print(id);
works is that the print() method has a form that specifically expects to take a String object as an argument.

Find address of constant in go

We have written one program by which we try to find an address of a constant. Is it possible to do it like that?
package main
func main() {
const k = 5
address := &k
}
It gives an error, can anyone tell how can we find the address of a constant?
In short: you can't.
The error message says:
cannot take the address of k
There are limitations on the operand of the address operator &. Spec: Address operators:
For an operand x of type T, the address operation &x generates a pointer of type *T to x. The operand must be addressable, that is, either a variable, pointer indirection, or slice indexing operation; or a field selector of an addressable struct operand; or an array indexing operation of an addressable array. As an exception to the addressability requirement, x may also be a (possibly parenthesized) composite literal. If the evaluation of x would cause a run-time panic, then the evaluation of &x does too.
Constants are not listed as addressable, and things that are not listed in the spec as addressable (quoted above) cannot be the operand of the address operator & (you can't take the address of them).
It is not allowed to take the address of a constant. This is for 2 reasons:
A constant may not have an address at all.
And even if a constant value is stored in memory at runtime, this is to help the runtime to keep constants that: constant. If you could take the address of a constant value, you could assign the address (pointer) to a variable and you could change that (the pointed value, the value of the constant). Robert Griesemer (one of Go's authors) wrote why it's not allowed to take a string literal's address: "If you could take the address of a string constant, you could call a function [that assigns to the pointed value resulting in] possibly strange effects - you certainly wouldn't want the literal string constant to change." (source)
If you need a pointer to a value being equal to that constant, assign it to a variable of which is addressable so you can take its address, e.g.
func main() {
const k = 5
v := k
address := &v // This is allowed
}
But know that in Go numeric constants represent values of arbitrary precision and do not overflow. When you assign the value of a constant to a variable, it may not be possible (e.g. the constant may be greater than the max value of the variable's type you're assigning it to - resulting in compile-time error), or it may not be the same (e.g. in case of floating point constants, it may lose precision).
I often hit this problem when creating large, nested JSON objects during unit tests. I might have a structure where all the fields are pointers to strings/ints:
type Obj struct {
Prop1 *string
Prop2 *int
Status *string
}
and want to write something like:
obj := Obj{
Prop1: &"a string property",
Prop2: &5,
Status: &statuses.Awesome,
}
When I initialise it, but the language doesn't allow this directly. A quick way to bypass this is to define a function that takes a constant and returns its address:
s := func(s string) *string { return &s }
i := func(i int) *int { return &i }
obj := Obj{
Prop1: s("a string property"),
Prop2: i(5),
Status: s(statuses.Awesome)
}
This works due to the fact that when the constant is passed as a parameter to the function, a copy of the constant is made which means the pointer created in the function does not point to the address of the constant, but to the address of its copy, in the same way as when a constant value is assigned to a var. However, using a function to do this makes it more readable/less cumbersome IMO than having to forward declare large blocks of variables.
The AWS SDK uses this technique. I now find myself regularly adding a package to my projects that looks something like:
package ref
import "time"
func Bool(i bool) *bool {
return &i
}
func Int(i int) *int {
return &i
}
func Int64(i int64) *int64 {
return &i
}
func String(i string) *string {
return &i
}
func Duration(i time.Duration) *time.Duration {
return &i
}
func Strings(ss []string) []*string {
r := make([]*string, len(ss))
for i := range ss {
r[i] = &ss[i]
}
return r
}
Which I call in the following way:
func (t: Target) assignString(to string, value string) {
if to == tags.AuthorityId {
t.authorityId = ref.String(value)
}
// ...
}
You can also add a deref package, though I have generally found this to be less useful:
package deref
func String(s *string, d string) string {
if s != nil { return *s }
return d
}
// more derefs here.
EDIT April 2022:
With the release of go 1.18, it's now possible to define a single method to handle all conversions from constants into pointers:
package ref
func Of[E any](e E) *E {
return &e
}
I found another way to deal with this, which is using AWS API:
import "github.com/aws/aws-sdk-go/aws"
type Obj struct {
*int
}
x := aws.Int(16) // return address
obj := Obj{x} // work fine
this method is literally same as the answer above, but you dont have to write the whole functions on your own.
See: https://docs.aws.amazon.com/sdk-for-go/api/aws/
These 3 options could be helpful:
Using a helper function with generics. (Works for both primitive and custom types)
package main
import "fmt"
type Role string
const (
Engineer Role = "ENGINEER"
Architect Role = "ARCHITECT"
)
const (
EngineerStr string = "ENGINEER"
ArchitectStr string = "ARCHITECT"
)
func main() {
fmt.Println(PointerTo(Engineer)) // works for custom types
fmt.Println(PointerTo(EngineerStr)) // works for primitive types
}
func PointerTo[T any](v T) *T {
return &v
}
Try it on playground
Using pointy. (Works only for primitive types)
Using a ToPointer() method. (Works only for custom types)
package main
import "fmt"
type Role string
const (
Engineer Role = "ENGINEER"
Architect Role = "ARCHITECT"
)
func (r Role) ToPointer() *Role {
return &r
}
func main() {
fmt.Println(Engineer.ToPointer())
}
Try it on playground
What the constants section does not make very clear: Constants are, unlike variables, not present in the compiled code or running program. They are untyped and will only be in memory once they are assigned to a variable.
As a result, they seem1 to have infinite precision. If you look at this example, you can see that I can assign the constant to a variable without casting it, and the variable will hold as much of the constants precision as it can.
1 As the spec also points out, integers have at least 256 bits, floats at least 256 bits mantissa and 32 bits exponent, and the compiler will throw an error if its internal constructs cannot accurately store a constant.

