What's a good alternative to uint8_t when it is not provided by the compiler?

I'm using nvcc to compile a CUDA kernel. Unfortunately, nvcc doesn't seem to support uint8_t, although it does support int8_t (!). I'd just as soon not use unsigned char, for portability, readability, and sanity reasons. Is there another good alternative?

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Just to forestall any possible misunderstanding, here are some details.

$ nvcc --version
nvcc: NVIDIA (R) Cuda compiler driver
Copyright (c) 2005-2010 NVIDIA Corporation
Built on Mon_Jun__7_18:56:31_PDT_2010
Cuda compilat开发者_StackOverflowion tools, release 3.1, V0.2.1221

Code containing

int8_t test = 0;

is fine, but code containing

uint8_t test = 0;

throws an error message like

test.cu(8): error: identifier "uint8_t" is undefined

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C99 integer types are not "defined by the compiler" - they are defined in <stdint.h>.

Try:

#include <stdint.h>

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typedef unsigned char uint8_t;

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This is no different from what Mac OS X uses:

typedef unsigned char uint8_t;

What is your concern about the portability of unsigned char? If the concern is that a char might not represent 8 bits of storage, then you can include a static assertion along the lines of:

typedef int Assert8BitChar[(CHAR_BIT == 8)? 0 : -1];

This will cause compilation to error out when the assumption is violated.

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This seems to compile just fine with nvcc:

#include <stdint.h>

int main() {
    uint8_t x = 0;
    return (int) x;
}