How to declare a C function with an undetermined return type? - objective-c

Can I declare a C function with an undetermined return type (without C compiler warning)? The return type could be int, float, double, void *, etc.
undetermined_return_type miscellaneousFunction(undetermined_return_type inputValue);
And you can use this function in other functions to return a value (although that could be a run time error):
BOOL isHappy(int feel){
return miscellaneousFunction(feel);
};
float percentage(float sales){
return miscellaneousFunction(sales);
};
What I'm looking for:
To declare and to implement a C function (or Obj-C method) with an undefined-return-type could be useful for aspect-oriented programming.
If I could intercept Obj-C messages in another function in run time, I might return the value of that message to the original receiver or not with doing something else action. For example:
- (unknown_return_type) interceptMessage:(unknown_return_type retValOfMessage){
// I may print the value here
// No idea how to print the retValOfMessage (I mark the code with %???)
print ("The message has been intercepted, and the return value of the message is %???", retValOfMessage);
// Or do something you want (e.g. lock/unlock, database open/close, and so on).
// And you might modify the retValOfMessage before returning.
return retValOfMessage;
}
So I can intercept the original message with a little addition:
// Original Method
- (int) isHappy{
return [self calculateHowHappyNow];
}
// With Interception
- (int) isHappy{
// This would print the information on the console.
return [self interceptMessage:[self calculateHowHappyNow]];
}

You can use a void * type.
Then for example:
float percentage(float sales){
return *(float *) miscellaneousFunction(sales);
}
Be sure not to return a pointer to a object with automatic storage duration.

You may use the preprocessor.
#include <stdio.h>
#define FUNC(return_type, name, arg) \
return_type name(return_type arg) \
{ \
return miscellaneousFunction(arg); \
}
FUNC(float, undefined_return_func, arg)
int main(int argc, char *argv[])
{
printf("\n %f \n", undefined_return_func(3.14159));
return 0;
}

May be a union as suggested by thejh
typedef struct
{
enum {
INT,
FLOAT,
DOUBLE
} ret_type;
union
{
double d;
float f;
int i;
} ret_val;
} any_type;
any_type miscellaneousFunction(any_type inputValue) {/*return inputValue;*/}
any_type isHappy(any_type feel){
return miscellaneousFunction(feel);
}
any_type percentage(any_type sales){
return miscellaneousFunction(sales);
}
Here with ret_type you can know data type of return value and ret_type. i,f,d can give you corresponding value.
All elements will use same memory space and only one should be accessed.

Straight C doesn't support dynamically-typed variables (variants) since it is statically typed, but there might be some libraries that do what you want.

Related

How do I get the Return Stmt of objective-c through clang-3.9?

I am a rookie in doing the static-analysis of objective-c through clang now.
I face a problem that when I find the ReturnStmt through RecursiveASTVisitor ,clang sometimes can not find the ReturnStmt.
The RecursiveASTVisitor code like this:
class MyASTVisitor : public RecursiveASTVisitor<MyASTVisitor> {
public:
MyASTVisitor(Rewriter &R) : TheRewriter(R) {}
.........
else if(isa<ReturnStmt>(s)){
//The Return Stmt find block
ReturnStmt *returnStat = cast<ReturnStmt>(s);
TheRewriter.InsertText(returnStat->getLocStart(),"//the return stmt\n",true,true);
}
return true;
}}
And that's the result
The first result can find the return stmt
int main (int argc, const char* argv[]) {
#autoreleasepool {
//the func--->NSLog() begin called!
NSLog (#"Programming is fun!");
}
//the return stmt
return 0; }
But the second can not find it
int main(int argc, char * argv[]) {
#autoreleasepool {
return UIApplicationMain(argc, argv, nil, NSStringFromClass([AppDelegate class]));
}}
Let me explain how clang AST works with a small example. Assume this code exists in temp.cpp
int b()
{
return 0;
}
int main()
{
return b();
}
Now let's see what is the clang's AST representation of the above code is: (By the way, this is something you can do anytime you don't see your code doing what it should. Look at the raw AST to see what is wrong. To get the raw AST dump from clang we will run this.
clang -Xclang -ast-dump -fsyntax-only temp.cpp
This gives us this output:
|-FunctionDecl 0x5f952e0 <t.cpp:2:1, line:5:1> line:2:5 used b 'int (void)'
| `-CompoundStmt 0x5f95400 <line:3:1, line:5:1>
| `-ReturnStmt 0x5f953e8 <line:4:2, col:9>
| `-IntegerLiteral 0x5f953c8 <col:9> 'int' 0
`-FunctionDecl 0x5f95440 <line:7:1, line:10:1> line:7:5 main 'int (void)'
`-CompoundStmt 0x5f95620 <line:8:1, line:10:1>
`-ReturnStmt 0x5f95608 <line:9:2, col:11>
`-CallExpr 0x5f955e0 <col:9, col:11> 'int'
`-ImplicitCastExpr 0x5f955c8 <col:9> 'int (*)(void)' <FunctionToPointerDecay>
`-DeclRefExpr 0x5f95570 <col:9> 'int (void)' lvalue Function 0x5f952e0 'b' 'int (void)'
If you look at the first FunctionDecl for function b, it's a simple return statement which returns the integer value 0. But, now if you look at the FunctionDecl for main, you can see that returnStmt calls CallExpr which then gets the return from function b. This is exactly what is happening in your case. One return statement is getting detected and one is not.
What you can do in this case is call getRetValue() of ReturnStmt which will give you an Expr type and you need to resolve that for different possible return cases.

