Related
For example:
#include <stdio.h>
int main (void) /* Why int and not short int? - Waste of Memory */
{
printf("Hello World!");
return 0;
}
Why main() is conventional defined with int type, which allocates 4 bytes in memory on 32-bit, if it usually returns only 0 or 1, while other types such as short int (2 bytes,32-bit) or even char (1 byte,32-bit) would be more memory saving?
It is wasting memory space.
NOTE: The question is not a duplicate of the thread given; its answers only correspond to the return value itself but not its datatype at explicit focus.
The Question is for C and C++. If the answers between those alter, share your wisdom with the mention of the context of which language in particular is focused.
Usually assemblers use their registers to return a value (for example the register AX in Intel processors). The type int corresponds to the machine word That is, it is not required to convert, for example, a byte that corresponds to the type char to the machine word.
And in fact, main can return any integer value.
It's because of a machine that's half a century old.
Back in the day when C was created, an int was a machine word on the PDP-11 - sixteen bits - and it was natural and efficient to have main return that.
The "machine word" was the only type in the B language, which Ritchie and Thompson had developed earlier, and which C grew out of.
When C added types, not specifying one gave you a machine word - an int.
(It was very important at the time to save space, so not requiring the most common type to be spelled out was a Very Good Thing.)
So, since a B program started with
main()
and programmers are generally language-conservative, C did the same and returned an int.
There are two reasons I would not consider this a waste:
1 practical use of 4 byte exit code
If you want to return an exit code that exactly describes an error you want more than 8 bit.
As an example you may want to group errors: the first byte could describe the vague type of error, the second byte could describe the function that caused the error, the third byte could give information about the cause of the error and the fourth byte describes additional debug information.
2 Padding
If you pass a single short or char they will still be aligned to fit into a machine word, which is often 4 Byte/32 bit depending on architecture. This is called padding and means, that you will most likely still need 32 bit of memory to return a single short or char.
The old-fashioned convention with most shells is to use the least significant 8 bits of int, not just 0 or 1. 16 bits is increasingly common due to that being the minimum size of an int allowed by the standard.
And what would the issue be with wasting space? Is the space really wasted? Is your computer so full of "stuff" that the remaining sizeof(int) * CHAR_BIT - 8 would make a difference? Could the architecture exploit that and use those remaining bits for something else? I very much doubt it.
So I wouldn't say the memory is at all wasted since you get it back from the operating system when the program finishes. Perhaps extravagent? A bit like using a large wine glass for a small tipple perhaps?
1st: Alone your assumption/statement if it usually returns only 0 or 1 is wrong.
Usually the return code is expected to be 0 if no error occurred but otherwise it can return any number to represent different errors. And most (at least command line programs) do so. Many programs also output negative numbers.
However there are a few common used codes https://www.tldp.org/LDP/abs/html/exitcodes.html also here another SO member points to a unix header that contains some codes https://stackoverflow.com/a/24121322/2331592
So after all it is not just a C or C++ type thing but also has historical reasons how most operating systems work and expect the programs to behave and since that the languages have to support that and so at least C like languages do that by using an int main(...).
2nd:
your conclusion It is wasting memory space is wrong.
Using an int in comparison to a shorter type does not involve any waste.
Memory is usually handled in word-size (that that mean may depend from your architecture) anyway
working with sub-word-types involves computation overheand on some architecture (read: load, word, mask out unrelated bits; store: load memory, mask out variable bits, or them with the new value, write the word back)
the memory is not wasted unless you use it. if you write return 0; no memory is ever used at this point. if you return myMemorySaving8bitVar; you only have 1 byte used (most probable on the stack (if not optimized out at all))
You're either working in or learning C, so I think it's a Real Good Idea that you are concerned with efficiency. However, it seems that there are a few things that seem to need clarifying here.
First, the int data type is not an never was intended to mean "32 bits". The idea was that int would be the most natural binary integer type on the target machine--usually the size of a register.
