The following code compiles:
^{}();
And this compiles:
void (^x)();
(x = ^{})();
But this doesn't:
(void (^x)() = ^{})();
The error I get is Expected ')'. Is this a bug with llvm or something? It's totally holding me back from pretending Objective-C is JavaScript.
This wouldn't make sense in a C-like language. To see why, let's build the statement from the ground up.
First, we'll use your working declaration for x:
void (^x)();
Now let's initialize it in the same statement:
void (^x)() = ^{};
So far so good - x has been initialized with the correct block. So let's invoke x now. But where will the () go? Naturally, we need to place the () immediately after a block-valued expression. However, in C, declarations are statements, not expressions so
(void (^x)() = ^{})();
doesn't make sense. The only place the () can go is after the ^{}:
void (^x)() = ^{}();
But ^{}() has type void, not type void (^)().
To sum up: you can't declare a block variable and invoke it at the same time. you'll have to either declare and initialize the variable, and then call it
void (^x)() = ^{};
x();
or declare it and then assign and call it
void (^x)();
(x = ^{})();
or just separate all three:
void (^x)();
x = ^{};
x();
As a concluding thought, let's say it was desirable to declare and invoke blocks at the same time. If we decided to allow code like (void (^x)() = ^{})();, then for the sake of consistency, we would have to also allow code such as ++(void x = 4); or (void x = 1) + (void y = 2);. I hope you'll agree that these just look strange in C.
As an analogy, consider:
This compiles:
if (42) { }
And this compiles:
int x;
if (x = 42) { }
But this doesn't:
if (int x = 42) { }
Related
i was looking some examples of interactions with the keyboard and stumbled upon this code that i found interesting. But i'm having trouble understanding a certain part of it(it's marked down below).I don't get how all this whole ''boolean'' declaration, ''switch'' and ''CASE'' works, i tried to look in the reference but still. Could someone explain in a simple maner how these work?
float x = 300;
float y = 300;
float speed = 5;
boolean isLeft, isRight, isUp, isDown;
int i = 0;
void keyPressed() {
setMove(keyCode, true);
if (isLeft ){
x -= speed;
}
if(isRight){
x += speed;
}
}
void keyReleased() {
setMove(keyCode, false);
}
boolean setMove(int k, boolean b) {// <<<--- From this part down
switch (k) {
case UP:
return isUp = b;
case DOWN:
return isDown = b;
case LEFT:
return isLeft = b;
case RIGHT:
return isRight = b;
default:
return b; }
}
Questions like these are best answered by the reference:
Works like an if else structure, but switch() is more convenient when you need to select between three or more alternatives. Program controls jumps to the case with the same value as the expression. All remaining statements in the switch are executed unless redirected by a break. Only primitive datatypes which can convert to an integer (byte, char, and int) may be used as the expression parameter. The default is optional.
The rest of the code is setting the corresponding variable to whatever value you passed in as the b parameter, and then returning it.
You should get into the habit of debugging your code. Add print statements to figure out exactly what the code is doing.
Take a look at this C++ code:
#include <iostream>
using namespace std;
class B{
public:
int& f() {
int local_n = 447;
return local_n ;
} // local_n gets out of scope here
};
int main()
{
B b;
int n = b.f(); // and now n = 447
}
I don't understand why n = 447 at the end of main, because I tried to return a reference to a local_n, when it should be NULL;
Returning a reference to a local variable invokes undefined behavior - meaning you might get lucky and it might work... sometimes... or it might format your hard drive or summon nasal demons. In this case, the compiler generated code that managed to copy the old value off the stack before it got overwritten with something else. Oh, and references do not have a corresponding NULL value...
Edit - here's an example where returning a reference is a bad thing. In your example above, since you copy the value out of the reference immediately before calling anything else, it's quite possible (but far from guaranteed) that it might work most of the time. However, if you bind another reference to the returned reference, things won't look so good:
extern void call_some_other_functions();
extern void lucky();
extern void oops();
int& foo()
{ int bar = 0;
return bar;
}
main()
{ int& x = foo();
x = 5;
call_some_other_functions();
if (x == 5)
lucky();
else
oops();
}
I apologize if this was asked many times.
I'm trying to understand why both of this works fine without any warnings or other visible issues (in Xcode):
int testFunctionAcceptingIntPointer(int * p) {
return *p = *p +5;
}
void test() {
int testY = 7;
typedef int (*MyPointerToFunction)(int*);
// Both this (simply a function name):
MyPointerToFunction functionPointer = testFunctionAcceptingIntPointer;
// And this works (pointer to function):
MyPointerToFunction functionPointer = &testFunctionAcceptingIntPointer;
int y = functionPointer(&testY);
}
The code works fine without warnings both ways because a function designator is converted to a function pointer
MyPointerToFunction functionPointer = testFunctionAcceptingIntPointer;
unless it is the operand of the address operator
MyPointerToFunction functionPointer = &testFunctionAcceptingIntPointer;
(or sizeof and _Alignof).