Why should constructor of Go return address?

I understand that Go doesn't have any constructors and a New func is used in its place, but according to this example.
func NewFile(fd int, name string) *File {
if fd < 0 {
return nil
}
f := File{fd, name, nil, 0}
return &f
}
They always return &f. Why just simply returning File isn't suffice?
Update
I've tried returning the created object for a simple struct and it's fine. So, I wonder if returning an address is a standard way of constructor or something.
Thanks.
As mentioned, yes, the spec allows you to return either values (as non-pointers) or pointers. It's just a decision you have to make.
When to return pointer?
Usually if the value you return is "more useful" as a pointer. When is it more useful?
For example if it has many methods with pointer receiver. Yes, you could store the return value in a variable and so it will be addressable and you can still call its methods that have pointer receivers. But if a pointer is returned right away, you can "chain" method calls. See this example:
type My int
func (m *My) Str() string { return strconv.Itoa(int(*m)) }
func createMy(i int) My { return My(i) }
Now writing:
fmt.Println(createMy(12).Str())
Will result in error: cannot call pointer method on createMy(12)
But if works if you return a pointer:
func createMy(i int) *My { return (*My)(&i) }
Also if you store the returned value in a data structure which is not addressable (map for example), you cannot call methods on values by indexing a map because values of a map are not addressable.
See this example: My.Str() has pointer receiver. So if you try to do this:
m := map[int]My{0: My(12)}
m[0].Str() // Error!
You can't because "cannot take the address of m[0]". But the following works:
m := map[int]*My{}
my := My(12)
m[0] = &my // Store a pointer in the map
m[0].Str() // You can call it, no need to take the address of m[0]
// as it is already a pointer
And another example for pointers being useful is if it is a "big" struct which will be passed around a lot. http.Request is a shining example. It is big, it is usually passed around a lot to other handlers, and it has methods with pointer receiver.
If you return a pointer, that usually suggests that the returned value is better if stored and passed around as a pointer.
Pointer receiver accepts both pointer and value types, as long as it matches the data type.
type User struct {
name string
email string
age int
}
// NewUserV returns value ... ideally for a User we should not be
// returning value
func NewUserV(name, email string, age int) User {
return User{name, email, age}
}
// NewUserP returns pointer ...
func NewUserP(name, email string, age int) *User {
return &User{name, email, age}
}
// ChangeEmail ...
func (u *User) ChangeEmail(newEmail string) {
u.email = newEmail
}
func main() {
// with value type
usr1 := NewUserV("frank", "frank#camero.com", 22)
fmt.Println("Before change: ", usr1)
usr1.ChangeEmail("frank#gmail.com")
fmt.Println("After change: ", usr1)
// with pointer type
usr2 := NewUserP("john", "john#liliput.com", 22)
fmt.Println("Before change: ", usr2)
usr2.ChangeEmail("john#macabre.com")
fmt.Println("After change: ", usr2)
}
In addition to what icza mentioned about the big struct being passed around. Pointer values are a way of saying that pointer semantics are at play and who ever uses the particular type should not make copy of the value which is being shared by the pointer.
If you look at the struct of File or http type, it maintains channels or some other pointer types which is unique to that value. Make a copy of the value (given to you by the pointer) would lead to hard to find bugs since the copied value might end up writing or reading to the pointer types of the original value.

