Array of Constants Initialization

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08-29-2008
danlee58

Array of Constants Initialization

I am trying to initialize an array of Constants for an embedded computer. It's a lookup table for a non-linear function. I can't seem to get it to compile correctly. Here is the simplified Syntax:

Code:

const int junk [3][2] = {10, 2. 5, 4, 1, 6};

This is how it's used:

Code:

int trsh; int trsh2; trsh = 3; //This value is input from an A/D conveter. if (trsh == junk[0][0]) (trsh2 = junk[0][1]; else if (trsh >= junk[1][0]) (trsh2 = junk[1][1]; else trsh2 = junk[2][1];

I actually extrapolate between the elements to determine the value of trsh2, but I didn't do it here for simplification.
08-29-2008
matsp

Your posted code will not compile due to several syntax errors (parenthesis where I beleive you mean braces, a dot instead of comma in the intializer list for the array).

Also, a 2D array would normally have two levels of braces, e.g.

Code:

const int array[3][2] = { { 0, 1 } , { 2, 3 }, {3, 4} };

At the very least, gcc will give you a warning if you don't have the extra braces.

--
Mats
08-29-2008
QuantumPete

You are also not indenting your code and your if/else statements are missing braces.

QuantumPete
08-29-2008
Elysia

Code:

if (trsh == junk[0][0]) trsh2 = junk[0][1]; else { if (trsh >= junk[1][0]) trsh2 = junk[1][1]; else trsh2 = junk[2][1]; }

This is more like how it should look like.
08-29-2008
kcpilot

why the extra braces for the this:

Code:

else { if (trsh >= junk[1][0]) trsh2 = junk[1][1]; else trsh2 = junk[2][1]; }

you don't need that!
08-29-2008
laserlight

Quote:

why the extra braces

To improve clarity.
08-29-2008
kcpilot

what clarity???
08-29-2008
laserlight

The clarity of the code, of course, in addition to what the indentation provides. Put it another way: why the indentation? You don't need that!

Code:

if (trsh == junk[0][0]) trsh2 = junk[0][1]; else if (trsh >= junk[1][0]) trsh2 = junk[1][1]; else trsh2 = junk[2][1];

EDIT:
On second thought, perhaps what you mean is that we can indent like this:

Code:

if (trsh == junk[0][0]) trsh2 = junk[0][1]; else if (trsh >= junk[1][0]) trsh2 = junk[1][1]; else trsh2 = junk[2][1];

Admittedly, in answering your question, this did not cross my mind since your focus was on the braces, but we certainly could write:

Code:

if (trsh == junk[0][0]) { trsh2 = junk[0][1]; } else if (trsh >= junk[1][0]) { trsh2 = junk[1][1]; } else { trsh2 = junk[2][1]; }
08-29-2008
kcpilot

EXACTLY! And you don't need the braces there either (not even for clarity!)
08-29-2008
tabstop

I'm guessing the braces are there because the actual code is bigger than one line (as hinted at by OP).
08-29-2008
laserlight

Quote:

Originally Posted by kcpilot

And you don't need the braces there either (not even for clarity!)

I agree, but of course it may be advised by a style guideline motivated by clarity and future proofing against misleading additions of statements, or simply for consistency.
08-29-2008
Elysia

Quote:

Originally Posted by kcpilot

why the extra braces for the this:

Code:

else { if (trsh >= junk[1][0]) trsh2 = junk[1][1]; else trsh2 = junk[2][1]; }

you don't need that!

Because it's more than one line.
When it is, I usually add them for more clarity.
It's purely a style question.
08-29-2008
vart

I always add braces, It helps to avoid erros in the code like

Code:

if(x) if(y) statement1; else statement2;

as well as prevents adding errors when the additional statements are added, or when the statement is the macro resolved to some construction like
if(p) puts(p);
08-29-2008
danlee58

Thanks for the replys. The dot inplace of the comma was a typo in my example. My ANSI C book does not show two levels of braces for multi dimensional arrays. I will try that and let you know.
08-29-2008
whoie

You don't need all those extra braces for the initializer though. They are entirely optional, although gcc does complain about it from what I remember. Your compiler shouldn't be bombing for that, especially since you don't have too many.
08-29-2008
danlee58

Yes, it did compile when I got all the syntax right, but it gave me warnings about the array initialization. I put the internal braces betrween the array elements and now I have no warnings or errors.

