我怎样才能避免海湾合作委员会的警告为纯＆QUOT; char的＆QUOT;于：＆QUOT; unsigned char型＆QUOT;或QUOT;符号字符＆QUOT;转换？ [英] How can I avoid gcc warning for plain "char" to : "unsigned char" OR "signed char" conversion?

问题描述

我的默认字符类型是无符号的字符作为在海湾合作委员会选项（-funsigned-CHAR GCC）的设置。所以，可以说，当我需要无符号的字符，在code，我可以使用字符。但我得到了（字符*）和（无符号字符*或符号字符*）之间的转换警告：

的错误：在通过测试2的参数1指向目标符号不一致。

我如何才能避免警告当我通过无符号字符*变量为char *（知道我syetem具有默认的无符号的字符由编译器选项设置）？

 静态无效测试2（字符*一）// char是deafult无符号由-funsigned-CHAR gcc的选项设置
{
}TEST1无效（无效）
{
        //这个传球，但如果我将其更改为无符号的字符（或符号字符'）失败
        //我不希望它失败无符号的字符C，因为默认情况下char是无符号的。
        焦C = 65;
        TEST2（和C）;
}
 

解决方案

开关 -funsigned-CHAR 和 -fsigned-CHAR 不引用的char * 。

您可以使用 -Wno指针符号来关闭您将收到警告。

My default char type is "unsigned char" as set in the gcc option (-funsigned-char gcc). So arguably I can use "char" when I need "unsigned char" in the code. But i am getting warning for conversion between (char*) and (unsigned char* or signed char*):

"error: pointer targets in passing argument 1 of 'test2' differ in signedness" .

How can I avoid warning when I pass unsigned char* variable to char* (knowing that my syetem has default unsigned char as set by compiler option)?

static void test2(char* a)      //char is unsigned by deafult as set by -funsigned-char gcc option
{
}

void    test1(void)
{
        // This passes, but if i change it to unsigned char (or 'signed char') it fails   
        // I dont want it to fail for "unsigned char c" since default char is unsigned.
        char    c = 65; 
        test2(&c);
}

解决方案

The switches -funsigned-char and -fsigned-char do not refer to char *.

You might use -Wno-pointer-signto switch off the warning you receive.

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