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BernardIE531730-Nov-20 5:50

30-Nov-20 5:50

As I understand it declarations are written as if operators e.g. you would write "name[123]" to access the 123'rd element in the array named "name" so that's how it is declared. Maybe it should be "[123]name".

honey the codewitch30-Nov-20 5:51

30-Nov-20 5:51

You seem to be talking about accessing the array, not declaring it. I'm talking about declaring it.

Edit: Or if you are, then I think get your point? My response would be that pointers aren't declared like pointer operations.

Real programmers use butterflies

BernardIE531730-Nov-20 6:02

30-Nov-20 6:02

They aren't? I would declare a pointer using the rules of operator precedence always starting with the identifier e.g. "char* name" and utilize it as an operator as "*name" so except for the white space and the casting operator "char" which is no longer needed they are the same. Cheerios

honey the codewitch30-Nov-20 6:04

30-Nov-20 6:04

I think we're misunderstanding each other. My OP maybe wasn't as clear as it should have been.

I've always declared pointers with the * next to the type. You CANNOT declare arrays that way. Hence my complaint.

Real programmers use butterflies

BernardIE531730-Nov-20 6:22

30-Nov-20 6:22

Greetings My point is that if you accept the rules of operator precedence and associativity which I am assuming you do than you must accept the rules of declaration since they are identical. Otherwise you must argue against both and insist e.g. that array elements be referenced as "[123]name" or pointer targets as "name*". Or perhaps you would argue that the rules of declaration be different from the rules of operation but that would complicate things. Kind Regards Cheerios

honey the codewitch30-Nov-20 6:24

30-Nov-20 6:24

Oh, I get you. I guess using operator precedence on type modifiers kind of threw me, but i can see why you look at it that way. Smile | :)

Real programmers use butterflies

BernardIE531730-Nov-20 8:38

30-Nov-20 8:38

Greetings and Kind Regards Just in passing I wish to mention in case you do not already know about it is that the C text by Harbison & Steele is what the K&R text wished it was. H&S is a beautiful text and is where I learned that declarations and operations follow the same rules. Why K&R didn't explain this is as simply is difficult to understand. Best Wishes Cheerios

Stuart Dootson30-Nov-20 5:53

Stuart Dootson

30-Nov-20 5:53

You could use 123[name] as an alternative - that matches the original PDP addressing mode syntax better as well, I suspect!

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Stuart Dootson30-Nov-20 5:52

Stuart Dootson

30-Nov-20 5:52

honey the codewitch wrote:
I think it's inconsistent, and I think the array specifier should have been declared with the type since it's essentially a type modifier like * and &

One thing it's not is inconsistent - the pointer modifier belongs with the variable, not the type. For example, the following fragment declares a pointer to integer variable and an integer variable.

int *pa, a;

See this Godbolt...

One option, were you using C++...

template<class T, size_t N>
using Array = T[N];

template<class T>
using Ptr = T*;

Array<int, 10> test_array;

Ptr<int> test_pointer;

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

honey the codewitch30-Nov-20 5:54

30-Nov-20 5:54

Stuart Dootson wrote:
One thing it's not is inconsistent - the pointer modifier belongs with the variable, not the type.

Yet it's a type modifier. A pointer to an int is a different type than an int.

Real programmers use butterflies

Member 1330167930-Nov-20 8:18

Member 13301679

30-Nov-20 8:18

Quote:
Does it bother anyone else that you declare a pointer like:
char* sz; // pointer type, pointer declared *with* type

It only looks odd if you use it like that. If you use it like the way it was meant to be used:

char *sz;

It makes sense. The way you write it is not consistent and lends itself to errors.

char* szA, szB, szC;   // Not what you intended, and even if you did intend that
                       // the **reader** isn't sure you intended to do that!

char *szD, *szE, *szF; // Anyone reading this knows exactly what was intended.

Putting the "*" next to the typename is logically inconsistent - you still have to put the "*" in the correct place for other pointer types:

void (*fptr) (void);     // See?  The "*" 'binds' to the variable, not the type.

So instead of doing it one way for some variables and the correct way for others, just do it the correct way for all.

honey the codewitch30-Nov-20 8:26

30-Nov-20 8:26

I suppose then that I do not like that type modifiers are not declared with the type.