Determine types from a variadic function's arguments in C

I'd like a step by step explanation on how to parse the arguments of a variadic function
so that when calling va_arg(ap, TYPE); I pass the correct data TYPE of the argument being passed.
Currently I'm trying to code printf.
I am only looking for an explanation preferably with simple examples but not the solution to printf since I want to solve it myself.
Here are three examples which look like what I am looking for:
https://stackoverflow.com/a/1689228/3206885
https://stackoverflow.com/a/5551632/3206885
https://stackoverflow.com/a/1722238/3206885
I know the basics of what typedef, struct, enum and union do but can't figure out some practical application cases like the examples in the links.
What do they really mean? I can't wrap my brain around how they work.
How can I pass the data type from a union to va_arg like in the links examples? How does it match?
with a modifier like %d, %i ... or the data type of a parameter?
Here's what I've got so far:
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include "my.h"
typedef struct s_flist
{
char c;
(*f)();
} t_flist;
int my_printf(char *format, ...)
{
va_list ap;
int i;
int j;
int result;
int arg_count;
char *cur_arg = format;
char *types;
t_flist flist[] =
{
{ 's', &my_putstr },
{ 'i', &my_put_nbr },
{ 'd', &my_put_nbr }
};
i = 0;
result = 0;
types = (char*)malloc( sizeof(*format) * (my_strlen(format) / 2 + 1) );
fparser(types, format);
arg_count = my_strlen(types);
while (format[i])
{
if (format[i] == '%' && format[i + 1])
{
i++;
if (format[i] == '%')
result += my_putchar(format[i]);
else
{
j = 0;
va_start(ap, format);
while (flist[j].c)
{
if (format[i] == flist[j].c)
result += flist[i].f(va_arg(ap, flist[i].DATA_TYPE??));
j++;
}
}
}
result += my_putchar(format[i]);
i++;
}
va_end(ap);
return (result);
}
char *fparser(char *types, char *str)
{
int i;
int j;
i = 0;
j = 0;
while (str[i])
{
if (str[i] == '%' && str[i + 1] &&
str[i + 1] != '%' && str[i + 1] != ' ')
{
i++;
types[j] = str[i];
j++;
}
i++;
}
types[j] = '\0';
return (types);
}
You can't get actual type information from va_list. You can get what you're looking for from format. What it seems you're not expecting is: none of the arguments know what the actual types are, but format represents the caller's idea of what the types should be. (Perhaps a further hint: what would the actual printf do if a caller gave it format specifiers that didn't match the varargs passed in? Would it notice?)
Your code would have to parse the format string for "%" format specifiers, and use those specifiers to branch into reading the va_list with specific hardcoded types. For example, (pseudocode) if (fspec was "%s") { char* str = va_arg(ap, char*); print out str; }. Not giving more detail because you explicitly said you didn't want a complete solution.
You will never have a type as a piece of runtime data that you can pass to va_arg as a value. The second argument to va_arg must be a literal, hardcoded specification referring to a known type at compile time. (Note that va_arg is a macro that gets expanded at compile time, not a function that gets executed at runtime - you couldn't have a function taking a type as an argument.)
A couple of your links suggest keeping track of types via an enum, but this is only for the benefit of your own code being able to branch based on that information; it is still not something that can be passed to va_arg. You have to have separate pieces of code saying literally va_arg(ap, int) and va_arg(ap, char*) so there's no way to avoid a switch or a chain of ifs.
The solution you want to make, using the unions and structs, would start from something like this:
typedef union {
int i;
char *s;
} PRINTABLE_THING;
int print_integer(PRINTABLE_THING pt) {
// format and print pt.i
}
int print_string(PRINTABLE_THING pt) {
// format and print pt.s
}
The two specialized functions would work fine on their own by taking explicit int or char* params; the reason we make the union is to enable the functions to formally take the same type of parameter, so that they have the same signature, so that we can define a single type that means pointer to that kind of function:
typedef int (*print_printable_thing)(PRINTABLE_THING);
Now your code can have an array of function pointers of type print_printable_thing, or an array of structs that have print_printable_thing as one of the structs' fields:
typedef struct {
char format_char;
print_printable_thing printing_function;
} FORMAT_CHAR_AND_PRINTING_FUNCTION_PAIRING;
FORMAT_CHAR_AND_PRINTING_FUNCTION_PAIRING formatters[] = {
{ 'd', print_integer },
{ 's', print_string }
};
int formatter_count = sizeof(formatters) / sizeof(FORMAT_CHAR_AND_PRINTING_FUNCTION_PAIRING);
(Yes, the names are all intentionally super verbose. You'd probably want shorter ones in the real program, or even anonymous types where appropriate.)
Now you can use that array to select the correct formatter at runtime:
for (int i = 0; i < formatter_count; i++)
if (current_format_char == formatters[i].format_char)
result += formatters[i].printing_function(current_printable_thing);
But the process of getting the correct thing into current_printable_thing is still going to involve branching to get to a va_arg(ap, ...) with the correct hardcoded type. Once you've written it, you may find yourself deciding that you didn't actually need the union nor the array of structs.