There is another problem. I get a data abort message when the programs runs. I may have some illegal math operation. I have to look into that before I can ask for help.

Thanks for your help.
08-30-2008
danlee58

I got rid of the data abort message, it was due to a while(1) being cleared. It is valid in the actual code, but not for my testing.

My original example was simplified. My actual program statements are more complex, and I am having a difficult time with Braces in:

Code:

int x; int y; const int p[4][2] = {(1,2), (3,4), (5,6), (7,8)}; if (x >= p[0][0]) (y=p[0][1]); else {if (x>=p[1][0]) (y=(p[1][1] * (1+(x/(p[1][0]-p[0][0]))))); else {if (x>=p[2][0]) (y=(p[2][1] * (1+(x/(p[2][0]-p[1][0]))))); else {if (x>=p[3][0]) (y=(p[3][1] * (1+(x/(p[3][0]-p[2][0]))))); else {if (x>=p[4][0]) (y=(p[4][1] * (1+(x/(p[4][0]-p[3][0]))))); else (x = 154);}}}}
08-30-2008
Elysia

Quote:

Originally Posted by danlee58

I got rid of the data abort message, it was due to a while(1) being cleared. It is valid in the actual code, but not for my testing.

My original example was simplified. My actual program statements are more complex, and I am having a difficult time with Braces in:

Code:

int x; int y; const int p[4][2] = {(1,2), (3,4), (5,6), (7,8)}; if (x >= p[0][0]) (y=p[0][1]); else {if (x>=p[1][0]) (y=(p[1][1] * (1+(x/(p[1][0]-p[0][0]))))); else {if (x>=p[2][0]) (y=(p[2][1] * (1+(x/(p[2][0]-p[1][0]))))); else {if (x>=p[3][0]) (y=(p[3][1] * (1+(x/(p[3][0]-p[2][0]))))); else {if (x>=p[4][0]) (y=(p[4][1] * (1+(x/(p[4][0]-p[3][0]))))); else (x = 154)}}}};

It's really easy.

Code:

int x; int y; const int p[4][2] = { {1, 2}, {3, 4}, {5, 6}, {7, 8} }; if (x >= p[0][0]) y=p[0][1]; else { if (x >= p[1][0]) { y = ( p[1][1] * ( 1 + ( x / (p[1][0] - p[0][0]) ) ) ); } else { if (x >= p[2][0]) { y = ( p[2][1] * ( 1 + ( x / (p[2][0] - p[1][0]) ) ) ); } else { if (x >= p[3][0]) { y = ( p[3][1] * ( 1 + ( x / (p[3][0] - p[2][0]) ) ) ); } else { if (x >= p[4][0]) { y = ( p[4][1] * ( 1 + ( x / (p[4][0] - p[3][0]) ) ) ); } else x = 154; } } } }

Every IF begins with { and ends with }.
Every line must be terminated with ;. Use more whitespace to make it more visible.
This may seem like a huge amount of whitespace, but it's necessary to make your convulted ifs readable.
08-31-2008
danlee58

Elysia,

Thanks for the help. That worked. Part of my difficulty is that I have a hard time telling the difference between a Brace and a Parenthesis.

I had some other problems in the code, but I was able to fix them. I had to declare the const array a floating point, since the results of a calculation is fractional, and would return zero and have no effect if left as an integer.

Now that section of code is working, I will go on to the next, but much of what I have to do is similar to this section, but with larger arrays and with more dimensions.
08-31-2008
matsp

Quote:

Originally Posted by Elysia

Every line must be terminated with ;

Wilst most of the time a line and a statement is the same thing, it is not a line that must end with a ; - a statment ends with ;

There are several examples in the posted code where the same statement covers more than one line - if-statements are commonly written over two or more lines, where the first line is the if (condition), and the second line contains the "then do" part of the if-statement. Note that an "else" is a separate statement [but it must follow immediately after the if-"then-do" block.

--
Mats