Second, the return value from main() is meant to accommodate a wide range of implementations on different operating systems. A POSIX system uses an unsigned 8-bit return code. Windows uses 32-bits that are interpreted by the CMD shell as 2's complement signed. Another OS might choose something else.
And finally, if you're worried about memory "waste", that's an implementation issue that isn't even an issue in this case. Return codes from main are typically returned in machine registers, not in memory, so there is no cost or savings involved. Even if there were, saving 2 bytes in the run of a nontrivial program is not worth any developer's time.
The answer is "because it usually doesn't return only 0 or 1." I found this thread from software engineering community that at least partially answers your question. Here are the two highlights, first from the accepted answer:
An integer gives more room than a byte for reporting the error. It can be enumerated (return of 1 means XYZ, return of 2 means ABC, return of 3, means DEF, etc..) or used as flags (0x0001 means this failed, 0x0002 means that failed, 0x0003 means both this and that failed). Limiting this to just a byte could easily run out of flags (only 8), so the decision was probably to use an integer.
An interesting point is also raised by Keith Thompson:
For example, in the dialect of C used in the Plan 9 operating system main is normally declared as a void function, but the exit status is returned to the calling environment by passing a string pointer to the exits() function. The empty string denotes success, and any non-empty string denotes some kind of failure. This could have been implemented by having main return a char* result.
Here's another interesting bit from a unix.com forum:
(Some of the following may be x86 specific.)
Returning to the original question: Where is the exit status stored? Inside the kernel.
When you call exit(n), the least significant 8 bits of the integer n are written to a cpu register. The kernel system call implementation will then copy it to a process-related data structure.
What if your code doesn't call exit()? The c runtime library responsible for invoking main() will call exit() (or some variant thereof) on your behalf. The return value of main(), which is passed to the c runtime in a register, is used as the argument to the exit() call.
Related to the last quote, here's another from cppreference.com
5) Execution of the return (or the implicit return upon reaching the end of main) is equivalent to first leaving the function normally (which destroys the objects with automatic storage duration) and then calling std::exit with the same argument as the argument of the return. (std::exit then destroys static objects and terminates the program)
Lastly, I found this really cool example here (although the author of the post is wrong in saying that the result returned is the returned value modulo 512). After compiling and executing the following:
int main() {
return 42001;
}
on a POSIX compliant my* system, echo $? returns 17. That is because 42001 % 256 == 17 which shows that 8 bits of data are actually used. With that in mind, choosing int ensures that enough storage is available for passing the program's exit status information, because, as per this answer, compliance to the C++ standard guarantees that size of int (in bits)
can't be less than 8. That's because it must be large enough to hold "the eight-bit code units of the Unicode UTF-8 encoding form."
EDIT:
*As Andrew Henle pointed out in the comment:
A fully POSIX compliant system makes the entire int return value available, not just 8 bits. See pubs.opengroup.org/onlinepubs/9699919799/basedefs/signal.h.html: "If si_code is equal to CLD_EXITED, then si_status holds the exit value of the process; otherwise, it is equal to the signal that caused the process to change state. The exit value in si_status shall be equal to the full exit value (that is, the value passed to _exit(), _Exit(), or exit(), or returned from main()); it shall not be limited to the least significant eight bits of the value."
I think this makes for an even stronger argument for the use of int over data types of smaller sizes.
A c++ specific question. So i read a question about what makes a program 32 bit/64 bit, and the anwser it got was something like this (sorry i cant find the question, was somedays ago i looked at it and i cant find it again:( ): As long as you dont make any "pointer assumptions", you only need to recompile it. So my question is, what are pointer assumtions ? To my understanding there is 32 bit pointer and 64 bit pointers so i figure it is something to do with that . Please show the diffrence in code between them. Any other good habits to keep in mind while writing code, that helps it making it easy to convert between the to are also welcome :) tho please share examples with them
Ps. I know there is this post:
How do you write code that is both 32 bit and 64 bit compatible?
but i tougth it was kind of to generall with no good examples, for new programmers like myself. Like what is a 32 bit storage unit ect. Kinda hopping to break it down a bit more (no pun intended ^^ ) ds.
