Thread: Other then uint8_t what is the other alternative we can use in c compller

I need to assign the port pins to 0 to 15 binary bits,as i'm using c compiler it's not compiling the following code,even though i have added the stdint.h header file,so is any other way i can compile this...
uint8_t v

uint8_t is a 8 bit number, not 16. You probably want uint16_t

Originally Posted by rcgldr

What machine and what compiler are you using? 0 to 15 bits seems like you'd want a 16 bit unsigned integer, not an 8 bit one. On your compiler, how many bits are there in an "unsigned short"?

I'm using Keil IDE...Maximum 16 bits,and i mean to say 0 to 15 binary dig[0000-1111]..

Assuming compilation under C99 or later, support of uintN_t (N = 8, 16, 32, 64) types by an implementation is optional.

It strikes me as strange trying to use uint8_t to represent something that is 16 bits, rather than a uint16_t. Presumably an array of two of them .....


Assuming a digital target machine (), the unsigned char type will always be implemented with at least eight bits.

I think you're all misinterpreting him. He seems to mean a 4-bit quantity. I.e enough to represent 0 to 15.

Just add this to an appropriate header file:

Code:

typedef unsigned char uint8_t;

Even better, use preprocessor macros which determine whether the #include stdint.h, or just use this typedef, so that it works everywhere. This is a common way to deal with code that needs to run across multiple compilers.

Originally Posted by iMalc

I think you're all misinterpreting him. He seems to mean a 4-bit quantity. I.e enough to represent 0 to 15.

The wording "I need to assign the port pins to 0 to 15 binary bits" articulates a need for 16 bits. If that is not the meaning, it is because the OP misstated the goal.

Right yeah sorry, I do think the goal certainly was misstated originally. It is hard to interpret strangely worded descriptions with the correct intent.
Post #5 clarified it somewhat though.