How to make a pointer into an integer?

How to make a pointer into an integer?

so the right way to store a pointer as an integer is to use the uintptr_t or intptr_t types. (See also in cppreference integer types for C99). these types are defined in for C99 and in the namespace std for C++11 in (see integer types for C++).

Can Void be converted to int?

3 Answers. You’re return ing the value of int sum by setting a void * address to it. In this case, the address is not valid. But, if you keep that in mind and get the value of sum by casting a void * to int it will work.

How to convert pointer to int in Cpp?

An object pointer (including void* ) or function pointer can be converted to an integer type using reinterpret_cast . This will only compile if the destination type is long enough. The result is implementation-defined and typically yields the numeric address of the byte in memory that the pointer pointers to.

What is uint8_ t in Cpp?

Side note: the “uint8_t” and “int16_t” types are commonly used in C/C++ to indicate precisely what the type is, i.e. unsigned single-byte and signed double-byte in this case. They are defined in a standard C header file called “stdint. h”. A quick count tells us that this data structure uses 6 bytes of memory.

Can a pointer point to itself?

Yes, a pointer can contain the position of a pointer to itself; even a long can contain the position of a pointer to itself.

How do you turn a pointer into a variable?

The “address of” operator This is the best way to attach a pointer to an existing variable: int * ptr; // a pointer int num; // an integer ptr = &num // assign the address of num into the pointer // now ptr points to “num”!

How do I print an int pointer?

Printing pointers. You can print a pointer value using printf with the %p format specifier. To do so, you should convert the pointer to type void * first using a cast (see below for void * pointers), although on machines that don’t have different representations for different pointer types, this may not be necessary.

How do I cast a void pointer to integer?

Example program:- ptr=&a // Assigning address of integer to void pointer. printf(“The value of integer variable is= %d”,*( (int*) ptr) );// (int*)ptr – is used for type casting.

What is pointer to void?

A pointer to void means a generic pointer that can point to any data type. We can assign the address of any data type to the void pointer, and a void pointer can be assigned to any type of the pointer without performing any explicit typecasting.

Why is namespace std bad?

The compiler may detect this and not compile the program. In the worst case, the program may still compile but call the wrong function, since we never specified to which namespace the identifier belonged. Namespaces were introduced into C++ to resolve identifier name conflicts. The std namespace is huge.

What is the difference between uint32 and uint32_t?

2 Answers. uint32_t is standard, uint32 is not. That is, if you include or

Does a pointer have an address?

The main feature of a pointer is its two-part nature. The pointer itself holds an address. The pointer also points to a value of a specific type – the value at the address the point holds.

How to return an array of uint8 _ T?

Im returning a pointer to an array of uint8_t from my function. So what am I getting wrong here? The compiler told you that return &buffer is returning a pointer to an array of 7 uint8_t (spelled uint8_t (*) [7] ), but you said the function returns uint8_t *, and these are different and incompatible pointer types.

How to convert a pointer to a normal integer?

– Arduino Stack Exchange How do I properly convert a pointer (uint8_t* or void*) to a “normal” integer? I’m writing a program for arduino that needs dynamic memory allocation frequently, but as I don’t want the memory to get fragmented, I’m also writing a “memory handler” for it.

Is the uint8 _ T A byte or a octet?

uint8_t is not a byte it’s a octet, char is a byte, this is important because you are talking about C compliant code. In C, byte != octet, char could be greater than 8 bits, but uint8_t if defined has exactly 8 bits. Plus, uint8_t is not a character type so strict aliasing exception with character type theoretically don’t work with uint8_t.

Is the statement & buffer of type uint8 _ T?

The statement &buffer is of type uint8_t** – a pointer to a pointer. In C and C++ an array name is the same as a pointer. The second thing is that you’re returning a pointer to an array CREATED ON STACK. This array is not valid after function exit. In practice you’re returning a dangling pointer. But there is another problem.