Category Archives: C Programming

Function names and “Clean Code”, part 2

See also: Part 1

Oh, the frustrations of obvious function names that don’t actually help.


First, it’s quite obvious that LED1_On() probably turns LED #1 on, and LED1_Off() turns it off. Simple.

But what is LED #1?

For most of my programming career, I have coded this way. You can always tell folks what that LED is in the commands… But then, of course, you don’t always do that everywhere it is used, and then if it is ever changed, all the old comments lie. (Thank you, Clean Code book, for making me fear comments.)

Today, I treat this kind of like how the OS-9 operating system treated devices… There is a device description (the thing that you are trying to access by name) and a device driver (the thing that controls the thing you are actually trying to access). Under OS-9, a hard disk might have been called “/h0” or “/h1”. A disk drive might have been “/d0” or “/d1”. The device drivers may have been some cryptic name with a chip number in it, such as “fd507” or “rb1003”. But that didn’t matter, because the used didn’t have to know and didn’t have to care. “/d0” was probably the main disk drive on any OS-9 system with a disk drive, regardless of what the actual underlying hardware was.

So for coding these days, I tend to write “low level” driver type functions like LED1_On() and LED2_Off(), and then wrap them in some clean code like:

void StatusLEDOn()

void StatusLEDOff()

That adds a “useless” layer of code, but it keeps things nice and separate. If I see “StatusLEDOn()” in code, I can assume it probably turns on whatever the Status LED is. And if I see that function, I can determine that the Status LED must be LED #1.

And a good optimizer will likely just optimize these “useless” functions away, anyway.

Thanks to that stupid book, now when I see “ReadAI(2)” or “ToggleDO(5)” I just want to dive in and write wrapper functions to make that code look like “GetCurrentTemperature()” or “ToggleStatusLight()”.

I really need to go back and update all my code on GitHub some day…

Function names and “Clean Code”

At a previous job, I was introduced to the concept of “clean code.” As an embedded programmer, where we are often trying to squeeze extra bytes out of already optimized code, many of the principals of clean code do not apply.

But, in general, I like what I have read so far in this recommended book on the subject: (affiliate link)

With this in mind, I wanted to ask a question about function names (whether they be in C, Perl, Java, or whatever).

In the past, I would write my functions using verb/noun, like this:


But after my exposure to object oriented programming, I learned about how classes and methods are used:


This approach makes it very clean which class a function (method?) belongs to.

I started doing my C code similarly, letting the noun of the code lead:


With all the Userlog functions starting with Userlog, I think it makes it clearer where things come from, and helps avoid collisions with outside code (or just code from another engineer on the same team).

With clean code, code should read more like English. Thus, using “InitializeUserlog()” is probably cleaner than “UserlogInit()”. But since you can’t do that with object oriented languages, perhaps writing backwards (Userlog.init(), UserlogInit(), etc.) is accepted due to the limitation of the language.

Indeed, “Userlog.lookupName()” seems less clean than “LookupNameInUserlog()”.

Perhaps I shouldn’t be doing this in C, since I *can* write the functions out more like proper English.

What do you think? Comments appreciated.

C and the dangers of memcpy()


  • 2019-08-12 – Added note about using unions.

In the C programming language, memcpy (memory copy) is a function used to copy a range of bytes from one location in memory to another. I have used it often. Today, my boss mentioned something about not liking memcpy() because of all the dangers of using it. I understand that incorrect addresses, bad calculations, and simple omissions from a copy/paste can cause a memcpy() to have bad results. But, what’s the alternative?

He mentioned that he prefers to use C structures to copy data around, as opposed to using memcpy() to copy blocks of data.

Instead of doing this…

char Buffer1[64];
char Buffer2[64];

// Load Buffer1 with something.
strncpy (Buffer1, "Copy this string over", 64);

// Copy Buffer1 into Buffer2.
memcpy (Buffer2, Buffer1, 64);

That’s a simple, straight-forward example that seems hard to mess up. But doing something like this would be impossible to mess up:

typedef struct
    char Buffer[64];
} CopyStruct64;

CopyStruct64 Buffer1;
CopyStruct64 Buffer2;

// Load Buffer1 structure with something.
strncpy (Buffer1.Buffer, "Copy this string over", 64);

// Copy Buffer1 structure into Buffer2 structure.
Buffer2 = Buffer1;

Interesting. With properly defined structures, copying a block of data from one place to the other is done using code generated by the compiler.