In general it means that your program behavior should never depend on the sizeof() of any types (that are not made to be of some exact size), neither explicitly nor implicitly (this includes possible struct alignments as well).
Pointers are just a subset of them, and it probably also means that you should not try to rely on being able to convert between unrelated pointer types and/or integers, unless they are specifically made for this (e.g. intptr_t).
In the same way you need to take care of things written to disk, where you should also never rely on the size of e.g. built in types, being the same everywhere.
Whenever you have to (because of e.g. external data formats) use explicitly sized types like uint32_t.
For a well-formed program (that is, a program written according to syntax and semantic rules of C++ with no undefined behaviour), the C++ standard guarantees that your program will have one of a set of observable behaviours. The observable behaviours vary due to unspecified behaviour (including implementation-defined behaviour) within your program. If you avoid unspecified behaviour or resolve it, your program will be guaranteed to have a specific and certain output. If you write your program in this way, you will witness no differences between your program on a 32-bit or 64-bit machine.
A simple (forced) example of a program that will have different possible outputs is as follows:
int main()
{
std::cout << sizeof(void*) << std::endl;
return 0;
}
This program will likely have different output on 32- and 64-bit machines (but not necessarily). The result of sizeof(void*) is implementation-defined. However, it is certainly possible to have a program that contains implementation-defined behaviour but is resolved to be well-defined:
int main()
{
int size = sizeof(void*);
if (size != 4) {
size = 4;
}
std::cout << size << std::endl;
return 0;
}
This program will always print out 4, despite the fact it uses implementation-defined behaviour. This is a silly example because we could have just done int size = 4;, but there are cases when this does appear in writing platform-independent code.
So the rule for writing portable code is: aim to avoid or resolve unspecified behaviour.
Here are some tips for avoiding unspecified behaviour:
Do not assume anything about the size of the fundamental types beyond that which the C++ standard specifies. That is, a char is at least 8 bit, both short and int are at least 16 bits, and so on.
Don't try to do pointer magic (casting between pointer types or storing pointers in integral types).
Don't use a unsigned char* to read the value representation of a non-char object (for serialisation or related tasks).
Avoid reinterpret_cast.
Be careful when performing operations that may over or underflow. Think carefully when doing bit-shift operations.
Be careful when doing arithmetic on pointer types.
Don't use void*.
There are many more occurrences of unspecified or undefined behaviour in the standard. It's well worth looking them up. There are some great articles online that cover some of the more common differences that you'll experience between 32- and 64-bit platforms.
"Pointer assumptions" is when you write code that relies on pointers fitting in other data types, e.g. int copy_of_pointer = ptr; - if int is a 32-bit type, then this code will break on 64-bit machines, because only part of the pointer will be stored.
So long as pointers are only stored in pointer types, it should be no problem at all.
Typically, pointers are the size of the "machine word", so on a 32-bit architecture, 32 bits, and on a 64-bit architecture, all pointers are 64-bit. However, there are SOME architectures where this is not true. I have never worked on such machines myself [other than x86 with it's "far" and "near" pointers - but lets ignore that for now].
Most compilers will tell you when you convert pointers to integers that the pointer doesn't fit into, so if you enable warnings, MOST of the problems will become apparent - fix the warnings, and chances are pretty decent that your code will work straight away.
There will be no difference between 32bit code and 64bit code, the goal of C/C++ and other programming languages are their portability, instead of the assembly language.
The only difference will be the distrib you'll compile your code on, all the work is automatically done by your compiler/linker, so just don't think about that.
But: if you are programming on a 64bit distrib, and you need to use an external library for example SDL, the external library will have to also be compiled in 64bit if you want your code to compile.
One thing to know is that your ELF file will be bigger on a 64bit distrib than on a 32bit one, it's just logic.
What's the point with pointer? when you increment/change a pointer, the compiler will increment your pointer from the size of the pointing type.