Static Parameter Function Specialization in D

I've read somewhere that D supports specialization of functions to calls where arguments are compile-time constants. Typical use of this is in matrix power functions (if exponent is 2 x*x is often faster than the general case).
I want this in my member function
bool opIndexAssign(bool b, size_t i) #trusted pure nothrow in {
assert(i < len); // TODO: Add static assert(i < len) when i is constant
} body {
b ? bts(ptr, i) : btr(ptr, i);
return b;
}
of a statically sized BitSet struct I'm writing. This in order to, when possible, get compile-time bounds checking on the index variable i. I thought
bool opIndexAssign(bool b, const size_t i) #trusted pure nothrow in {
static assert(i < len);
} body {
b ? bts(ptr, i) : btr(ptr, i);
return b;
}
would suffice but then DMD complains as follows
dmd -debug -gc -gs -unittest -D -Dd/home/per/.emacs.d/auto-builds/dmd/Debug-Boundscheck-Unittest/home/per/Work/justd/ -w -main ~/Work/justd/bitset.d /home/per/Work/justd/assert_ex.d -of/home/per/.emacs.d/auto-builds/dmd/Debug-Boundscheck-Unittest/home/per/Work/justd/bitset
/home/per/Work/justd/bitset.d(58): Error: bitset.BitSet!2.BitSet.opIndexAssign called with argument types (bool, int) matches both:
/home/per/Work/justd/bitset.d(49): opIndexAssign(bool b, ulong i)
and:
/home/per/Work/justd/bitset.d(65): opIndexAssign(bool b, const(ulong) i)
/home/per/Work/justd/bitset.d(66): Error: variable i cannot be read at compile time
/home/per/Work/justd/bitset.d(66): while evaluating: static assert(i < 2LU)
/home/per/Work/justd/bitset.d(58): Error: bitset.BitSet!2.BitSet.opIndexAssign called with argument types (bool, int) matches both:
/home/per/Work/justd/bitset.d(49): opIndexAssign(bool b, ulong i)
Do I have to make parameter i a template parameter, say using type U, and then use static if someTypeTrait!U. I tried this but isMutable!Index always evaluates to true.
import std.traits: isIntegral;
bool opIndexAssign(Index)(bool b, Index i) #trusted pure nothrow if (isIntegral!Index) in {
import std.traits: isMutable;
// See also: http://stackoverflow.com/questions/19906516/static-parameter-function-specialization-in-d
static if (isMutable!Index) {
assert(i < len);
} else {
import std.conv: to;
static assert(i < len,
"Index " ~ to!string(i) ~ " must be smaller than BitSet length " ~ to!string(len));
}
} body {
b ? bts(ptr, i) : btr(ptr, i);
return b;
}
What you're trying to do doesn't really work. You can do a template value parameter:
void foo(int i)() { /* use i at compile time */ }
but then you can't pass a runtime value to it, and it has different call syntax: foo!2 vs foo(2).
The closest you can get is is CTFE:
int foo(int i) { return i; }
enum something = foo(2); // works, evaluated at compile time
int s = foo(2); // also works, but runs at runtime.
Inside the function, there is a magic if(__ctfe) { running at compile time } else { at runtime}, but again, this isn't if there's a literal, it is if the function is run in a CT context, e.g., assigning the result to an enum constant.
But, otherwise, an int literal is still a mutable int as far as the function is concerned. So what you're specifically trying to do won't work in D as it is right now. (There's been some talk about wanting a way to tell if it is a literal, but as far as I know, there's no plan to actually do it.)

