2 bytes (16 bits) datatype, anyone?

Question

I'm using OnlineGDB compiler to test routines for embedded micros, where usual datatypes are:

'int8_t' and 'uint8_t', AKA 'char' and 'unsigned char', 1 byte-wide...

'int16_t' and 'uint16_t', AKA 'int' and 'unsigned int', 2 byte-wide...

'int32_t' and 'uint32_t', AKA 'double' and 'unsigned double' (sometimes referred to as 'long' and 'unsigned long'...), 4 byte-wide...

Apparently, OnlineGDB in "C" mode only supports 1 byte-wide (as 'char/uchar') and 4 byte-wide (as 'int/uint') datatypes (probably because aimed at the PC world) so before creating my own structure-based datatype, I would like to know if I'm missing something...

Any help appreciated, thanks.

Answer 1

I think you might have some confusion related to memory allocation for data types.

The C standard does not specify how many bits a specific data type is; thus, it is essentially left up to the operating system.

A long on a Windows 64-bit system is 4 bytes.

A long on Unix / Linux 64-bit system is typically 8 bytes.

This is why we have cstdint.h; it standardizes the number of bits for a particular data type regardless of your operating system enabling applications to be portable between different operating systems.

OnlineGDB is a Linux based system.

Answer 2

int takes 2/4 bytes depending upon memory

float - 4 bytes

char - 1 byte

double - 8 bytes

Answer 3

Here are your standard integral data types and how many bytes they take up on the OnlineGDB machine:

#include <stdio.h>

int main()
{
    printf("sizeof(char): %lu\n", sizeof(char)); // 1
    printf("sizeof(short int): %lu\n", sizeof(short int)); // 2
    printf("sizeof(int): %lu\n", sizeof(int)); // 4
    printf("sizeof(long int): %lu\n", sizeof(long int)); // 8
    printf("sizeof(long long int): %lu\n", sizeof(long long int)); // 8

    return 0;
}

As others have already mentioned, these sizes are specific for the OnlineGDB machine and you may have different results on Windows or on your microcontroller.