The contained type size is defined by your processor's register size/the distrib your working on.
But you just don't have to care about this, the compilation will do everything for you.
Sum: That's why you can't execute a 64bit ELF file on a 32bit distrib.
Typical pitfalls for 32bit/64bit porting are:
The implicit assumption by the programmer that sizeof(void*) == 4 * sizeof(char).
If you're making this assumption and e.g. allocate arrays that way ("I need 20 pointers so I allocate 80 bytes"), your code breaks on 64bit because it'll cause buffer overruns.
The "kitten-killer" , int x = (int)&something; (and the reverse, void* ptr = (void*)some_int). Again an assumption of sizeof(int) == sizeof(void*). This doesn't cause overflows but looses data - the higher 32bit of the pointer, namely.
Both of these issues are of a class called type aliasing (assuming identity / interchangability / equivalence on a binary representation level between two types), and such assumptions are common; like on UN*X, assuming time_t, size_t, off_t being int, or on Windows, HANDLE, void* and long being interchangeable, etc...
Assumptions about data structure / stack space usage (See 5. below as well). In C/C++ code, local variables are allocated on the stack, and the space used there is different between 32bit and 64bit mode due to the point below, and due to the different rules for passing arguments (32bit x86 usually on the stack, 64bit x86 in part in registers). Code that just about gets away with the default stacksize on 32bit might cause stack overflow crashes on 64bit.
This is relatively easy to spot as a cause of the crash but depending on the configurability of the application possibly hard to fix.
Timing differences between 32bit and 64bit code (due to different code sizes / cache footprints, or different memory access characteristics / patterns, or different calling conventions ) might break "calibrations". Say, for (int i = 0; i < 1000000; ++i) sleep(0); is likely going to have different timings for 32bit and 64bit ...
Finally, the ABI (Application Binary Interface). There's usually bigger differences between 64bit and 32bit environments than the size of pointers...
Currently, two main "branches" of 64bit environments exist, IL32P64 (what Win64 uses - int and long are int32_t, only uintptr_t/void* is uint64_t, talking in terms of the sized integers from ) and LP64 (what UN*X uses - int is int32_t, long is int64_t and uintptr_t/void* is uint64_t), but there's the "subdivisions" of different alignment rules as well - some environments assume long, float or double align at their respective sizes, while others assume they align at multiples of four bytes. In 32bit Linux, they align all at four bytes, while in 64bit Linux, float aligns at four, long and double at eight-byte multiples.
The consequence of these rules is that in many cases, bith sizeof(struct { ...}) and the offset of structure/class members are different between 32bit and 64bit environments even if the data type declaration is completely identical.
Beyond impacting array/vector allocations, these issues also affect data in/output e.g. through files - if a 32bit app writes e.g. struct { char a; int b; char c, long d; double e } to a file that the same app recompiled for 64bit reads in, the result will not be quite what's hoped for.
The examples just given are only about language primitives (char, int, long etc.) but of course affect all sorts of platform-dependent / runtime library data types, whether size_t, off_t, time_t, HANDLE, essentially any nontrivial struct/union/class ... - so the space for error here is large,
And then there's the lower-level differences, which come into play e.g. for hand-optimized assembly (SSE/SSE2/...); 32bit and 64bit have different (numbers of) registers, different argument passing rules; all of this affects strongly how such optimizations perform and it's very likely that e.g. SSE2 code which gives best performance in 32bit mode will need to be rewritten / needs to be enhanced to give best performance 64bit mode.
There's also code design constraints which are very different for 32bit and 64bit, particularly around memory allocation / management; an application that's been carefully coded to "maximize the hell out of the mem it can get in 32bit" will have complex logic on how / when to allocate/free memory, memory-mapped file usage, internal caching, etc - much of which will be detrimental in 64bit where you could "simply" take advantage of the huge available address space. Such an app might recompile for 64bit just fine, but perform worse there than some "ancient simple deprecated version" which didn't have all the maximize-32bit peephole optimizations.
