在初始化签名值时避免来自我的C ++编译器的截断警告 [英] Avoiding truncation warnings from my C++ compiler when initializing signed values

问题描述

我想将 short 初始化为十六进制值，但是我的编译器会给出截断警告。显然，它认为我试图将 short 设置为正值。

I would like to initialize a short to a hexadecimal value, but my compiler gives me truncation warnings. Clearly it thinks I am trying to set the short to a positive value.

short my_value = 0xF00D; // Compiler sees "my_value = 61453"

如何避免此警告？我只能使用负值，

How would you avoid this warning? I could just use a negative value,

short my_value = -4083; // In 2's complement this is 0xF00D

但在我的代码中，使用十六进制更容易理解。

but in my code it is much more understandable to use hexadecimal.

推荐答案

强制转换常量。

short my_value =（short）0xF00D;

编辑：初始解释在我的头上有意义，但是进一步的反思是有点错误。但是，这应该抑制警告，并给你你期望的。

Initial explanation made sense in my head, but on further reflection was kind of wrong. Still, this should suppress the warning and give you what you expect.

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