Avoiding truncation warnings from my C++ compiler when initializing signed values

Question

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"

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

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

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

If you're trying to assign something with the top bit set, is it really a short? or should you be using an unsigned short? — The Archetypal Paul, Nov 18 '10 at 20:22

score 3 · Answer 1 · answered Nov 18 '10 at 20:23

3

Cast the constant.

short my_value = (short)0xF00D;

EDIT: 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.

answered Nov 18 '10 at 20:23

Donnie

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I am using C++ so I tried a static_cast, but I still get the warning. Maybe I'm supposed to use reinterpret_cast? – sourcenouveau Nov 18 '10 at 20:24
Using a C-style cast got rid of the warning, thank you. Now I need to read up on why there isn't a C++-style cast to handle this scenario. – sourcenouveau Nov 18 '10 at 20:27
1

You could `short(0xF00D)`, which is a little more C++-style. The problem is that there isn't an integer literal to specify a `short int` like there is for a `long`. – Travis Gockel Nov 18 '10 at 20:33

score 0 · Answer 2 · answered Nov 18 '10 at 21:03

You are assigning an int value that can not fit short, hence the warning. You could silent the compiler by using a c-type cast, but it is usually a bad practice.

A better way is not to do that, and to assign a negative value. Or, if the hexadecimal value makes more sense, to switch to unsigned short.

Avoiding truncation warnings from my C++ compiler when initializing signed values

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