So, ultimately, it's also about enhancements / gains, and that's where more work, partly in programming, partly in design/requirements comes in. Even if your app cleanly recompiles both on 32bit and 64bit environments and is verified on both, is it actually benefitting from 64bit ? Are there changes that can/should be done to the code logic to make it do more / run faster in 64bit ? Can you do those changes without breaking 32bit backward compatibility ? Without negative impacts on the 32bit target ? Where will the enhancements be, and how much can you gain ?
For a large commercial project, answers to these questions are often important markers on the roadmap because your starting point is some existing "money maker"...
I'm trying to figure out how it is that two variable types that have the same byte size?
If i have a variable, that is one byte in size.. how is it that the computer is able to tell that it is a character instead of a Boolean type variable? Or even a character or half of a short integer?
The processor doesn't know. The compiler does, and generates the appropriate instructions for the processor to execute to manipulate bytes in memory in the appropriate manner, but to the processor itself a byte of data is a byte of data and it could be anything.
The language gives meaning to these things, but it's an abstraction the processor isn't really aware of.
The computer is not able to do that. The compiler is. You use the char or bool keyword to declare a variable and the compiler produces code that makes the computer treat the memory occupied by that variable in a way that makes sense for that particular type.
A 32-bit integer for example, takes up 4 bytes in memory. To increment it, the CPU has an instruction that says "increment a 32-bit integer at this address". That's what the compiler produces and the CPU blindly executes it. It doesn't care if the address is correct or what binary data is located there.
The size of the instruction for incrementing the variable is another matter. It may very well be another 4 or so bytes, but instructions (code) are stored separately from data. There may be many instructions generated for a program that deal with the same location in memory. It is not possible to formally specify the size of the instructions beforehand because of optimizations that may change the number of instructions used for a given operation. The only way to tell is to compile your program and look at the generated assembly code (the instructions).
Also, take a look at unions in C. They let you use the same memory location for different data types. The compiler lets you do that and produces code for it but you have to know what you're doing.
Because you specify the type. C++ is a strongly typed language. You can't write $x = 10. :)
It knows
char c = 0;
is a char because of... well, the char keyword.
The computer only sees 1 and 0. You are in command of what the variable contains.
you can cast that data also into what ever you want.
char foo = 'a';
if ( (bool)(foo) ) // true
{
int sumA = (byte)(foo) + (byte)(foo);
// sumA == (97 + 97)
}
Also look into data casting to look at the memory location as different data types. This can be as small as a char or entire structs.
In general, it can't. Look at the restrictions of dynamic_cast<>, which tries to do exactly that. dynamic_cast can only work in the special case of objects derived from polymorphic base classes. That's because such objects (and only those) have extra data in them. Chars and ints do not have this information, so you can't use dynamic_cast on them.
Background Information: Ultimately, I would like to write an emulator of a real machine such as the original Nintendo or Gameboy. However, I decided that I need to start somewhere much, much simpler. My computer science advisor/professor offered me the specifications for a very simple imaginary processor that he created to emulate first. There is one register (the accumulator) and 16 opcodes. Each instruction consists of 16 bits, the first 4 of which contain the opcode, the rest of which is the operand. The instructions are given as strings in binary format, e.g., "0101 0101 0000 1111".
My Question: In C++, what is the best way to parse the instructions for processing? Please keep my ultimate goal in mind. Here are some points I've considered:
I can't just process and execute the instructions as I read them because the code is self-modifying: an instruction can change a later instruction. The only way I can see to get around this would be to store all changes and for each instruction to check whether a change needs to be applied. This could lead to a massive amounts of comparisons with the execution of each instruction, which isn't good. And so, I think I have to recompile the instructions in another format.
Although I could parse the opcode as a string and process it, there are instances where the instruction as a whole has to be taken as a number. The increment opcode, for example, could modify even the opcode section of an instruction.
