What is C++ equipvalent of Java Long?

[ictoan]

Hello!

I need to convert the following Java code to C++ and have trouble figuring out how to represent Java Long in C++.

Here's the Java code:

String function(Number n){
long l = n.longValue();
return Long.toString(l, 16);
}

This is what I'm thinking of doing in C++
string function(long l){
stringstream ss;
ss << l;
return ss.str();
}

Would that work?
In C++, there are different types of longs...
http://home.att.net/~jackklein/c/inttypes.html#long

Is long returned by n.longValue() the same as long (signed) in C++?

[D_Drmmr]

In C++, there are different types of longs...
http://home.att.net/~jackklein/c/inttypes.html#long

Is long returned by n.longValue() the same as long (signed) in C++?

As you can read on the link you provided, there are really only two types of long in C++, signed and unsigned. The C++ standards guarantees a long to be at least 32-bit. So the range of a (signed) long is from -2147483648 to 2147483647.

[Paul McKenzie]

Hello!

I need to convert the following Java code to C++ and have trouble figuring out how to represent Java Long in C++.

Your problem is that you're trying to equate Java to C++ and variable types.

In Java, there is a standard size for each type, not so in C++. A "long" in C++ could be any size -- it is up to the compiler as to what size any simple data type is.

The only requirement for C++ is that long is at least as big as an "int". The same rules apply to the other simple types in C++ -- there is no concrete size (except for char, which must be 1 byte). In C++, simple type sizes are determined by what type must be at least as big as another type, etc.

So you tell us -- what is a Java long? What size is it? Whatever it is, you have to make sure that your compiler has an integer type of that size.

Regards,

Paul McKenzie

[Paul McKenzie]

The C++ standards guarantees a long to be at least 32-bit.

Are you sure about that?

Regards,

Paul McKenzie

[laserlight]

Are you sure about that?

Yes, by virtue of the minimum range required.

[ictoan]

Hi guys,

Thanks for the quick reply.

This is what Java doc says...

http://java.sun.com/docs/books/tutor...datatypes.html
long: The long data type is a 64-bit signed two's complement integer. It has a minimum value of -9,223,372,036,854,775,808 and a maximum value of 9,223,372,036,854,775,807 (inclusive). Use this data type when you need a range of values wider than those provided by int.

So in this case.. use long long?

[monarch_dodra]

(except for char, which must be 1 byte).

However, the standard doesn't say how many bits a byte should be. Only that it should hold at least the values 0 to 255 (ie At least 8 bits).

It also states that for chars, all possible binary values of the char be a valid value. If you read a RAW byte, you know it is a valid value. This is usually not the case for floating type numbers, for example. Certain binary values simply represent nothing in floating point, not even NAN.

[monarch_dodra]

Hi guys,

Thanks for the quick reply.

This is what Java doc says...

http://java.sun.com/docs/books/tutor...datatypes.html
long: The long data type is a 64-bit signed two's complement integer. It has a minimum value of -9,223,372,036,854,775,808 and a maximum value of 9,223,372,036,854,775,807 (inclusive). Use this data type when you need a range of values wider than those provided by int.

So in this case.. use long long?

long long is not (yet) part of standard C++. It is standard C99 though. All of the compilers I have seen support it though, you my say is you are 99% safe using it.

However, you have no guarantees on the size of your long long. It is at least 64 bits, but could be 128.

[nuzzle]

[QUOTE=ictoan;1894891]
I need to convert the following Java code to C++ and have trouble figuring out how to represent Java Long in C++.

C++ doesn't specify the exact byte size of primitives so what you do will be implementation dependent. The easiest way to handle this is to use a define and then change it to the 64 bit signed type supported by the specific compiler. For example If you use MS Visual C++ you can study this table,

http://msdn.microsoft.com/en-us/libr...tz(VS.80).aspx

and use,

#define MY_LONG __int64 // VC++ specific

[D_Drmmr]

Are you sure about that?

99%, since I don't have a copy of the standard handy.

The only requirement for C++ is that long is at least as big as an "int".

I thought the rule was:

1 = sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long)
Moreover, a char is at least 8-bit and a long is at least 32-bit.

Indeed, the standard doesn't talk about bits, but guarantees a valid range instead. However, assuming a computer works with bits (which most of them do) that translates into these requirements.

[nuzzle]

and use,

#define MY_LONG __int64 // VC++ specific

A somewhat more general approach would be to use,

#define MY_LONG long long // biggest chance of being a Java long

Then you put in this check somewhere at the beginning of your program,

if (sizeof(MY_LONG) < 8) {
// long isn't long enougth
}

[Paul McKenzie]

A somewhat more general approach would be to use,

#define MY_LONG long long // biggest chance of being a Java long

Then you put in this check somewhere at the beginning of your program,

if (sizeof(MY_LONG) < 8) {
// long isn't long enougth
}

Better yet, do a static assert:

char dummy[ sizeof(MY_LONG) == 8 ];

Then the code will not compile if the long isn't exactly 8 bytes.

Regards,

Paul McKenzie

[couling]

Its always funny how simple questions can spark such debates.

Anyway the standard does not dictate sizes, there are some conventions which are very common (but not universal).

char 1 byte
short 2 bytes
int 4 bytes (or 2 bytes when compiling 16 bit code).
long 4 bytes

Since 64 bit integers have been a problem for a while its no surprise that some compilers allow you to use "long long" for this. But this is not officially a type before the most recent standard (which I believe is still in draft) and not all compilers support it.

A very common addition to the standard are the types:

__int8 1 byte
__int16 2 bytes
__int32 4 bytes
__int64 8 bytes

Most compilers support these.

[JohnW@Wessex]

However, the standard doesn't say how many bits a byte should be. Only that it should hold at least the values 0 to 255 (ie At least 8 bits).

I worked with a C compiler for TMS320C50 that had 16 bit char. The processor was unable to manipulate anything less than 16 bits at a time.

[monarch_dodra]

I worked with a C compiler for TMS320C50 that had 16 bit char. The processor was unable to manipulate anything less than 16 bits at a time.

Isn't that the definition of a byte though? The smallest element a processor can manipulate?

Since your processor cannot manipulate anything smaller than 16 bits => (byte size = 16) => char is 16 bits?