On an 8 … Also, in the BCL, binary data is more often given as a byte[] (array of octets) rather than an int[] or uint[], so byte seems to be preferred in that situation. An int is a format likewise you get it as value from the accumulator. This yields a range of -2,147,483,648 to 2,147,483,647 (minimum value of -2^31 and a maximum value of (2^31) - 1). The _t means it's a typedef. If doing math with integers at least one of the values must be of type long, either an integer constant followed by an L or a variable of type long, forcing it to be a long. A uint16_t is an unsigned 16 bit value, so it takes 2 bytes (16/8 = 2) The only fuzzy one is int. A word is 2 bytes (except on Due, Zero and similar boards where it is 4 bytes), which already gets you a huge range of 65536 (256 2). In the Arduino, the only place that this is likely to occur is in sections of code associated with interrupts, called an interrupt service routine. For X64 that is Int64. For compatibility the "Integer" is kept currently on Int32, the register format from the X86 computers. The int data type is a word and Arduino comes with highByte() and lowByte() to extract the left and right byte from a word. Older computers likewise the 8088 had an 8 bit Int and therefore that was the same as the byte. You may not see that savings in the program memory, because changeable data are stored in RAM. The 80286 had a 16 bits Integer. int myInputpin = 3; void setup() What we've done here is to use 2 bytes of RAM, which can be a scarce resource on a microcontroller, in order to store a value that couldn't possible be more than 28, and worse one than that it will never change! Figure 2: How data is stored in memory cells: 1 byte, 2 byte integer and a string Volatile vs Non-volatile Memory There are different types of memory devices available, and the application area of each memory device depends on the architecture of the memory system. On the Arduino Due, for example, an int stores a 32-bit (4-byte) value. Now, load the code onto your Arduino board. 2. 5 thoughts on “ Arduino: Difference in “Byte” VS “uint8_t” VS “unsigned char” ” ujjwal sikriwal 7th January 2017 at 6:20 am. On the Arduino, an "unsigned int" is a 16-bit quantity so a "word" is also a 16-bit quantity. This makes it really easy to encode and decode. Of course if you use int[] you restrict yourself to the case where the length of the data is an integral multiple of 32 bits (as … int or long volatiles If the volatile variable is bigger than a byte (e.g. That is "a signed integer value at the native size for the compiler". This yields a range of -32,768 to 32,767 (minimum value of -2^15 and a maximum value of (2^15) - 1). The header file for the ESP8266 corrects for the change in the size of integers by equating "word" with "uint16_t" but no such adjustment has been made for the ESP32. Long variables are extended size variables for number storage, and store 32 bits (4 bytes), from -2,147,483,648 to 2,147,483,647. is it possible to convert byte value into integer… and can we compare two byte values if yes then how we will do that from initialising to execution… a 16 bit int or a 32 bit long), then the microcontroller can not read it in one step, because it is an 8 bit microcontroller. So a uint8_t is an unsigned 8 bit value, so it takes 1 byte. The int size varies from board to board. On the ESP8266 and ESP32, an "unsigned int" is a 32-bit quantity. Yes, on Arduino (but not on many other computers) an int is by default 16 bits (2 bytes). Check the compile size: 2488 bytes for int versus 2458 bytes for byte.Not a lot bigger, but it IS bigger. 1. Yes, a byte requires 8 bits of storage. Example int counter = 32 ;// declaration of variable with type int and initialize it with 32 Unsigned int There's 8 bits to the byte. On all Arduinos (ATMega and ARM based) a short stores a 16-bit (2-byte) value.