If I were to convert the instructions to integers, I'm not sure then how I could parse just the opcode or operand section of the int. Even if I were to recompile each instruction into three parts, the whole instruction as an int, the opcode as an int, and the operand as an int, that still wouldn't solve the problem, as I might have to increment an entire instruction and later parse the affected opcode or operand. Moreover, would I have to write a function to perform this conversion, or is there some library for C++ that has a function convert a string in "binary format" to an integer (like Integer.parseInt(str1, 2) in Java)?
Also, I would like to be able to perform operations such as shifting bits. I'm not sure how that can be achieved, but that might affect how I implement this recompilation.
Thank you for any help or advice you can offer!
Parse the original code into an array of integers. This array is your computer's memory.
Use bitwise operations to extract the various fields. For instance, this:
unsigned int x = 0xfeed;
unsigned int opcode = (x >> 12) & 0xf;
will extract the topmost four bits (0xf, here) from a 16-bit value stored in an unsigned int. You can then use e.g. switch() to inspect the opcode and take the proper action:
enum { ADD = 0 };
unsigned int execute(int *memory, unsigned int pc)
{
const unsigned int opcode = (memory[pc++] >> 12) & 0xf;
switch(opcode)
{
case OP_ADD:
/* Do whatever the ADD instruction's definition mandates. */
return pc;
default:
fprintf(stderr, "** Non-implemented opcode %x found in location %x\n", opcode, pc - 1);
}
return pc;
}
Modifying memory is just a case of writing into your array of integers, perhaps also using some bitwise math if needed.
I think the best approach is to read the instructions, convert them to unsigned integers, and store them into memory, then execute them from memory.
Once you've parsed the instructions and stored them to memory, self-modification is much easier than storing a list of changes for each instruction. You can just change the memory at that location (assuming you don't ever need to know what the old instruction was).
Since you're converting the instructions to integers, this problem is moot.
To parse the opcode and operand sections, you'll need to use bit shifting and masking. For example, to get the op code, you mask off the upper 4 bits and shift down by 12 bits (instruction >> 12). You can use a mask to get the operand too.
You mean your machine has instructions that shift bits? That shouldn't affect how you store the operands. When you get to executing one of those instructions, you can just use the C++ bit-shifting operators << and >>.
Just in case it helps, here's the last CPU emulator I wrote in C++. Actually, it's the only emulator I've written in C++.
The spec's language is slightly idiosyncratic but it's a perfectly respectable, simple VM description, possibly quite similar to your prof's VM:
http://www.boundvariable.org/um-spec.txt
Here's my (somewhat over-engineered) code, which should give you some ideas. For instance it shows how to implement mathematical operators, in the Giant Switch Statement in um.cpp:
http://www.eschatonic.org/misc/um.zip
You can maybe find other implementations for comparison with a web search, since plenty of people entered the contest (I wasn't one of them: I did it much later). Although not many in C++ I'd guess.
If I were you, I'd only store the instructions as strings to start with, if that's the way that your virtual machine specification defines operations on them. Then convert them to integers as needed, every time you want to execute them. It'll be slow, but so what? Yours isn't a real VM that you're going to be using to run time-critical programs, and a dog-slow interpreter still illustrates the important points you need to know at this stage.
It's possible though that the VM actually defines everything in terms of integers, and the strings are just there to describe the program when it's loaded into the machine. In that case, convert the program to integers at the start. If the VM stores programs and data together, with the same operations acting on both, then this is the way to go.
The way to choose between them is to look at the opcode which is used to modify the program. Is the new instruction supplied to it as an integer, or as a string? Whichever it is, the simplest thing to start with is probably to store the program in that format. You can always change later once it's working.
In the case of the UM described above, the machine is defined in terms of "platters" with space for 32 bits. Clearly these can be represented in C++ as 32-bit integers, so that's what my implementation does.
I created an emulator for a custom cryptographic processor. I exploited the polymorphism of C++ by creating a tree of base classes:
struct Instruction // Contains common methods & data to all instructions.
