why no overflow warning when converting int to char

Question

int i=9999;  

char c=i;

gives no overflow warning, While

char c=9999;

gives,

warning C4305 initializing truncation from int to char

why no overflow warning when converting int to char?

You're right @KeithThompson, thank you. I deleted the wrong comment. — pmg
– pmg, Commented May 11, 2013 at 18:35
@pmg That comment of yours confused me so much that I had to post it in a separate question.This is what you had said--It's not the same thing. The first is an assignment, the second is an initialization.And here's the link stackoverflow.com/questions/16500028/… — Rüppell's Vulture
– Rüppell's Vulture, Commented May 12, 2013 at 3:31
@pmg Actually,StackOverflow plebs like me don't expect ANY mistakes from the aces like you,with 25k+ reputation..LOL.. — Rüppell's Vulture
– Rüppell's Vulture, Commented May 12, 2013 at 9:21

James McNellis · Accepted Answer · 2012-05-08 17:28:07Z

9

You'll get warning C4244 when compiling with /W4 (which you should always do).

warning C4244: 'initializing' : conversion from 'int' to 'char', possible loss of data

answered May 8, 2012 at 17:28

James McNellis

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2 Comments

Vince Over a year ago

:( I forgot to enable the highest warning. Thanks

Keith Thompson Over a year ago

Note that that warning doesn't take into account the fact that the value of i happens to be 9999. The compiler would probably issue the same warning for int i = 42; char c = i;

Community · Accepted Answer · 2017-05-23 12:13:04Z

1

Whether any code construct produces a warning is up to the cleverness of the compiler and the choices made by its authors.

char c=9999;

9999 is a constant expression. The compiler can determine, just by analyzing the declaration with no additional context, that it's going to overflow. (Presumably plain char is signed; if it's unsigned, the conversion is well defined -- but a compiler could still choose to warn about it.)

int i=9999;  

char c=i;

This has the same semantics, but for a compiler to warn about the initialization of c, it would have to know that i has the value 9999 (or at least a value outside the range of char) when it analyzes that declaration. Suppose you instead wrote:

int i = 9999;
i = 42;
char c = i;

Then clearly no warning would be necessary or appropriate.

As James McNellis's answer indicates, a suffiently clever compiler can warn about either case if it performs additional analysis of what's going to happen during the execution of the program. For some compiler, it helps to enable optimization, since the analysis required to optimize code (without breaking it) can also reveal this kind of potential run-time error.

I'll note that this is an answer to the question you asked: why is there no warning. The answer you accepted is to the implied question: "I want a warning here; how can I enable it?". I'm not complaining, just observing.

edited May 23, 2017 at 12:13

CommunityBot

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answered May 11, 2013 at 18:15

Keith Thompson

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1 Comment

Rüppell's Vulture Over a year ago

+1. pmg too admitted the comment about one being an initialization and other being an assignment was wrong.Now the whole thing is very clear.Thanks for the time it took to come up with 2 fine answers.

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