C:警告的解:表达式中的整数溢出?-C:Acureforthewarning:integeroverflowinexpression?

22  

Integer overflow means that you have exceeded the upper limit of an int value, which is likely to be 32767 if you are getting this error. It has nothing to do with floating point; the operations you have specified are in fact integer math operations, so the fractional part of the division is discarded anyway.

整数溢出意味着您已经超过了int值的上限，如果您得到这个错误，它很可能是32767。它与浮点数无关;您指定的操作实际上是整数数学运算，因此该除法的小数部分无论如何都会被丢弃。

Try something like this:

试试这样:

#define FOSC        8000000L
#define BAUDRATE    9600L
#define BRGVAL      ((unsigned int)((FOSC/2)/(16*BAUDRATE)-1))

void uart_init(){
   U1BRG = BRGVAL;
}

The L suffix turns these constants into long type instead of int type. The (unsigned int) cast converts to U1BRG's type, and lets the compiler know that you understand that the long value will fit into an unsigned int and thus hide any warnings it may throw at you.

L后缀将这些常量变为long类型而不是int类型。(unsigned int)转换为U1BRG的类型，并让编译器知道您知道长值将与未签名的int相匹配，从而隐藏它可能向您抛出的任何警告。

Normally, it's bad practice to silence compiler warnings, but in this case, it's clear that although you need long to store intermediate values in the calculation, the final result will fit into an unsigned int.

通常，对编译器警告进行静默是很糟糕的做法，但是在这种情况下，很明显，尽管您需要很长时间来存储中间值，但最终结果将与无符号整数相匹配。

8  

I like Philip's answer, but I think a better solution is to reduce the formula and change your macro to:

我喜欢Philip的答案，但我认为一个更好的解决方案是减少公式，把你的宏改为:

#define BRGVAL (FOSC/32/BAUDRATE-1)

In doing so, you eliminate the cast so the compiler can continue to warn you if you choose a low baud rate that would result in a divider value too large for a 16-bit int.

在这样做时，您消除了cast，这样编译器就可以继续警告您，如果您选择一个低的波特率，将导致一个16位整数的分隔值太大。

0  

I would probably use this:

我可能会用这个:

#define BRGVAL      ((int)(FOSC/2)/(16*BAUDRATE)-1)

0  

It's not clear from your example whether U1BRG is a global variable or a #define'ed constant. In any case, simply casting to an integer should work:

从你的例子来看，U1BRG是一个全局变量还是#define'ed常数，这一点还不清楚。在任何情况下，简单地选择一个整数就可以了:

 U1BRG = (int)BRGVAL;

0  

You have failed to point this out, What's the data type for U1BRG? If it's an int, cast it like as shown

你没有指出这一点，U1BRG的数据类型是什么?如果它是一个int类型，就像这样cast它。

#define FOSC        8000000
#define BAUDRATE    9600
#define BRGVAL      ((long)(FOSC/2)/(16*BAUDRATE)-1)

void uart_init(){
   U1BRG = BRGVAL;
}

Edit: Amended this to take into consideration of Adam Liss's comment that an unsigned int is too small to hold the result of the macro, I have changed it to make it a long...Thanks Adam for the headsup...

编辑:修改了这一点，以考虑到Adam Liss的评论，一个未签名的int数太小，无法保存宏的结果，我已经更改了它，使它成为一个长……感谢亚当的领导…

Hope this helps, Best regards, Tom.

希望这对你有帮助，最好的问候，汤姆。