Pointer won't return with assigned address

I'm using Qt Creator 4.5 with GCC 4.3 and I'm having the following problem that I am not sure is Qt or C++ related: I call a function with a char * as an input parameter. Inside that function I make a dynamic allocation and I assign the address to the char *. The problem is when the function returns it does not point to this address anymore.
bool FPSengine::putData (char CommandByte , int Index)
{
char *msgByte;
structSize=putDatagrams(CommandByte, Index, msgByte);
}
int FPSengine::putDatagrams (char CommandByte, int Index, char *msgByte)
{
int theSize;
switch ( CommandByte ) {
case (CHANGE_CONFIGURATION): {
theSize=sizeof(MsnConfigType);
msgByte=new char[theSize];
union MConfigUnion {
char cByte[sizeof(MsnConfigType)];
MsnConfigType m;
};
MConfigUnion * msnConfig=(MConfigUnion*)msgByte;
...Do some assignments. I verify and everything is OK.
}
}
return theSize;
}
When I return the pointer it contains a completely different address than the one assigned in putDatagrams(). Why?
...
Ok thx I understand my mistake(rookie mistake :( ). When sending a pointer as an input parameter to the function you send the address of your data but not the address of your pointer so you cant make the pointer point somewhere else...it is actually a local copy like Index. The only case the data would of been returned succesfully with the use of a char * is by allocating the memory before the function call:
bool FPSengine::putData (char CommandByte , int Index)
{
char *msgByte;
msgByte=new char[sizeof(MsnConfigType)];
structSize=putDatagrams(CommandByte, Index, msgByte);
}
int FPSengine::putDatagrams (char CommandByte, int Index, char *msgByte)
{
int theSize;
switch ( CommandByte ) {
case (CHANGE_CONFIGURATION): {
theSize=sizeof(MsnConfigType);
union MConfigUnion {
char cByte[sizeof(MsnConfigType)];
MsnConfigType m;
};
MConfigUnion * msnConfig=(MConfigUnion*)msgByte;
...Do some assignments. I verify and everything is OK.
}
}
return theSize;
}
There are two ways. The pass-by-value way (C style):
int FPSengine::putDatagrams (char CommandByte, int Index, char **msgByte)
Note the second * for msgByte. Then inside of putDatagrams(), do:
*msgByte = new char[theSize];
In fact, anywhere in that function where you currently have msgByte, use *msgByte. When calling putDatagrams(), do:
structSize=putDatagrams(CommandByte, Index, &msgByte);
And the second way, since you're in C++, you could use pass-by-reference. Just change the signature of putDatagrams() to:
int FPSengine::putDatagrams (char CommandByte, int Index, char * &msgByte)
And you should be good. In this case, you shouldn't need to modify the caller or anything inside of your putDatagrams() routine.
Well, yes. Everything in C++ is, by default, passed by value. Parameters in the call putDatagrams(a, b, c) are sent by value - you wouldn't expect assigning to index in the code to change the value of b at the call site. Your msgByte=new char[theSize]; is just assigning to the local variable msgByte, overwriting the value passed in.
If you want to change a passed parameter such that the call site variable changes, you'll need to either pass by reference, or (in this case) pass a "pointer to a pointer` (and deference away the first pointer, assigning to the actual pointer).

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