pointer to an array

This is a discussion on pointer to an array within the Linux Programming forums, part of the Platform Specific Boards category; Why Doesnt This code work in Linux?? Code: main() { int a[3][4] = {1,2,3,4, 5,6,7,8, 0,9,1,6 }; show(a,3,4); } show(int ...

  1. #1
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    Question pointer to an array

    Why Doesnt This code work in Linux??

    Code:
    main()
    {
    	int a[3][4] = {1,2,3,4,
              	       5,6,7,8,
    			0,9,1,6
    				  };
    	show(a,3,4);
    }
    
    show(int (*q)[4],int row, int col)
    {
    	int i,j,*p;
    	for(i=0;i<row;i++)
    	{
    	p = q+i;
    	for(j=0;j<col;j++)
    	printf("%d",*(p+j));
    	printf("\n");
    	}
    	printf("\n");
    }

  2. #2
    and the hat of wrongness Salem's Avatar
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    Well that depends on how you define "doesn't work".

    Perhaps it's the lack of a prototype?
    Perhaps it's the broken array notation in the function?
    Perhaps it's Penry, the mild mannered janitor.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
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  3. #3
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    If you enable warnings (with -Wall) you will get a heap of warnings for your code.

    I managed to make it work with (on Windows, but using gcc which is the same compiler as you'd use on Linux, so the code generated should match) a few minor changes.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  4. #4
    Dragon Rider jas_atwal's Avatar
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    Following version of your program is working fine with gcc( if we ignore a few warnings that it through):
    Code:
    #include<unistd.h>
    #include<stdio.h>
    #include<stdlib.h>
    
    int main()
    {
            int a[3][4] = {1,2,3,4,
                           5,6,7,8,
                                            0,9,1,6
                                      };
            show(a,3,4);
            return(0);
    }
    
    show(int (*q)[4],int row, int col)
    {
            int i,j,*p;
            for(i=0;i<row;i++)
            {
            p = q+i;
            for(j=0;j<col;j++)
            printf("&#37;d",*(p+j));
            printf("\n");
            }
            printf("\n");
    }
    Folloing is the output that i get:
    Code:
    1234
    5678
    0916
    What do you get when you try to compile it?
    -------------------------------------------------------------------
    There are 10 kinds of people in this world....Those who understand binary and those who don't

  5. #5
    C++まいる!Cをこわせ! Elysia's Avatar
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    Quote Originally Posted by jas_atwal View Post
    Following version of your program is working fine with gcc( if we ignore a few warnings that it through):
    Code:
    #include<unistd.h>
    #include<stdio.h>
    #include<stdlib.h>
    
    void show(int (*q)[4],int row, int col);
    
    int main()
    {
    	int a[3][4] = {1,2,3,4,
    				5,6,7,8,
    				0,9,1,6
    				};
    	show(a,3,4);
    	return(0);
    }
    
    void show(int (*q)[4],int row, int col)
    {
    	int i,j;
    	int (*p)[4];
    	for(i=0;i<row;i++)
    	{
    		p = &q[i];
    		for(j=0;j<col;j++)
    			printf("&#37;d",p[j]);
    		printf("\n");
    	}
    	printf("\n");
    }
    Folloing is the output that i get:
    Code:
    1234
    5678
    0916
    What do you get when you try to compile it?
    You realize it's much easier to read when indented properly, like above?
    And x+y and *(x+y) is harder to read than &x[y] and x[y].

    And jas_atwal, you just had to go pad with spaces when the OP used tabs, didn't you? Sheesh.
    Other errors are show should have a return value, and if you return nothing, it should be void. Main should be int as pointed out.
    p should be declared as int (*)[4]. Because int* is not the type of your pointer q!
    I also modified the arrays to use [] instead of x + y and *(x + y). It's easier to read.

    ...And don't mix tabs and spaces! Use one, stick with one.

    And I did not test the code, of course.
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.
    For information on how to enable C++11 on your compiler, look here.
    よく聞くがいい!私は天才だからね! ^_^

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