gnu linux 上ARM默认char是不带符号的。微软wince上char又默认是signed。

C标准表示char类型可以带符号也可以不带符号，由具体的编译器、处理器或由它们两者

共同决定到底char是带符号合适还是不带符号合适。

大部分体系结构上，char默认是带符号的，它可以自-128到127之间取值。而也有一些例

外，比如ARM体系结构上，char就是不带符号的，它的取值范围是0～255

举例来说，在默认char不带符号，下面的代码实际会把255而不是-1赋予i：

char i = -1;

而另一种机器上，默认char带符号，就会确切地把-1赋予i。如果程序员本意是把－1保

存在i中，那么前面的代码就该修改成：

    signed char i = -1;

    另外，如果程序员确实希望存储255，那么代码应该如下：

     unsigned char = 255;   

       如果你在自己的代码中使用了char类型，那么你要保证在带符号和不带符号的情况下

代码都没问题。如果你能明确要用的是哪一个，那么就直接声明它。

The following email fragment appeared on the linux-arm mailing list recently: consider this simple program:

int main(void){    char i = -1;    printf("%d\n", i);    return 0;}

The print out is 255 in stead of -1, unless I define i as signed char i;then I get the "-1" print out. The above code is actually buggy in that it assumes that the type "char" is equivalent to "signed char". The C standards do say that "char" may either be a "signed char" or "unsigned char" and it is up to the compilers implementation or the platform which is followed. As the poster points out, the above code does not work as expected if "char" is "unsigned". It is difficult to detect this code at compile time, since GCC does not issue any warnings. The only way to detect it is either by visual examination of the code, or by actually running it and finding a problem.This causes problems on ARM based machines since "char" is of the "unsigned" variety, which allows the compiler to generate faster, more efficient code. ARM is not alone in this - SGI Mips running IRIX also encounters this problem.However, dispite the lack of warning for the above case, GCC does warn with the following code:

{    char foo;    foo = bar();    if (foo == -1) {        ...    }}

Code like the above will generate a compiler warning, which will be one of the following depending on the actual test used: 
warning: comparison is always 0 due to limited range of data type
warning: comparison is always 1 due to limited range of data type
Please note however that the above warnings are not issued if "char" is "signed" and therefore can be difficult to pick up when compiling in such an environment. The following table lists the four types of code which cause problems when "c" is declared as just "char", and the most likely correct method of fixing the code.