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  3. C declarations are half backward

C declarations are half backward

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  • H honey the codewitch

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

    Real programmers use butterflies

    U Offline
    U Offline
    User 13269747
    wrote on last edited by
    #47

    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.

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    • H honey the codewitch

      Does it bother anyone else that you declare a pointer like:

      char* sz; // pointer type, pointer declared *with* type

      But an array is declared like this:

      char sz[1024];// array type, array declared *after* var name

      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 & Maybe it's just me?

      Real programmers use butterflies

      L Offline
      L Offline
      Lost User
      wrote on last edited by
      #48

      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](http://c-faq.com/decl/spiral.anderson.html). 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).

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      • L Lost User

        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](http://c-faq.com/decl/spiral.anderson.html). 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).

        H Offline
        H Offline
        honey the codewitch
        wrote on last edited by
        #49

        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:

        Real programmers use butterflies

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        • S swampwiz

          What has bothered me more is the fact that: int* pa; int *pa; are the same. Had I been Bjorn, I wouldn't have allowed the latter.

          J Offline
          J Offline
          James Lonero
          wrote on last edited by
          #50

          The latter may be the better notation since I can declare; int *pa, a; where pa is a pointer to an int and a is an int. If I declare int* pa, a; this is also legal, where pa is still a point to an int and a is an int.

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          • H honey the codewitch

            Does it bother anyone else that you declare a pointer like:

            char* sz; // pointer type, pointer declared *with* type

            But an array is declared like this:

            char sz[1024];// array type, array declared *after* var name

            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 & Maybe it's just me?

            Real programmers use butterflies

            J Offline
            J Offline
            Julie777
            wrote on last edited by
            #51

            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.

            B 1 Reply Last reply
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            • L Lost User

              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](http://c-faq.com/decl/spiral.anderson.html). 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).

              B Offline
              B Offline
              BernardIE5317
              wrote on last edited by
              #52

              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

              L 1 Reply Last reply
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              • J Julie777

                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.

                B Offline
                B Offline
                BernardIE5317
                wrote on last edited by
                #53

                Greetings and Kind Regards May I please direct you to my previous post. Cheerios The Lounge[^]

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                • B BernardIE5317

                  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

                  L Offline
                  L Offline
                  Lost User
                  wrote on last edited by
                  #54

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

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                  • L Lost User

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

                    B Offline
                    B Offline
                    BernardIE5317
                    wrote on last edited by
                    #55

                    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

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                    • H honey the codewitch

                      Does it bother anyone else that you declare a pointer like:

                      char* sz; // pointer type, pointer declared *with* type

                      But an array is declared like this:

                      char sz[1024];// array type, array declared *after* var name

                      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 & Maybe it's just me?

                      Real programmers use butterflies

                      M Offline
                      M Offline
                      mike codeproject
                      wrote on last edited by
                      #56

                      C declarations are fine. The problem is in C pointer expressions, where two unfortunate changes were made. First, Ritchie (I presume) chose to make * the pointer dereference operator and either chose or had forced upon him by the * choice the need to make it a *left* unary operator. Had he followed Wirth's prior example in Pascal (using p^ to dereference p), then your backwards issue is automatically solved. Why? Because C declarators are based on how the variable is used in an expression. So int *p; has the * first because you use *p in an expression to make use of the pointer. The array dimension come after the variable name: int a[5]; because you use a[index] in an expression to access a member of an array. If the pointer dereference was on the right, then you wouldn't need the quirky -> operator that only exists to cut down on parentheses, where p->member exists only to avoid typing (*p).member. By the way, the declaration should be "char *p;" instead of the awful "char* p;" that revisionists like to type. The * says that p is a pointer, not that "char" is a pointer. To see the difference, try using: int* p1, p2; /* this will NOT declare two integer pointers! */ The correct syntax is: int *p1, *p1; ...since the * says that what's on the right is a pointer. Again, this misunderstanding wouldn't even come up with a right-unary dereference operator. Most of the C language is admirable, particularly as a product of the early '70s, but this (along with allowing the and parts of libraries to become de facto standards) get my votes for Dennis Ritchie's biggest mistakes.

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                      • J James Lonero

                        The latter may be the better notation since I can declare; int *pa, a; where pa is a pointer to an int and a is an int. If I declare int* pa, a; this is also legal, where pa is still a point to an int and a is an int.

                        J Offline
                        J Offline
                        Jeff Clausius SG
                        wrote on last edited by
                        #57

                        I was just to reply with the same. I always put the * modifier RIGHT in front of the variable. And it makes sense since, "char *" is not really the type. A lot of beginners are confused when something like the following: char* psz, pszHi, pszBye; and they discover pszHi / pszBye are just character variables. Code like the following helps those initiates: char *psz, *pszHi, *pszBye, chA, chB, *pszString3; Just my $0.02.

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