In C, How does one set the value at one address to the value at another address?

Question

I'm trying to make some UART code which will put a small string on the stack to be transmitted via a system of interrupts. This is specifically for an Atmel SAM microcontroller

#define UART_BUF_LEN 16

//GPS SERIAL INSTANCE
struct usart_module usart_instance; 
char buffer[UART_BUF_LEN];
uintptr_t bufferPtr = (uintptr_t)buffer;

bool transmitting(){
    return (uintptr_t)buffer == bufferPtr; //If the read head is at the top of the array, then we are not transmitting.
}

bool transmit(char* c, uint len){
    if(!transmitting() && len<=UART_BUF_LEN){
        bufferPtr= (uintptr_t)buffer + len;//set write head to top of the len stack
        while(transmitting()){
            *bufferPtr = *c; //Set the value at the address of bufferPtr to the value at the address of c.
            bufferPtr --; //Move the buffer closer to the head of the array
            c++; //Move the head of the array down some.
        }
    }else{
        return false;
    }
    bufferPtr= (uintptr_t)buffer + len; //reset the read head so that our transmit code knows where to read from.
    return true;
}

Trouble here is the line *bufferPtr = *c;. When I build the solution, the bufferptr seems to not be dereferencable. I get the following error:

invalid type argument of unary '*' (have 'uintptr_t {aka unsigned int}')

I've looked online, and all sources tell me that I must cast uintptr_t back into a pointer of the native datatype which the memory address points to. I am not sure how to do this, as using the typecast (char *), indicating that the buffer pointer is a character pointer does not compile, giving me the same error as above.

Now the rabbit hole goes one level deeper when I change the line to \*(char\*)bufferPtr = \*c;, which does not give me any error. What does this line mean?

I expect that it means that the value at the address of the typecast of bufferPtr to a char pointer is set to the value at the address of c. Is this correct?

Answer 1

Since bufferPtr is a uintptr_t, which is an integral type, not a pointer type, you will have to cast it back to the pointer type before dereferencing it.

*(char *)bufferPtr = *c;

This is fine, since the source of bufferPtr's value came from a char array.

char buffer[UART_BUF_LEN];
uintptr_t bufferPtr = (uintptr_t)buffer;