(Update) And it seems I misspoke. He was actually talking about using unions, not structures. I’ve used unions as unions, but never for something like this. I did a similar test. (I realize there are several ways to declare a union, but I am using typedef so it looks similar to the above structure example.)

typedef union
    char Buffer[64];
} CopyUnion64;

CopyUnion64 Buffer1;
CopyUnion64 Buffer2;

// Load Buffer1 structure with something.
strncpy (Buffer1.Buffer, "Copy this string over", 64);

// Copy Buffer1 structure into Buffer2 structure.
Buffer2 = Buffer1;

Using union appears to work the same way, and since my example was not using structure elements, it seems to make more sense to do it that way. (End of Update)

I’m just now wrapping my brain around this, trying to figure out how this approach could be used to avoid using memcpy() and pointers.

What do you think? Am I on the right track to what he was talking about?

Until next time…

Braces! Foiled again!

Just a quick rant, based on how example code from a new compiler I am using is presented.

In C (and similar languages), it is very common to see simple logic presented in one line, such as:

if (AlertLevel == RED) TurnOnSiren();

That is simple to understand, and only takes up one line on the screen or a printout. However, at a previous job, our coding standard forbid that, and always required the use of braces around any actions of an “if”.

if (AlertLevel == RED)

Now, ignoring style preferences of where the braces should go, the use of braces for something like this has some advantages when it comes to preventing potential bugs if this code is changed in the future.

In the case of the compiler examples I have been seeing, they show them without braces, such as:

if (AlertLevel == RED)

For simple and short logic, this is no different than my first example — it just places the “what to do” stuff on an indented line after the “if”.

However, a common bug I see occurs when someone expands upon that like this:

if (AlertLevel == RED)

In some languages, where program flow is determined by indentation, this would work. But in C, indentation does not matter. The above code would actually be processed as:

if (AlertLevel == RED) TurnOnSiren();


The red lights would flash every time through, regardless of the status of AlertLevel.

Unintended consequences of not using braces, and using indents to format each statement on its own line.

By always using braces (even on simple one-statement things like this example), you avoid that:

if (AlertLevel == RED)

Now we have the desired result, as adding new code or function calls inside the braces will work as we intended.

It’s 2019. The C Programming Language came out 47 years ago, and I still occasionally run into bugs caused by the lack of braces.

Thanks to a former job, I no longer make this mistake. I always use braces.

This has been a public service announcement of Sub-Etha Software.

C musing of the day: signed ints

I ran across some code today that puzzled me. It was an infinite loop that used a counter to determine if things took too long. Something like this:

int main()
  int count;
  int status;

  count = 0;

    status = GetStatus();


  } while( status == 0 );

  if (count < 0)
    printf("Time out! count = %d\n", count);

    return EXIT_FAILURE;

  printf("Done. count = %d\n", count);

  return EXIT_SUCCESS;

The code would read some hardware status (“GetStatus() in this example)  and drop out of the do/while loop once it had a value. Inside that loop, it increments a count variable. After done, it would check for “count < 0” and exit with a timeout if that were true.

Count is only incremented, so the only way count could ever be less than zero is if it incremented so high it rolled over. With a signed 8-bit value (int8_t), you count from 0 to 127, then it rolls over to -128 and counts back up to 0.

So with an “int” being a 32-bit value (on the system I am using), the rollover happens at 2147483647.

And that seems like quite a weird value.

But I suppose it it took that long, it certainly timed out.

I think if was going to do it, I would have probably used an unsigned integer, and just specified an appropriate value:

unsigned int count;


if (count > 10000)
  printf("Time out! count = %u\n", count);
  return EXIT_FAILURE;

What says you?

C warnings, %d versus %u and more C fun.

Code cleanup on aisle five…

I recently spent two days at work going through projects to clean up compiler warnings. In GNU C, you can enable options such as “-Wall” (all warnings), “-Wextra” (extra warnings) and “-Werror” (warnings as errors). By doing steps like these, the compiler will scream at you and fail to build code that has warnings in it.

Many of these warnings don’t impact how your code runs. They just ask you “are you sure this is what you are meaning to do?”