Real programmers use butterflies

Member 1330167930-Nov-20 8:49

Member 13301679

30-Nov-20 8:49

Quote:
I suppose then that I do not like that type modifiers are not declared with the type.

Pointers aren't type modifiers.

You can tell by the way that actual type modifiers can be placed in any order ("short int" and "int short") while the pointer notation can only go before the variable name.

short int si1;     // Compiles
int short si2;     // Compiles
short int *psi3;   // Compiles
*short int psi4;   // Error, won't compile.

If you do not associate the '*' with the variable name, then everything looks very confusing and arbitrary and some things that should work won't. If you associate the '*' with the variable name then everything is logical and can be worked out - any "*symbol" means that symbol is a pointer to something, so things like this can be worked out:

const char *varname[100];

You cannot logically infer what that means if you think that the "*" is part of the typename. If the "*" is part of the type, that would mean that the pointer can not be changed. In reality, it is the individual chars that cannot be changed, while each of the pointers in the array can be changed.

There's a lot of misunderstanding that will happen when "typename* varname" is used in place of "typename *varname". A compiler won't catch all of it.

harold aptroot30-Nov-20 9:50

harold aptroot

30-Nov-20 9:50

It bothers me to no end. Actually what bothers me even more is that every time I mention it, I mostly get replies defending the stupid Spiral of Death. It's bad enough that it's bad, but worse that people feel this kind of Stockholm Syndrome towards a type syntax that just doesn't make sense. (in some sense it's not even a type syntax, because it's not just a type, there's a declaration stuck in the middle of it)

Anyway I'll show you something even worse, the syntax for returning a function pointer. Let's say you want to return a pointer to a function that takes two ints and returns an int, a function like int add(int a, int b) maybe. It would look like this:

int (*getFunc())(int, int) { … }

Unless you use a typedef of course (in C# that is essentially mandatory: you must declare a delegate with the signature first and then you can use that).

honey the codewitch30-Nov-20 10:10

30-Nov-20 10:10

harold aptroot wrote:
but worse that people feel this kind of Stockholm Syndrome towards a type syntax that just doesn't make sense

I think you're the first person on this thread to agree with me. Laugh | :laugh:

Real programmers use butterflies

BernardIE531730-Nov-20 21:46

30-Nov-20 21:46

Greetings but I must differ

int foobar(int, int) { return 0; }

// I merely followed the operator rules of precedence and associativity for:
// "f is a pointer to a function which takes two arguments of type int and int and returns an int"
// and voila though the return type doesn't seem to be an operator unless perhaps a cast operator
int (*f)(int, int) = foobar; // compiles ok

int (*getFunc())(int, int) = foobar; // compiles with syntax error

// Cheerios

harold aptroot1-Dec-20 1:33

harold aptroot

1-Dec-20 1:33

But that's not what I wrote. I wanted to return a function pointer from a function named getFunc.

BernardIE53171-Dec-20 7:30

1-Dec-20 7:30

Greetings and Kind Regards Please permit me to demonstrate the following: By merely following the rules of operator precedence and associativity I deduce the same declaration for getFunc as yourself. I thank Harbison & Steele for teaching me this in their fine C text. Why K&R don't do this is difficult to understand.
// "getFunc is a function which returns a pointer to a function which takes two int args and returns an int"
// getFunc is a function ...
getFunc()
// ... which returns a pointer ...
*getFunc()
// ... to a function which takes two int args ...
*getFunc()(int, int)
(*getFunc())(int, int) // added ()'s because function call (int, int) has higher precedence than indirection *
// ... and returns an int
int (*getFunc())(int, int)
// Voila No Spiral of Death is needed. Best Wishes Cheerios

Julie77730-Nov-20 13:36

Julie777

30-Nov-20 13:36

It's hard to believe that of all the replies no one has ever read K&R C.
A variable declaration consists of a type and name and possibly a type reference spec such as * or [].
Multiple variable declarations may be combined in a single statement (line) if they are the same type. this is why reference specs go with the name

char *sz, sz2[], sz3[1024];

Types and reference specs can also have modifiers which are can get very confusing with multiple declarations combined on a line.

static const char sz4, *sz5, const *sz6;

Add initializers and you will see why it's pretty standard now days to put one declaration per line.

BernardIE531730-Nov-20 22:41

30-Nov-20 22:41

Greetings and Kind Regards May I please direct you to my previous post. Cheerios
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