C/CAPL defining one array from another

Question

this is my first question here, please notice I'm very new to coding. Quick search doesn't helped me since I think the answer might be too simple.

Im writing some code in CAPL (a CANoe specific language based on C). Lets have my scenario simplified: I have to read 10 values Input, but I'm just able to read one by a time (due to HW reasons).

My value is stored in a buffer (rsBuf), now I'm trying to define a help-array for everytime I read the value (rsBuf1 ... 10). At the end I will create another array with added values of rsBuf1 ... rsBuf10.

for every "read-action", I want to define rsBuf1 = rsBuf; rsBuf2 = rsBuf; and so on...

error: for "rsBuf1 = rsBuf;" Error 1112 at (732,16): operand types are incompatible. Compilation failed -- 1 errors, 0 warnings

my "real" values:

variables
{
byte rsBuf[1024];
byte rsBuf1[1024];
}

is there an easy way to do this one-array-from-another? I also tried some other notations I found, like rsBuf1 = {rsBuf}, but wasn't helping. Of course I could define rsBuf1[1]=rsBuf[1]; ... rsBuf1[1024]=rsBuf[1024];

but that would be a waste of time I guess. Thanks in advance, cheers Robert

have you tried `memcpy()` ? :) something along the lines of `memcpy(rsBuf1, rsBuf, 1024);` — Shark, Mar 03 '16 at 10:33
Hi, thanks, I just tried http://www.tutorialspoint.com/c_standard_library/c_function_memcpy.htm: memcpy(rsBuf1, rsBuf, strlen(rsBuf)+1); // same error: Error 1154 at (732,31): types of parameters do not match. Error 1112 at (733,16): operand types are incompatible. ******************** thanks, now tried with 1024, its working! — Robertk2016, Mar 03 '16 at 10:39

score 2 · Accepted Answer · answered Mar 03 '16 at 10:40

You can't copy arrays through assignment in C, because the syntax does not allow it. The best solution is to use the memcpy function.

Alternatively, if it makes sense for the program design, you could put the arrays inside a wrapper struct:

typedef struct
{
  int array [5];
} array_wrapper_t;

array_wrapper_t this = {1,2,3,4,5};
array_wrapper_t that;
that = this;

This should yield identical machine code as a call to memcpy.

score 1 · Answer 2 · answered Mar 07 '16 at 16:47

Option 1

You can use memcpy as follows to copy the entire array:

memcpy(rsBuf1, rsBuf, elcount(rsBuf));

The following would copy the first four bytes:

memcpy(rsBuf1, rsBuf, 4);

Option 2

Optionally you could define your copy-function, see the copy_buffer() implementation:

variables
{
  byte rsBuf[1024] = {1, 2, 3, 4, 5, 6, 7, 8};
  byte rsBuf1[1024]; 
}

on start
{
  print_buffer(rsBuf, 8);
  print_buffer(rsBuf1, 8);

  copy_buffer(rsBuf1, rsBuf, elcount(rsBuf));

  print_buffer(rsBuf1, 8);
}

void copy_buffer(byte dest[], byte src[], long len)
{
  int i;
  for (i = 0; i < len; i++)
    dest[i] = src[i];
}

void print_buffer(byte buf[], byte len)
{
  int i;
  writeLineEx(0, 0, ">");
  for (i = 0; i < len; i++)
    writeEx(0, 0, " 0x%02x", buf[i]);
}

C/CAPL defining one array from another

2 Answers2

Option 1

Option 2