For example, if you leave out a “break” in a switch/case block, the compiler can warn you about that:

  x = 1;

  switch( x )
  case 1:
    printf("x is one\n");
    // did I mean to not have a break here?

  case 2:
    printf("x is two\n");

    printf("I don't know what X is\n");

This code would print:

x is one
x is two

…because without the “break” in the “case 1”, the code drops down to the following case. I found several places in our embedded TCP/IP stack where this was being done intentionally, and the author had left comments like “/* falls through below */” to let future observers know their intent. But, with warnings cranked up, it would no longer build for me, even though it was perfectly fine code working as designed.

I found there was a GCC thing you could do where you put in “//no break” as a comment and it removes that warning. I expect that are many more “yes, I really mean to do this” comments GCC supports, but I have not looked in to it yet.

Size (of int) matters

Another issue I would see would be warnings when you used the wrong specifier in a printf. Code might compile fine without warning on a PC, but generate all kinds of warnings on a different architecture where an “int” might be a different size. For example:

int answer = 42;
printf("The answer is %d\n", answer);

On my PC, “%d” can print an “int” type just fine. But, if I had used a “long” data type, it would error out:

long answer = 42;
printf("The answer is %d\n", answer);

This produces this warning/error:

error: format '%d' expects argument of type 'int', but argument 2 has type 'long int' [-Werror=format=]|

You need to use the “l” (long) specifier (“%ld”) to be correct:

long answer = 42;
printf("The answer is %ld\n", answer);

I found that code that compiled without warnings on the PC would not do the same on one of my embedded target devices.

%u versus %d: Fight!

Another warning I had to deal with was printf() and using “%d” versus “%u”. Most code I see always uses %d, which is for a signed value which can be positive or negative. It seems works just fine is you print an unsigned integer type:

unsigned int z;

z = 100;
printf("z is %d\n", z);

Even though the data type for z is unsigned, the value is positive so it prints out a positive number. After all, a signed value can be positive.

But, it is more correct to use “%u” when printing unsigned values. And, here is an example of why it is important to use the proper specifier… Consider this:

#include <limits.h> // for UINT_MAX

unsigned int x;

x = UINT_MAX; // largest unsigned int

printf("x using %%d is %d\n", x);
printf("x using %%u is %u\n", x);

This prints:

x using %d is -1
x using %u is 4294967295

In this case, %d is not giving you what you expect. For a 32-bit int (in this example), ULONG_MAX of 4294967295 is all bits set:

11111111 11111111 11111111 11111111

That represents a -1 if the value was a signed integer, and that’s what %d is told it is. Thus, while %d works fine for smaller values, any value large enough to set that end bit (that represents a negative value for a signed int) will produce incorrect results.

So, yeah, it will work if you *know* you are never printing values that large, but %u would still be the proper one to use when printing unsigned integers… And you won’t get that warning :)

C warning: comparison between signed and unsigned integer expressions [-Wsign-compare]

Trick C question time … what will this print?

#include <stdio.h>
#include <stdlib.h>

int main()
  int x;
  unsigned int y;

  x = -1;
  y = 2;

  printf("x = %d\n", x);
  printf("y = %u\n", y);

  if ( x > y )
    printf("x > y\n");
  else if (x < y)
    printf("x < y\n");
    printf("x == y\n");

  return EXIT_SUCCESS;

I recently began looking in to various compiler warnings in some code I am using, and I found quite a few warnings about comparing signed and unsigned values:

warning: comparison between signed and unsigned integer expressions [-Wsign-compare]

I thought I could safely ignore these, since it seems plausible to compare a signed value with an unsigned value. A signed value of -42 should be less than an unsigned value of 42, right?

In the above example, it will print the following:

x = -1
y = 2
x > y

Nope. I was wrong. According to C, -1 is greater than 2.

C does something that I either never knew, or knew and have long since forgotten. I guess I generally try to write code that has no warnings at all, so I’ve avoided doing this. And now I know (or re-know) the reason why.

When dealing with mis-matched comparisons, C makes them both unsigned. Thus, “-1” becomes whatever -1 would be for that data type.

char  achar  = -1;
short ashort = -1;
int   aint   = -1;
long  along  = -1;

printf("char  -1 as unsigned: %u\n", (unsigned char)achar);
printf("short -1 as unsigned: %u\n", (unsigned short)ashort);
printf("int   -1 as unsigned: %u\n", (unsigned int)aint);
printf("long  -1 as unsigned: %u\n", (unsigned long)along);

This outputs:

char  -1 as unsigned: 255
short -1 as unsigned: 65535
int   -1 as unsigned: 4294967295
long  -1 as unsigned: 4294967295

Thus, on a PC, an 8-bit signed value of -1 is treated as a 255 when comparing against an unsigned value, and a 16-bit as 65535. It seems an int and long as both 32-bits on my system, but these could all be different on other architectures (on Arduino, and int is 16-bits, I believe).

