Pointer to function problem

This is a discussion on Pointer to function problem within the C Programming forums, part of the General Programming Boards category; Hi, So I have three similar working functions: Code: char *ltrim(char *); char *rtrim(char *); char *trim(char *); And i ...

  1. #1
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    Pointer to function problem

    Hi,

    So I have three similar working functions:
    Code:
    char *ltrim(char *);
    char *rtrim(char *);
    char *trim(char *);
    And i want to write a function that can test these functions and takes a pointer to function as its argument.

    I tried this first
    Code:
    void test_func(char *(*func)(char *in))
    {
        printf("in = |%s|\nin.length = %d\nout = |%s|\nout.length = %d\n",
    	    in,
    	    strlen(in),
    	    (*func)(in),
    	    strlen((*func)(in)));
    }
    But this won't compile and the compiler says 'in' is undeclared.

    Then I changed it to this
    Code:
    void test_func(char *(*func)(char *),char *in)
    {
        printf("in = |%s|\nin.length = %d\nout = |%s|\nout.length = %d\n",
    	    in,
    	    strlen(in),
    	    (*func)(in),
    	    strlen((*func)(in)));
    }
    And it seemed to work since i could print out the expected results with this
    Code:
    test_func(ltrim, str);
    So my problem is
    1) why the first piece of code won't compile (to be more specific, I'd expected 'in' to be an argument passed to the function that is pointed to. What does the compiler *think* of this variable to be)
    2) Is the second piece of code the right way to do this?

    Thanks,

  2. #2
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    char *(*func)(char *) is a function pointer. It refers to a function by itself. It does not include any arguments. So you can use the same function pointer multiple times with different arguments.

    And yes, your second code is right.

    By the way, you don't need * when you call the function from the pointer. you can just call func(in). It's a syntax convenience.
    It is too clear and so it is hard to see.
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    Had he known what fire was,
    He could have cooked his rice much sooner.

  3. #3
    Guest Sebastiani's Avatar
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    1) In the first case 'in' is just part of the function prototype (you aren't actually invoking the function at that point, after all, so how could 'in' actually get passed in?). Think of it as if you had created a typedef:

    Code:
    typedef char* (* function )( char* in );
    
    void test( function fun )
    {
    	fun( "ok" );
    }
    2) That's fine, but you don't actually have to dereference the pointer (see example above). Either way works, though.

  4. #4
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    Thank you guys for reply. Still i have a question.
    Quote Originally Posted by King Mir View Post
    char *(*func)(char *) is a function pointer. It refers to a function by itself. It does not include any arguments. So you can use the same function pointer multiple times with different arguments.
    So King Mir what do you mean by "It does not include any arguments."?
    In my understanding char *(*func)(char *) is a pointer to function that takes type char * as argument and returns type char *.

    For the dereferencing part, yes, that's quite cool and also Sebastiani's typedef solution looks neat. I'll try those and thanks for these good suggestions.

  5. #5
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by vincent01910
    So King Mir what do you mean by "It does not include any arguments."?
    In my understanding char *(*func)(char *) is a pointer to function that takes type char * as argument and returns type char *.
    King Mir means that just because you have a function pointer does not mean you have an argument that can be passed to the function called via that function pointer. In other words, what Sebastiani was talking about in.
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  6. #6
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    Thanks laserlight. I get it now.

  7. #7
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    also, you can just do
    Code:
    void test_func(char *func(char *),char *in)

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