In the first assignment, you don't use &, so the automatic conversion is done, resulting in a function pointer of appropriate type, in the second, you explicitly take the address, resulting in a function pointer of the appropriate type.
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.
I'm experimenting with Obj-C blocks and trying to have a struct with two blocks in it where one block is to change what the other block does.
this is a really roundabout way to do something simple... and there may be better ways to do it, but the point of the exercise is for me to understand blocks. here's the code , it doesn't work, so what am I missing/not understanding and/or doing wrong?
//enumerate my math operation options so i can have something more understandable
//than 0, 1, 2, etc... also makes it easier to add operations, as opTypeTotal
//will be 1 plus the index of the operation before it.
typedef enum
{
opTypeAdd = 0,
opTypeSubtract = 1,
opTypeTotal
} opType;
//not sure if (struct someMathStruct)* is correct, probably is wrong
//the intent is to pass a pointer to someMathStruct, but the compiler
//won't know about its existance until a few lines later...
typedef (void)(^changeBlock)(opType,(struct someMathStruct)*);
typedef (void)(^mathBlock)(int,int,int*);
//hold two blocks, to be defined later at runtime
typedef struct someMathStruct{
mathBlock doMath;
changeBlock changeOperation;
} SomeMath;
//i want to declare an array of blocks of type mathBlock
//the intent is to have the array index to correspond with the opTypes enumerated above
//almost certain i'm doing this wrong
mathBlock *m[opTypeTotal] = malloc(sizeof(mathBlock *)*opTypeTotal);
//just two simple math operations as blocks
m[opTypeAdd] = ^(void)(int a,int b,int *result){*result = a+b;};
m[opTypeSubtract] = ^(void)(int a,int b,int *result){*result = a-b;};
//this block is what's supposed to change the other block in the struct
//it takes an opType, and a pointer to the SomeMath struct
//is this the right way to access the member variables of the struct?
changeBlock changeMe = ^(void)(opType a, SomeMath *b) {
//should make adding operations as easy as just adding cases
switch (a)
{
case opTypeAdd: *b.doMath=m[a]; break;
case opTypeSubtract:
default: *b.doMath=m[a]; //catch-all goes to subtracting
}
}
...
SomeMath mathFun;
int theTotal = 0; //a test int to work with
//do i need to copy the changeMe block?
//or can i just do what i'm doing here as the block itself isn't unique
mathFun.changeOperation = changeMe;
mathFun->changeOperation(opTypeAdd, &mathFun);
mathFun->doMath(theTotal,11,&theTotal); //result should be 11
mathFun->changeOperation(opTypeSubtract, &mathFun);
mathFun->doMath(theTotal,3,&theTotal); //result should be 8
NSLog(#"the result: %d",theTotal); //should output "the result: 8"
The code seems to work as you expect (the result is 8) once you fix the compilation errors:
Compile with: gcc -o test test.m -framework Foundation
#import <Foundation/Foundation.h>
//enumerate my math operation options so i can have something more understandable
//than 0, 1, 2, etc... also makes it easier to add operations, as opTypeTotal
//will be 1 plus the index of the operation before it.
typedef enum
{
opTypeAdd = 0,
opTypeSubtract = 1,
opTypeTotal
} opType;
struct someMathStruct; // Forward declare this as a type so we can use it in the
// changeBlock typedef
typedef void (^changeBlock) (opType,struct someMathStruct*);
typedef void (^mathBlock) (int,int,int*);
//hold two blocks, to be defined later at runtime
typedef struct someMathStruct{
mathBlock doMath;
changeBlock changeOperation;
} SomeMath;
int main()
{
//i want to declare an array of blocks of type mathBlock
//the intent is to have the array index to correspond with the opTypes
// enumerated above
mathBlock *m = calloc(opTypeTotal, sizeof(mathBlock *));
//just two simple math operations as blocks
m[opTypeAdd] = ^(int a,int b,int *result){*result = a+b;};
m[opTypeSubtract] = ^(int a,int b,int *result){*result = a-b;};
changeBlock changeMe = ^(opType a, SomeMath *b) {
//should make adding operations as easy as just adding cases
switch (a)
{
case opTypeAdd: b->doMath = m[a]; break;
case opTypeSubtract:
default: b->doMath = m[a]; //catch-all goes to subtracting
}
};
SomeMath mathFun;
int theTotal = 0; //a test int to work with
mathFun.changeOperation = changeMe;
mathFun.changeOperation(opTypeAdd, &mathFun);
mathFun.doMath(theTotal,11,&theTotal); //result should be 11
mathFun.changeOperation(opTypeSubtract, &mathFun);
mathFun.doMath(theTotal,3,&theTotal); //result should be 8
NSLog(#"the result: %d",theTotal); //should output "the result: 8"
}