Structure of a block declaration

When declaring a block what's the rationale behind using this syntax (i.e. surrounding brackets and caret on the left)?
(^myBlock)
For example:
int (^myBlock)(int) = ^(int num) {
return num * multiplier;
};
C BLOCKS: Syntax and Usage
Variables pointing to blocks take on the exact same syntax as variables pointing to functions, except * is substituted for ^. For example, this is a function pointer to a function taking an int and returning a float:
float (*myfuncptr)(int);
and this is a block pointer to a block taking an int and returning a float:
float (^myblockptr)(int);
As with function pointers, you'll likely want to typedef those types, as it can get relatively hairy otherwise. For example, a pointer to a block returning a block taking a block would be something like void (^(^myblockptr)(void (^)()))();, which is nigh impossible to read. A simple typedef later, and it's much simpler:
typedef void (^Block)();
Block (^myblockptr)(Block);
Declaring blocks themselves is where we get into the unknown, as it doesn't really look like C, although they resemble function declarations. Let's start with the basics:
myvar1 = ^ returntype (type arg1, type arg2, and so on) {
block contents;
like in a function;
return returnvalue;
};
This defines a block literal (from after = to and including }), explicitly mentions its return type, an argument list, the block body, a return statement, and assigns this literal to the variable myvar1.
A literal is a value that can be built at compile-time. An integer literal (The 3 in int a = 3;) and a string literal (The "foobar" in const char *b = "foobar";) are other examples of literals. The fact that a block declaration is a literal is important later when we get into memory management.
Finding a return statement in a block like this is vexing to some. Does it return from the enclosing function, you may ask? No, it returns a value that can be used by the caller of the block. See 'Calling blocks'. Note: If the block has multiple return statements, they must return the same type.
Finally, some parts of a block declaration are optional. These are:
The argument list. If the block takes no arguments, the argument list can be skipped entirely.
Examples:
myblock1 = ^ int (void) { return 3; }; // may be written as:
myblock2 = ^ int { return 3; }
The return type. If the block has no return statement, void is assumed. If the block has a return statement, the return type is inferred from it. This means you can almost always just skip the return type from the declaration, except in cases where it might be ambiguous.
Examples:
myblock3 = ^ void { printf("Hello.\n"); }; // may be written as:
myblock4 = ^ { printf("Hello.\n"); };
// Both succeed ONLY if myblock5 and myblock6 are of type int(^)(void)
myblock5 = ^ int { return 3; }; // can be written as:
myblock6 = ^ { return 3; };
source: http://thirdcog.eu/pwcblocks/
I think the rationale is that it looks like a function pointer:
void (*foo)(int);
Which should be familiar to any C programmer.

function with multiple arguments

how to pass multiple arguments in a single function in Objective-C? I want to pass 2 integer values and the return value is also integer. I want to use the new Objective-C syntax, not the old C/C++ syntax.
In objective-c it is really super easy. Here is the way you would do it in C:
int functName(int arg1, int arg2)
{
// Do something crazy!
return someInt;
}
This still works in objective-c because of it's compatibility with C, but the objective-c way to do it is:
// Somewhere in your method declarations:
- (int)methodName:(int)arg1 withArg2:(int)arg2
{
// Do something crazy!
return someInt;
}
// To pass those arguments to the method in your program somewhere:
[objectWithOurMethod methodName:int1 withArg2:int2];
Best of luck!
Since this is still google-able and there are better solutions than the accepted answer; there's no need for the hideous withArg2 – just use colons:
Declaration:
#interface
-(void) setValues: (int)v1 : (int)v2;
Definition:
#implementation
-(void) setValues: (int)v1 : (int)v2 {
//do something with v1 and v2
}
Like this:
int sum(int a, int b) {
return a + b;
}
Called like this:
int result;
result = sum(3, 5);
// result is now 8
More here
int add (int a, int b)
{
int c;
c = a + b;
return c;
}
link text