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.

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.

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.

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).

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.

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).

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 ?

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.

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)

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

im getting the same result with %lld

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.

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.

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.

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.

you should be using %lld not %LLd

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.

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..

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 ?

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);

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.

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.

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.

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);

Originally Posted by laserlight

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