char and int

BlaX · 07-10-2009, 10:32 AM

Hello everyone.Im having some problem understanding whats going on here ...

This code

Code:

unsigned char x;
x=-5;
printf("%d",x);

Gives an output of 5 .. no problemo

but

Code:

unsigned int x;
x=-5;
printf("%d",x);

prints out -5 ... why ?

As for my 2nd question,

Code:

long long x = 18232323232322;
printf("%d",x);

The compiler is generating a warning :
[Warning] integer constant is too large for "long" type and printing a value of 187060802.
Im on a 32-bit machine and using Dev C++ 4.9.9.2 , where the long long int is supported and is in the range –9,223,372,036,854,775,807 to 9,223,372,036,854,775,807.

Thank you.

**laserlight** · 07-10-2009, 10:44 AM

Quote:

Originally Posted by BlaX

Gives an output of 5 .. no problemo

hmm... are you sure you get 5 as the output? I am not sure how to explain 5, but I can explain 251, which I do get when running your sample code.

Quote:

Originally Posted by BlaX

The compiler is generating a warning :
[Warning] integer constant is too large for "long" type and printing a value of 187060802.

You should use the long long constant 18232323232322LL.

cas · 07-10-2009, 10:49 AM

I'm surprised the first one prints out -5. When you assign -5 to an unsigned char, it stores a positive value (in this case, if you have 8-bit chars, 251). In your call to printf(), x is promoted to int, which can store 251. Thus I would expect 251 to be printed.

The second case is, I believe, technically undefined behavior. %d expects an int, and you're passing an unsigned int. But what's probably happening is this: -5 gets converted to unsigned int, and if you have 32-bit ints, that means 4294967291. It so happens that -5 and 4294967291 look the same to the computer, and which you get depends on how you interpret it. %u says to interpret it as an unsigned int, %d as a signed int. So you're asking printf() to print out -5, essentially.

As for 18232323232322, you want to print out a long long with %lld, not %d. %d means int. The warning you're getting is just informational, and the initialization is correct (in C99); but you can silence it by using 18232323232322LL to explicitly tell what type of constant it is (LL meaning long long, obviously).

Elysia · 07-10-2009, 11:01 AM

Quote:

Originally Posted by cas

...The warning you're getting is just informational...

I am not so sure.
Usually the compiler truncates the number and spits out a warning for you.
Be on the safe side - use the proper postfix to make the compiler understand.

**laserlight** · 07-10-2009, 11:04 AM

Quote:

Originally Posted by Elysia

Usually the compiler truncates the number and spits out a warning for you.

I double checked, and apparently that is not true for a standard conforming compiler (other than the warning part).

BlaX · 07-10-2009, 11:05 AM

Ohh .. sorry sorry my mistake.its printing 251.(I know the reason no need to explain.thank you)
But why is the unsigned int not printing 4294967291 and is printing -5 ?

Quote:

You should use the long long constant 18232323232322LL.

Why should i use the long long constant when im declaring x as a long long integer and assigning a value within its range.

Even trying

Code:

long long x = 18232323232322LL;
printf("%LLd",x);

Im getting 187060802 but without a warning.

Edit:

@cas:Sorry, i didnt see your post because i was writing a reply when u had already answered.

Quote:

It so happens that -5 and 4294967291 look the same to the computer, and which you get depends on how you interpret it

The same thing is for the 251 and -5 in the first part.But why is it printing 251 during char (the +ve value) and -5 for the int(the -ve value)

Quote:

you want to print out a long long with %lld, not %d

im getting the same result with %lld

Elysia · 07-10-2009, 11:10 AM

Quote:

Originally Posted by laserlight

I double checked, and apparently that is not true for a standard conforming compiler (other than the warning part).

Hmm. Still, I would not rely on it, though.
And it does no harm in silencing a warning.

Quote:

Originally Posted by BlaX

Why should i use the long long constant when im declaring x as a long long integer and assigning a value within its range.

Any number is treated as int by default, and thus it tries to assign an int to a long long, hence the warning.
Just as you can prefix 0x to make it hexadecimal, you can postfix numbers with with things to make it a different type. For example, .0f for float, .0 for double, etc.

Quote:

Even trying

Code:

long long x = 18232323232322LL;
printf("%LLd",x);

Im getting 187060802 but without a warning.

Use %lld. There is a difference between %LLd and %lld. C is case sensitive.

**laserlight** · 07-10-2009, 11:16 AM

Quote:

Originally Posted by BlaX

But why is the unsigned int not printing 4294967291 and is printing -5 ?

Implementation defined behaviour. cas has outlined the reasons.

itCbitC · 07-10-2009, 11:16 AM

you should be using %lld not %LLd

BlaX · 07-10-2009, 11:22 AM

Quote:

It so happens that -5 and 4294967291 look the same to the computer, and which you get depends on how you interpret it

That's true.The same thing is for the 251 and -5.But why is it printing 251 during char (the +ve value of the variable) and -5 for the int(the -ve value of it) while using the %d format specifier.

