C programming questions from a java programmer

Question

So I'm in the need to translate a C library to pure java, so far its looking good, but I'm stuck here.

Can someone explain to me what does the following pointer is for?

double *DTimeSigBuf[MAX_TIME_CHANNELS];

Ok I know it is a double type pointer named DTimeSigBuf, but whats that in the brackets? also MAX_TIME_CHANNELS is defined in the h file as:

 #define MAX_TIME_CHANNELS 2

then in the code this constant value changes, like its pointing somewhere else, but I dont know what does exactly means. is it equivalent to say:

double *DTimeSigBuf = MAX_TIME_CHANNELS;

if I recall well there was something similar in assembler, like: mov [BX], CL called Indirect addressing mode register, does this have anything to do with this? I know I might be completly lost! because as the title says, I'm a java programmer.

And the other question, what is the effect of doing this:

DTimeSigBuf[chanNum]            = (double*)malloc(block_size_samples*sizeof(double));

Where block_size_samples is int and chanNum is an for iterator variable?

Please help! I sware I've been googling the whole time.

Thanks folks :)

Doesn't Java also have brackets? They serve a similar purpose in C. — Kerrek SB
– Kerrek SB, Commented Mar 20, 2012 at 20:03
the questions are basic enough that instead of asking individual questions that you should look at a tutorial first - especially looking at arrays — mmmmmm
– mmmmmm, Commented Mar 20, 2012 at 20:03

piokuc · Accepted Answer · 2012-03-20 20:09:50Z

5

It is an array of pointers to double. MAX_TIME_CHANNELS is size of the array.

The effect of the statement with malloc is allocation of a block of memory large enough for block_size_samples double values; address of the block of memory is then assigned to chanNum element of the DTimeSigBuf array.

edited Mar 20, 2012 at 20:09

answered Mar 20, 2012 at 20:03

piokuc

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1 Comment

LuisEspinoza Over a year ago

By the way #define MAX_TIME_CHANNELS 2, will cause that the precompiler will replace with 2 every ocurrece of MAX_TIME_CHANNELS, so the declaration will be compiles as double *DTimeSigBuf[2];

qxn · Accepted Answer · 2012-03-20 21:37:05Z

3

DTimeSigBuf is an array of pointers to type double. This could be thought of as an array of arrays of type double.

double *DTimeSigBuf[MAX_TIME_CHANNELS];

could be thought of as

double DTimeSigBuf[MAX_TIME_CHANNELS][]

The line

DTimeSigBuf[chanNum] = (double*)malloc(block_size_samples*sizeof(double));

is allocating memory for block_size_samples number of variables of type double to be placed in the array pointed at by DTimeSigBuf[chanNum].

For example:

If block_size_samples is 4 and chanNum is 1, you could think of it this way:

DTimeSigBuf[1] = new double[4];

edited Mar 20, 2012 at 21:37

answered Mar 20, 2012 at 20:12

qxn

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1 Comment

John Bode Over a year ago

Nit: it's an array of pointer to double, not a pointer to an array of double.

Darren · Accepted Answer · 2012-03-20 20:07:06Z

3

Its an array pointer of type double. MAX_CHANNEL_TIMES is a constant and also the array size

answered Mar 20, 2012 at 20:07

Darren

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Java42 · Accepted Answer · 2012-03-20 20:19:45Z

3

If you have C code like:

#define MAX_TIME_CHANNELS 2 
double *DTimeSigBuf[MAX_TIME_CHANNELS];

In Java it looks like:

final static int MAX_TIME_CHANNELS = 2;
double DTimeSigBuf[][] = new double[MAX_TIME_CHANNELS][];

And this in C:

DTimeSigBuf[chanNum] = (double*)malloc(block_size_samples*sizeof(double));

is allocating space for the y dimension.

In Java it is:

DTimeSigBuf[chanNum] = new double[block_size_samples];

answered Mar 20, 2012 at 20:19

Java42

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Adrian McCarthy · Accepted Answer · 2012-03-20 20:04:34Z

2

DTimeSigBuf is an array of pointers to doubles.

The allocation is allocating an array of doubles. That is, the pointer that's returned is a pointer to the first double in an array of block_size_samples doubles.

answered Mar 20, 2012 at 20:04

Adrian McCarthy

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Comments

Adrian McCarthy · Accepted Answer · 2012-03-20 21:32:13Z

1

As previously answered, the first sections are declaring an array of pointers to doubles. Since declaration does not necessarily allocate memory in C, the third line is allocating space for a new row of doubles.

Breaking it down:

DTimeSigBuf[chanNum] // chanNum is the position in the array
= // equals
(double*) // memory address to a double
malloc(  // get some memory from the system
block_size_samples*sizeof(double)); // number of samples times memory needed for one double

edited Mar 20, 2012 at 21:32

Adrian McCarthy

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answered Mar 20, 2012 at 20:14

Spencer Rathbun

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