C语言 警告消息：大整数隐式截断为无符号类型 [-Woverflow]
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Warning message: large integer implicitly truncated to unsigned type [-Woverflow]
提问by Sha
Written a function with unsigned chararguments as shown below
编写一个带unsigned char参数的函数，如下所示
void Address_set(unsigned char x1,unsigned char y1,unsigned char x2,unsigned char y2)
The above function is called in main()function of C Code as
上述函数在main()C 代码的函数中调用为
Address_set(0,0,239,319);
I received a warning as
我收到了警告
large integer implicitly truncated to unsigned type [-Woverflow]
大整数隐式截断为 unsigned type [-Woverflow]
How to avoid this warning.
如何避免此警告。
回答by David Ranieri
The range of unsigned charis [0 ... 255], the value 319 is truncated (wrap-around) to 319 % 256 = 63
的范围unsigned char是 [0 ... 255]，值 319 被截断（环绕）为319 % 256 = 63
How to avoid this warning.
如何避免此警告。
The value is still truncated even if you can avoid this warning using a cast:
即使您可以使用强制转换避免此警告，该值仍会被截断：
Address_set(0,0,239,(unsigned char)319);
Use a longer type, i.e: unsigned short
使用更长的类型，即： unsigned short
![C语言 char *array 和 char array[]](/res/img/loading.gif)