Pointer declaration in C, whats the difference?

I thought #1

uint8_t* start_ptr, end_ptr;

and #2

uint8_t* start_ptr;
uint8_t* end_ptr;

are generally the same. It seems to me they are not.
Could somebody specify what the first one does other than the second?

What happened:

          if(strncmp(mseq,mseq_a,8) == 0){
              start_ptr = MY_UART_RingBuffer_getReadPointer();
              start_found = 1;

          }

          if(strncmp(mseq+3,mseq_z,5) == 0){
              end_ptr = MY_UART_RingBuffer_getReadPointer();

              if (start_found == 1){
                  if(!MY_UART_RingBuffer_getOverlap()){
                    end_ptr = end_ptr - 6;
                  }
                  else{
                      start_found = 0;
                      continue;
                  }

                ptrdiff_t length = end_ptr - start_ptr;
                  start_found = 0;
              }
          }

On using #1 the compiler will give me this for the length calculation:

../Core/Src/main.c:143:32: error: invalid operands to binary - (have 'int' and 'uint8_t * {aka unsigned char *}')
     ptrdiff_t length = end_ptr - start_ptr;

On using #2 everything works out fine.

I am somehow confused here. I see the solution but I don’t really get the issue.

Thanks a lot 😉

>Solution :

Grammatically, the type uint8_t is a declaration-specifier; whereas the pointer * is a declarator. According to C’s grammar declarators bind to the declared name rather than the declaration-specifier. Consequently your first line is parsed like:

uint8_t (*start_ptr), end_ptr;

i.e. only the first name becomes a pointer.

This is one reason some tend to put the space before the pointer * rather than after. The correct way to declare those in one line would be:

uint8_t *start_ptr, *end_ptr; // both are pointers to uint8_t.