Converting arrays/pointers from C to Java

Question

I am trying to re-write a program in C to java. I have no experience in C, but some in C++, so I understand some of the pointer/array things. I am slightly confused though...I am given the following code in C:

void ProcessStatus(Char *Stat){
    DWORD relayDate;
    DWORD APIDate;

    version3=false;
    version4=false;
    version9=false;
    tiltAngle = false;
    ver4features=0;
    tooNew=false;
    ProgRev=0;

    switch(Stat[14]){

From what I understand, the function ProcessStatus is passed a pointer to a char; and I'm assuming since in the last line of the code provided Stat[14] is called its within in array.

So what I'm confused about is how I would pass a pointer to a char within an array in Java.

Any help would be appreciated, even if its helping with my understanding of the C code. Thanks.

the Char* in C is String. So String should work for you in java. — mihail
– mihail, Commented Jul 16, 2012 at 13:23
Is the capital 'C' on 'Char' a typo or deliberate - i.e. is it the C built-in-type char or something else? What? — Rup
– Rup, Commented Jul 16, 2012 at 13:26
@mihail Not quite. A char * in C can be a string, but can also be a generic array to store binary data. Since char in C is 8 bits, and in java char is 16 bits, the correct conversion is a byte array in java 'byte[]'. — Filipe Palrinhas
– Filipe Palrinhas, Commented Jul 16, 2012 at 13:34
Sorry, I wasn't quite descriptive enough. The code actually uses Palm Pilot API/libraries, and the capital C is their definition I'm pretty sure. Either way, its true that the correct conversion is byte[] — JuiCe
– JuiCe, Commented Jul 16, 2012 at 13:37

Alexander Pogrebnyak · Accepted Answer · 2012-07-16 13:24:39Z

4

Hard to say whether it's a string or a raw data that has been passed.

In case of string, use Java's built-in String class.

void ProcessStatus( String stat)
{
  ...

  switch ( stat.charAt( 14 ) )
  {
  } 
}

In case of raw data array, use byte array

void ProcessStatus( byte[] stat)
{
  ...

  switch ( stat[ 14 ] )
  {
  } 
}

BTW, C's char data type is translated to byte type in Java. char type in Java denotes a UTF-16 character which is 2 bytes long. byte is exactly what it is (8-bits, signed).

answered Jul 16, 2012 at 13:24

Alexander Pogrebnyak

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3 Comments

Filipe Palrinhas Over a year ago

You mean stat[14] instead byte[14] in the second ProcessStatus() function, right?

JuiCe Over a year ago

Why would it be byte[14]? Isn't byte the type? I think he's correct calling stat[14]. But I'm the OP and really don't know haha, let me know.

Filipe Palrinhas Over a year ago

@JuiCe: He had "switch ( byte[ 14 ] )" instead of "switch ( stat[ 14 ] )", but already corrected the typo. Now is everything ok! :)

algorowara · Accepted Answer · 2012-07-16 13:22:10Z

2

Since C does not have a built-in String type, a String in C is represented as an array of chars. Passing a pointer to the zeroth element in an array is equivalent, in C, to passing a reference to the array, which is what I think this code is doing. Essentially, replace char* Stat with String Stat and switch(Stat[14]) with switch(Stat.charAt(14))

answered Jul 16, 2012 at 13:22

algorowara

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Comments

Ernest Friedman-Hill · Accepted Answer · 2012-07-16 13:23:33Z

2

First, just to clarify, it's not a char, but a Char; Char with a capital C is some user defined type. It could just be a typedef for char, or it could be something else; we don't know.

Secondly, what's being passed to the method is apparently a pointer to the first element of a block of Char objects. The closest equivalent is a Java array, but depending on how you do this, you may leave the data structure entirely in C, pass the pointer as a long to Java, and let Java manipulate it by calling native functions you provide. A Java array is quite different from an "array" or sequence of contiguously-allocated objects in C.

answered Jul 16, 2012 at 13:23

Ernest Friedman-Hill

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Comments

T.J. Crowder · Accepted Answer · 2012-07-16 13:23:34Z

2

Most of the time, when you see char * (I don't know what your initially-capped Char is) in C or C++, the Java equivalent would be String or char[]. The former is the string class, the latter is a character array.

The equivalent function signature using String would look like this; we use charAt to access the 15th character:

void ProcessStatus(String Stat){
    // ...
    switch (Stat.charAt(14))
    // ...
}

The equivalent with char[] would look like this (and as it's an array, we index into it as you do with char * or char[] in C):

void ProcessStatus(char[] Stat){
    // ...
    switch (Stat[14])
    // ...
}

answered Jul 16, 2012 at 13:23

T.J. Crowder

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4 Comments

Filipe Palrinhas Over a year ago

char in C is 8 bits wide and char in java is 16 bits wide. If char * is a string in C then you can use String in java, but if char * was used to store raw binary data, byte[] should be used instead of char[].

T.J. Crowder Over a year ago

@FilipePalrinhas: It depends on how the function is being called. If the entire codebase is being translated from C to Java, then you'd use char[]. If the function were going to be called from C, or using 8-bit data for any other reason, then of course you're right that it should be byte[]. From the accepted answer, looks like it's not actually string data, so byte[] would be correct for the OP.

Filipe Palrinhas Over a year ago

"If the entire codebase is being translated from C to Java, then you'd use char[]". If the data is a string you can use either char[] or String, but the correct and most natural way to do it in java is using String directly.

T.J. Crowder Over a year ago

@FilipePalrinhas: Er, yes, quite. I sort of meant char[] or String as the case may be.

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