{
virtual void execute(void) = 0;
virtual size_t get_instruction_size(void) const = 0;
virtual unsigned int get_opcode(void) const = 0;
virtual const std::string& get_instruction_name(void) = 0;
};
class Math_Instruction
: public Instruction
{
// Operations common to all math instructions;
};
class Branch_Instruction
: public Instruction
{
// Operations common to all branch instructions;
};
class Add_Instruction
: public Math_Instruction
{
};
I also had a couple of factories. At least two would be useful:
Factory to create instruction from
text.
Factory to create instruction from
opcode
The instruction classes should have methods to load their data from an input source (e.g. std::istream) or text (std::string). The corollary methods of output should also be supported (such as instruction name and opcode).
I had the application create objects, from an input file, and place them into a vector of Instruction. The executor method would run the 'execute()` method of each instruction in the array. This action trickled down to the instruction leaf object which performed the detailed execution.
There are other global objects that may need emulation as well. In my case some included the data bus, registers, ALU and memory locations.
Please spend more time designing and thinking about the project before you code it. I found it quite a challenge, especially implementing a single-step capable debugger and GUI.
Good Luck!
In C++, I'm wondering why the bool type is 8 bits long (on my system), where only one bit is enough to hold the boolean value ?
I used to believe it was for performance reasons, but then on a 32 bits or 64 bits machine, where registers are 32 or 64 bits wide, what's the performance advantage ?
Or is it just one of these 'historical' reasons ?
Because every C++ data type must be addressable.
How would you create a pointer to a single bit? You can't. But you can create a pointer to a byte. So a boolean in C++ is typically byte-sized. (It may be larger as well. That's up to the implementation. The main thing is that it must be addressable, so no C++ datatype can be smaller than a byte)
Memory is byte addressable. You cannot address a single bit, without shifting or masking the byte read from memory. I would imagine this is a very large reason.
A boolean type normally follows the smallest unit of addressable memory of the target machine (i.e. usually the 8bits byte).
Access to memory is always in "chunks" (multiple of words, this is for efficiency at the hardware level, bus transactions): a boolean bit cannot be addressed "alone" in most CPU systems. Of course, once the data is contained in a register, there are often specialized instructions to manipulate bits independently.
For this reason, it is quite common to use techniques of "bit packing" in order to increase efficiency in using "boolean" base data types. A technique such as enum (in C) with power of 2 coding is a good example. The same sort of trick is found in most languages.
Updated: Thanks to a excellent discussion, it was brought to my attention that sizeof(char)==1 by definition in C++. Hence, addressing of a "boolean" data type is pretty tied to the smallest unit of addressable memory (reinforces my point).
The answers about 8-bits being the smallest amount of memory that is addressable are correct. However, some languages can use 1-bit for booleans, in a way. I seem to remember Pascal implementing sets as bit strings. That is, for the following set:
{1, 2, 5, 7}
You might have this in memory:
01100101
You can, of course, do something similar in C / C++ if you want. (If you're keeping track of a bunch of booleans, it could make sense, but it really depends on the situation.)
I know this is old but I thought I'd throw in my 2 cents.
If you limit your boolean or data type to one bit then your application is at risk for memory curruption. How do you handle error stats in memory that is only one bit long?
I went to a job interview and one of the statements the program lead said to me was, "When we send the signal to launch a missle we just send a simple one bit on off bit via wireless. Sending one bit is extremelly fast and we need that signal to be as fast as possible."
Well, it was a test to see if I understood the concepts and bits, bytes, and error handling. How easy would it for a bad guy to send out a one bit msg. Or what happens if during transmittion the bit gets flipped the other way.
Some embedded compilers have an int1 type that is used to bit-pack boolean flags (e.g. CCS series of C compilers for Microchip MPU's). Setting, clearing, and testing these variables uses single-instruction bit-level instructions, but the compiler will not permit any other operations (e.g. taking the address of the variable), for the reasons noted in other answers.
Note, however, that std::vector<bool> is allowed to use bit-packing, i.e. to store the bits in smaller units than an ordinary bool. But it is not required.