So, without this warning enabled, any comparison that looks correct might be doing something quite wrong.

Warnings are our friends. Even if we hate them and want them to go away.


const-ant confusion in C.


  • 2017-11-30 – Fixing description of MyStructure example. Thanks, Lost Wiz!

Embedded Life

I currently make my living doing embedded C programming. I am not quite sure how to define what “embedded” programming is other than to say: you probably don’t have everything you expect.

You often program on systems without file systems, without gigabytes of RAM and without an operating system to do most of your work for you. For example, at my previous job one of my main platforms (a TI MSP430 processor) had only 10K of RAM and the program was limited to 40K of flash storage. At my current job, I work on several variations of ARM processors, with one configured to give me only 7K of RAM and 36K of program space.

These systems are much closer to an Arduino UNO, which has 32K of flash and 2K of RAM, than a desktop Windows or Linux machine.

Not all embedded systems are this tiny, of course. There are many embedded systems that run Linux, but once you have a full operating system and a file system, the definition of “embedded” seems to be used to just mean “smaller than Windows”.

But I digress…

Over the past six years, I’ve worked on code that was created and maintained by many different programmers before I worked on it. I have learned some cool tricks and also seen some very un-cool tricks (i.e., just wrong). I expect I will be leaving my own cool/un-cool bits of code for future programmers to find.

With that said, there is one item that keeps turning up repeatedly that many of us C programmers don’t seem to really understand because we keep misusing it. And by “we” I include myself.


There is a keyword in C called “const” which, according to Wikipedia, “indicates that the data is read only.” For example, suppose you wrote a function that accepts a string (actually, pointer to a bunch of characters) like this:

void PrintErrorMessage( char *message )
  fputs( "ERROR: ", stderr );
  fputs( message, stderr );

In this example, whatever string passed in will be displayed with “ERROR: ” prepended to it.

int main( int argc, char **argv )
  PrintErrorMessage("File Not Found");

  return EXIT_SUCCESS;

That would display a message to standard error output:

ERROR: File Not Found

But, there is nothing preventing the function from trying* to modify the string that was passed in.

* Key word is “trying”… If that string were embedded in the binary and it was running out of ROM or Flash, attempts to modify it would be rejected by the “hardware can’t do that” exception ;-)

void PrintErrorMessage( char *message )
  fputs( "ERROR: ", stderr );

  // Convert message to uppercase
  for (int i=0; i&lt;strlen(message); i++)
    message[i] = toupper(message[i]);
  fputs( message, stderr );

The intent here would be to display the error message in uppercase, such as “ERROR: FILE NOT FOUND”. This would work if the string being passed in was modifiable, such as:

char message[80];

strcpy(message, "File Not Found");
PrintErrorMessage( message );

…but after returning from that call, the string would have been modified by the function and would now be “FILE NOT FOUND” in memory. This is fine, if that is the intent of the function, but if you do not intend the string to be modified, you can take steps to prevent the function from being able to do it.

Only read this…

“const” will tell the compiler to not allow code to be built that modifies the variable. You see it used all the time by standard C library functions that take strings, such as puts(), strcpy(), etc.

int puts ( const char * str );

For puts() and similar functions, the use of const disallows modifying that string within the function. In the earlier example, we could make the passed-in string “read only” like this:

void PrintErrorMessage( const char *message )
  int i;
  fputs( "ERROR: ", stderr );

  // Convert message to uppercase
  for (int i=0; i&lt;strlen(message); i++)
    message[i] = toupper(message[i]);

  fputs( message, stderr );

With that change made, the compiler now should issue warnings (or errors) on attempts to modify the “message” string inside the function:

error: assignment of read-only location '*(message + (sizetype)((unsigned int)i * 1u))'|

The moment the function tries to modify “message[i]” causes a problem, because “const” has told the compiler whatever is passed in should not be modified.

Because of the usefulness of this extra compile-time error checking, const is a good thing to use.

And many of us do.


