manipulating arg in the: function(char* arg); without returning it[C,freebsd7.1,gcc]

This is a discussion on manipulating arg in the: function(char* arg); without returning it[C,freebsd7.1,gcc] within the C Programming forums, part of the General Programming Boards category; As the title of the topic is not so clear, now here i can clear my problem... I have a ...

  1. #1
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    manipulating arg in the: function(char* arg); without returning it[C,freebsd7.1,gcc]

    As the title of the topic is not so clear, now here i can clear my problem...

    I have a program source code, main.c . In which i m intending to do following thing with out changing the source code...
    ----------------------------------------------------------------------------------------------------
    Code:
    #include <stdio.h>
    int main(int argc,char **argv)
    {
    char* backup;
    char* arg = argv && argc > 0 ? argv[1] : NULL;
    backup=arg;
    
    foo(arg);
    
    if (strcmp(arg, backup) != 0) 
      {
       printf("Yeah! u success; as u have manipulated arg");
       }   
    else
       printf("You Failed!; no change in arg; arg and backup both are same");
    exit(0);
    }
    
    void foo(char *arg)
    {
    arg="what can i do in it :( ??????? " ;
    }
    -----------------------------------------------------------------------------------------------------
    I am beginner in C, I want to do some thing in this funtion so i would be able to change the arg, without returning arg from this method, as i m not supposed to change in the main method, i can only do things in foo function.
    waiting for ur suggestions..
    thanks for your concentration.
    I m using freebsd7.1 and compiler is gcc.

  2. #2
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    I doubt you can do that. arg and backup point to the same bit of memory. You have no way to modify arg inside foo(), so you can't make it point to something else.

    Maybe you have misunderstood the task?

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    Quote Originally Posted by matsp View Post
    I doubt you can do that. arg and backup point to the same bit of memory. You have no way to modify arg inside foo(), so you can't make it point to something else.

    Maybe you have misunderstood the task?

    --
    Mats
    Thanks for your reply - my actual code is this... but it is assumes to be do in this way....

    Code:
    static void* function(char* arg){
        fprintf(stderr, "int 3h\n");
        printf("You've made it passed the first obstacle!\n");
        printf("-----\nStep 2. Case of the missing dynlib\n");
     
        char* backup = strdup(arg);
        void* handle = dlopen("mylib.so", RTLD_NOW);
        voidfun hsym = dlsym(handle, "NULL");
        hsym(arg);
        if (strcmp(arg, backup) != 0) cake(arg);    
    }
    Now i want to make this condition true , so cake function will parsed by compiler...
    that method which i have described above is supposed to be done in the shared library which i have created....

  4. #4
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    If you pass a char * to a function in C, you can not change it to point somewhere else.

    I'm not sure what you are supposed to do, but it's either breaking some rules, or you have misunderstood what you are supposed to do.

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  5. #5
    Deathray Engineer MacGyver's Avatar
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    If you know about the details of the implementation, you could theoretically alter the value of arg in main() by determining its location in relation to the copy of arg being passed into it (ie. calculations of the stack, but assuming a heck of a lot of things, most of which can't really be determined easily). This isn't really in the domain of standard C, though, as this requires one to deviate into, at the very least, a murky area of implementation defined behavior, if not outright undefined behavior.

    To make sure there is no mistake, I will state again that this task, as I understand it, is impossible in standard C. I don't even understand its purpose.

  6. #6
    and the hat of wrongness Salem's Avatar
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    > char* arg = argv && argc > 0 ? argv[1] : NULL;
    Your strcmp() will blow up anyway if arg ends up being NULL.

    As would your modified strdup() variant.

    Why does this look like an exercise in stack smashing to get the code to do something non-obvious based on it's intention.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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    Quote Originally Posted by matsp View Post
    I doubt you can do that. arg and backup point to the same bit of memory. You have no way to modify arg inside foo(), so you can't make it point to something else.

    Maybe you have misunderstood the task?

    --
    Mats
    Quote Originally Posted by MacGyver View Post
    If you know about the details of the implementation, you could theoretically alter the value of arg in main() by determining its location in relation to the copy of arg being passed into it (ie. calculations of the stack, but assuming a heck of a lot of things, most of which can't really be determined easily). This isn't really in the domain of standard C, though, as this requires one to deviate into, at the very least, a murky area of implementation defined behavior, if not outright undefined behavior.

    To make sure there is no mistake, I will state again that this task, as I understand it, is impossible in standard C. I don't even understand its purpose.
    hmm... the purpose: "It is my assignment given by the prof in my university", I am sure there is no mistake...the above hsym(arg) got the reference from dynamic shared lib, and my purpose to doing it is to make the if condition true, so cake function will parsed.

  8. #8
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    Quote Originally Posted by mali2
    the purpose: "It is my assignment given by the prof in my university", I am sure there is no mistake.
    Maybe you should provide some context: what module is this prof teaching?
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    Only way a variable can be altered in the calling function is by passing its address as an argument to the callee.

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    . . . for example:
    Code:
    char *rating = "Good";
    modify_rating(&rating);
    
    void modify_rating(char **rating) {
        *rating = "Bad";
    }
    dwk

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  11. #11
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    Perhalps the code is ment to be
    Code:
    backup= malloc(strlen(arg)+1);
    strcpy(backup, arg);
    Instead of
    Code:
    backup=arg;
    Then you can change the string passed to foo without effecting the backup variable. (Note: there is still the restriction that the new string must be shorter than the old one)
    It is too clear and so it is hard to see.
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