Adding to char*

[seforo]

Hi Guys,
I have a segment of code here that I am not able to analyze

template <typename FieldType>
int CheckHeaderField(
                FieldType& field, const char* buffer, unsigned long bufferSize,
                bool continueParse
        )
{
        const char* fieldptr = reinterpret_cast<const char*>(&field);
        const char* endptr = buffer + bufferSize;
        AliHLTUInt32_t bufferRemaining = endptr > fieldptr ? endptr - fieldptr : 0;
        /*....And the rest of the code here*/
}
 

What is the meaning of adding a variable of unsigned long to a variable of char*
I printed endptr but the output is just funny characters.

[weaknessforcats]

When you add to a pointer, the compiler does a calculation to get the correct address as a result:

int arr[5];
int* ptr= arr;
      ptr = ptr + 3;
 

Assume the array is located at address 1000:

1000 arr[0]
1004 arr[1]
1008 arr[2]
1012 arr[3]
1016 arr[4]

So, ptr is initialized to the address of arr[0].

When you add to a pointer, the address in the pointer is increased by the sizeof the type it points at. Here ptr is an int*. So, ptr + 3, will be
ptr + 3 * sizeof(int). sizeof(int) is 4, so ptr is increased by 3*4, or 12.

Location 1012 is &arr[3]. It is also &ptr[3]. It is also ptr + 3.

To use the int at that location you can use arr[3], ptr[3], or *(ptr+3).