There is a bit of confusion in how const works. In the above example, we pass in the pointer to some memory that contains a string. We do not want the memory that is being passed in to be modified, so we use “const char *message”. According to the “C Gibberish” website, that means:

declare message as pointer to const char

We might also want to prevent the pointer itself from being modified by using “const char const * message” … and that would be incorrect. That is not the proper syntax for “const”:

declare message as pointer to const const char

The confusion comes from const allowing two ways of doing the same thing. Did you know that:

const char

…is the same as:

char const

In C, the true syntax seems to be using “const” after the thing you are declaring, like “char is a constant” or “* is a constant”. But, at the start of that line, const can be used at the beginning to mean the same thing, and since we see that all the time, many of us seem to think that const describes what comes after it. Which it doesn’t.

To properly declare that the pointer and the data it points to should be read-only, it should be:

char const * const message;

C Gibberish agrees:

declare message as const pointer to const char

We need to learn this double “before or after is the same thing” use, or only use the “after” use and be consistent.

// declare message as const pointer to const char
const char * const message;

is the same as

// declare message as const pointer to const char
char const * const message;

My most recent encounter with this was in code at work that did something like this:

void Initialize(const MyStructure *config);

They must have intended this to mean “I’m passing in a constant MyStructure pointer which cannot be modified” but, in reality, what they were getting was:

declare config as pointer to const struct MyStructure

They were telling the compiler that the structure being pointed to was read-only and should not be modified. But the function’s purpose was to modify elements of that structure:

config->type = 10;
config->foo = 'a';

Because of this misuse of const, many compiler warnings were generated.

||=== Build: Debug in Const (compiler: GNU GCC Compiler) ===|
main.c||In function 'Initialize':|
main.c|16|error: assignment of member 'type' in read-only object|
main.c|17|error: assignment of member 'foo' in read-only object|
||=== Build failed: 2 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===|

The fix was to correct the prototype and function to do what was actually intended:

// declare config as const pointer to struct MyStructure
void Initialize(MyStructure * const config);

This now disallows the “config” pointer from being changed, but not the structure it points to. Thus, this would not work:

config++; // Increment config pointer.

…but the intended structure modifications would:

config->type = 10;
config->foo = 'a';

I’m sad to say I’ve been using “const” incorrectly as long as I can remember using “const.”

For a future article, I may dive in to some deep const-ant confusion I recently found myself in, and see if someone out there can tell me if I am finally doing it correctly or not.

Until then…

Nested ternary operators in C

I started learning C programming back in the late 1980s. I was using the original Microware OS-9 C compiler on a Radio Shack Color Computer 3. It was a K&R compiler, meaning it was the original version of C that was before the ANSI-C standard. Back then, I recall reading a magazine article that claimed C would be “the language of the 80s” so I decided to see what the fuss was about.

A Commodore-using friend of mine, Mark, was helping me learn C. He loaned me a Pocket C reference guide. That, and the Microware documentation, was all I had for reference material. At the time, Mark had moved from his Commodore 64 to a powerful Amiga computer. He would dial in to my OS-9 BBS system and upload source code he wrote on his Amiga and then compile it on my machine to see if it ran there, too. It was amazing that this was even possible, considering how non-portable BASIC programs were from machine to machine.

It was a fun time.

Over the years, I learned much about C from books and friends and just general experimentation. One thing I learned was this weird conditional assignment operation:

a = (b == 10 ? 100 : 200);

It is basically doing this:

condition ? value_if_true : value_if_false

The value of a would be set to 100 if b was 10, otherwise it would be set to 200. It was a shortcut to writing the code like this:

if (b == 10)
  a = 100;
  a = 200


switch( b )
  case 10:
    a = 100;
    a = 200;

I have used this many times over the years, but don’t even know what it’s called. I asked a coworker, and they told me it was a “ternary operator“. Here is the Wikipedia entry on how it works in C:

It is a great shortcut for response strings. For instance, turning a boolean true/false result in to a string:

printf( "Status: %s\n", (status == true ? "Enabled" : "Disabled" );

This will print “Status: Enabled” if status==true, or “Status: Disabled” if status==false. What a neat shortcut.

Recently, I saw a nested use of this ternary operator. It never dawned on me that you could do this. It was something like this:

char *colorStr = (value == RED) ? "Red" : (value == BLUE) ? "Blue" : "Unknown";

It was using a second ternary operator for the “value_if_false” condition, allowing it to have three conditions rather than just two. I realized you could nest these in many different ways to create rather complex things… Though, readability would likely suffer. I think I’d just stick with simple if/then or switch/case things for anything more than two choices, but in this case it seemed simple enough.

I thought it was neat, and decided I’d share it here in this quick article.