What is C++ equipvalent of Java Long?

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++?

Quote:

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Originally Posted by ictoan

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++?

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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.

Quote:

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Originally Posted by ictoan

Hello!

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

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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

Quote:

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Originally Posted by D_Drmmr

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

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Are you sure about that?

Regards,

Paul McKenzie

Quote:

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Originally Posted by Paul McKenzie

Are you sure about that?

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Yes, by virtue of the minimum range required.

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?

Quote:

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Originally Posted by Paul McKenzie

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

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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.

Quote:

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Originally Posted by 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?

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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.

[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

Quote:

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Originally Posted by Paul McKenzie

Are you sure about that?

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

Quote:

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Originally Posted by Paul McKenzie

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

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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.

Quote:

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Originally Posted by nuzzle

and use,

#define MY_LONG __int64 // VC++ specific

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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
}

Quote:

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Originally Posted by nuzzle

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
}

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Better yet, do a static assert:

Code:

---

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

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

Regards,

Paul McKenzie

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).

Code:

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

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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:

Code:

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

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Most compilers support these.

Quote:

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Originally Posted by monarch_dodra

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).

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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.

Quote:

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Originally Posted by JohnW@Wessex

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.

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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?