Quote:

Any number is treated as int by default, and thus it tries to assign an int to a long long, hence the warning.

Where's the problem ? It is trying converting from a lower type to a higher type..

Quote:

Use %lld. There is a difference between %LLd and %lld. C is case sensitive.

Code:

long long x = 18232323232322ll;
printf("%lld",x);

Getting same answer for both(187060802) ...How the heck should i print out a long long variable ?

itCbitC · 07-10-2009, 11:22 AM

Perhaps this might help to understand what laserlight and cas mean by "implementation defined behavior".

Code:

unsigned int x;
x=-5;
printf("%d  %u", x, x);

Elysia · 07-10-2009, 11:25 AM

Quote:

Originally Posted by BlaX

Where's the problem ? It is trying converting from a lower type to a higher type..

That's not it.
Your constant 18232323232322 is too large for an integer. It cannot be stored in 32 bits. Thus, it is not an integer type; it cannot be, for it is too large to be held in an integer.
It's like saying to the compiler: "Hi, this is my number and its type is an integer." And the compiler replies: "But your number is too large to be of type integer, because it will not fit."
This is before it is assigned to the long long type, note.

tabstop · 07-10-2009, 11:25 AM

Quote:

Originally Posted by BlaX

Getting same answer for both(187060802) ...How the heck should i print out a long long variable ?

Using MinGW (dev-c++/code::blocks on windows)? If so as I recall you have to use Microsoft's %I64d.

**laserlight** · 07-10-2009, 11:27 AM

Quote:

Originally Posted by BlaX

But why is it printing 251 during char (the +ve value of the variable) and -5 for the int(the -ve value of it)

Because 251 as a char can be represented as an int, but -5 as an unsigned int cannot be represented as an int, hence implementation defined behaviour is involved.

Quote:

Originally Posted by BlaX

How the heck should i print out a long long variable ?

Oh yeah, you're using MinGW, so I believe it should be:

Code:

long long x = 18232323232322ll;
printf("%I64d",x);

tabstop · 07-10-2009, 11:36 AM

Quote:

Originally Posted by laserlight

BlaX is online now Add Infraction for BlaX Report Post IP Edit/Delete Message

Now we know what godlike powers the moderators have! :P

07-10-2009, 10:32 AM	#1
BlaX Registered User Join Date: Feb 2008 Posts: 36	char and int Hello everyone.Im having some problem understanding whats going on here ... This code Code: unsigned char x; x=-5; printf("%d",x); Gives an output of 5 .. no problemo but Code: unsigned int x; x=-5; printf("%d",x); prints out -5 ... why ? As for my 2nd question, Code: long long x = 18232323232322; printf("%d",x); The compiler is generating a warning : [Warning] integer constant is too large for "long" type and printing a value of 187060802. Im on a 32-bit machine and using Dev C++ 4.9.9.2 , where the long long int is supported and is in the range –9,223,372,036,854,775,807 to 9,223,372,036,854,775,807. Thank you.
BlaX is offline

07-10-2009, 10:49 AM	#3
cas Registered User Join Date: Sep 2007 Posts: 372	I'm surprised the first one prints out -5. When you assign -5 to an unsigned char, it stores a positive value (in this case, if you have 8-bit chars, 251). In your call to printf(), x is promoted to int, which can store 251. Thus I would expect 251 to be printed. The second case is, I believe, technically undefined behavior. %d expects an int, and you're passing an unsigned int. But what's probably happening is this: -5 gets converted to unsigned int, and if you have 32-bit ints, that means 4294967291. It so happens that -5 and 4294967291 look the same to the computer, and which you get depends on how you interpret it. %u says to interpret it as an unsigned int, %d as a signed int. So you're asking printf() to print out -5, essentially. As for 18232323232322, you want to print out a long long with %lld, not %d. %d means int. The warning you're getting is just informational, and the initialization is correct (in C99); but you can silence it by using 18232323232322LL to explicitly tell what type of constant it is (LL meaning long long, obviously).
cas is offline

07-10-2009, 11:16 AM	#9
itCbitC Registered User Join Date: Oct 2008 Location: TX Posts: 1,262	you should be using %lld not %LLd Last edited by itCbitC; 07-10-2009 at 11:19 AM.
itCbitC is offline

07-10-2009, 11:22 AM	#11
itCbitC Registered User Join Date: Oct 2008 Location: TX Posts: 1,262	Perhaps this might help to understand what laserlight and cas mean by "implementation defined behavior". Code: unsigned int x; x=-5; printf("%d %u", x, x);
itCbitC is offline

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