Basic Questions

This is a discussion on Basic Questions within the C Programming forums, part of the General Programming Boards category; Hallo Guys I have some basic questions about C Programming, 'cause i'm not very good in it. Why is it ...

  1. #1
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    Basic Questions

    Hallo Guys

    I have some basic questions about C Programming, 'cause i'm not very good in it.

    Why is it impossible for an Array-initializing function to do the following ?
    Code:
     int[] init() { return int[32]; }
    Thanks for your help.

    Greez

  2. #2
    and the hat of sweating
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    You can't return an array, you can only return a pointer.
    Also, you can't return a pointer to a variable that is local to the function, since that variable disappears when the function ends.
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

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  3. #3
    spurious conceit MK27's Avatar
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    It's not impossible, it's just a matter of doing it properly. Your first problem is basic syntax, and your second is that you need to allocate heap memory for the array -- you cannot/should not return a "local" stack variable as those only last as long as the function call. So:
    Code:
    #include <stdlib.h>  /* for malloc() */
    #include <stdio.h> /* for printf() */
    
    int *rayfunc() {int *this=malloc(sizeof(int)*32); return this; }
    
    int main() {
    	int *ray=rayfunc(), i;
    	ray[31]=13;
    	for (i=0;i<32;i++) printf("%d\n",ray[i]);	
    	free(ray);  /* deallocate memory */
    	return 0;
    }
    Last edited by MK27; 05-17-2009 at 11:26 AM.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

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    Thanks.

    It's never ever possible to return an array?

    And another question:

    Why do the following ones do not return the same result?

    1. int *a = address; a++
    2. char *a = address; a++
    Last edited by Panda_; 05-17-2009 at 11:29 AM.

  5. #5
    Registered User whiteflags's Avatar
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    > It's never ever possible to return an array?
    You shouldn't need to return an array anyway since arrays can be accessed by other functions than the one the array is declared in by passing pointers. Try to design your functions to accept a pointer to the first element as an argument, instead.


    > Why do the following ones do not return the same result?
    The char type is smaller than int, so ++ moves the address by a different number of bytes.

  6. #6
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    Thanks guys for your help.

    If you don't mind there's one more question:

    which type do variables from
    "int * a,b"
    have?

  7. #7
    spurious conceit MK27's Avatar
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    Quote Originally Posted by Panda_ View Post
    It's never ever possible to return an array?
    Nope, because that would have to be a stack variable (get it?). You can only return a single value, either an address (ie a pointer, like to an array) or plain int/char/datatype.

    So one thing you can do to get around this is typedef a custom datatype:
    Code:
    #include <stdio.h>
    
    typedef struct {
    	int ray[32]; 
    } intray; 
    
    intray rayfunc() {
    	intray this = {{0}};
    	this.ray[31]=13;
    	return this; 
    }
    
    int main() {
    	int i;
    	intray example=rayfunc();
    	for (i=0;i<32;i++) printf("%d\n",example.ray[i]);
    	return 0;
    }
    However, the typedef has to be a struct or the array rules apply anyway.

    Also, if you want to pass an existing array to a function for modification, use a pointer.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  8. #8
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Panda_
    which type do variables from
    "int * a,b"
    have?
    It would be wise to ask such questions with a code snippet posted in bbcode tags. If you are talking about the type of a and of b as declared by this line of code:
    Code:
    int * a,b;
    then the answer is that a is of type int* (pointer to int) and b is of type int. Because of a potential misinterpretation (one could accidentally or ignorantly consider b as being of type int*, I try to declare variables separately, especially if a pointer is involved.
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    Ok i got it.... thanks a lot.

    Does anybody know in which order this expression gets evaluated?
    Code:
    *&a++

  10. #10
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Panda_
    Does anybody know in which order this expression gets evaluated?
    With some exceptions, the order of evaluation of an expression is unspecified (but in this case there is nothing to vary). If we talk about the grouping of operators with respect to precedence, then it would be:
    Code:
    *(&(a++));
    Out of curiosity, but what motivated you to ask this question?
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  11. #11
    spurious conceit MK27's Avatar
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    Quote Originally Posted by laserlight View Post
    Out of curiosity, but what motivated you to ask this question?
    Yeah, *& is kind of wacky or superfluous, no?
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  12. #12
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    Quote Originally Posted by laserlight View Post
    With some exceptions, the order of evaluation of an expression is unspecified (but in this case there is nothing to vary). If we talk about the grouping of operators with respect to precedence, then it would be:
    Code:
    *(&(a++));
    Out of curiosity, but what motivated you to ask this question?
    Sounds like homework.
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

    "the internet is a scary place to be thats why i dont use it much." - billet, 03/17/2010

  13. #13
    DESTINY BEN10's Avatar
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    Quote Originally Posted by laserlight View Post
    With some exceptions, the order of evaluation of an expression is unspecified (but in this case there is nothing to vary). If we talk about the grouping of operators with respect to precedence, then it would be:
    Code:
    *(&(a++));
    I dont understand what is the problem in evaluating *&a++; As i see it, '*' and '++' has the same priority, with RtoL associativity thus a++ will be done first and as a result the whole expression should give the value at address '&a'. But when i run this code it doesn't even compiles. Why is it so?
    Code:
    #include<stdio.h>
    int main(void)
    {
    int a=10;
    printf("%d",*&a++);
    }
    Error 1 error C2102: '&' requires l-value
    When i tried to print
    Code:
    printf("%d",*&a);
    it gave the answer 10.
    HOPE YOU UNDERSTAND.......

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  14. #14
    and the Hat of Guessing tabstop's Avatar
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    You can't take the address of a++, seeing it is not stored in memory anywhere. (Edit: If a is a pointer, and a++ is valid to dereference, you should be able to do &*a++.)

  15. #15
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by BEN10
    As i see it, '*' and '++' has the same priority
    Postfix operator++ has a higher precedence than unary operator*.

    Quote Originally Posted by BEN10
    the whole expression should give the value at address '&a'
    As I illustrated and as tabstop implied, the expression dereferences the address of the result of a++, but the result of a++ is an rvalue, which is why your compiler complained that "'&' requires l-value".
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