Segmentation fault in C while using pointers

Question

I don't know what's happening here:

#include<stdio.h>
int main() 
{
    int i, j, *k, x,array[]={5,3,4,1,8,9,2,7,6,0};
    int *ptr=array;

    for(j=1;j<10;j++) {
        printf("---------iteration %d--------------\n",j);
        *k=*(ptr+j);   // the segmentation error is occurring here at this line
        printf("key=%d\n",*k);
        i=j-1;

        while( i>=0 && *k < *(ptr+i)) {
            *(ptr+i+1)=*(ptr+i);
            i--;
        }

        *(ptr+i+1) = *k;
        printf("%d\n",*(ptr+i+1));

        for( x=0;x<10;x++)
            printf("%d,",*(ptr+x));

        printf("\n");
    }

    for( i=0;i<10;i++)
        printf("%d,",*ptr++);

    printf("\n");
}

The error is occurring right after the printf statement in the for loop and when I remove the * from the both the sides it works but the answer is wrong.

This is an insertion sort using pointers in C.

Welcome to Stack Overflow. Do not use void main() in code posted here unless you like being shouted at for doing so. And please, please, please indent your code, 4 spaces per level (no tabs). — Jonathan Leffler
– Jonathan Leffler, Commented Jun 15, 2012 at 0:18
My apologies for that. could you please direct me to an article or could you tell me about the rules of formatting while writing a code. — hunterr986
– hunterr986, Commented Jun 15, 2012 at 12:42
And also, I have come across few variables and structures and function names that start with __ (two under scores) and also ones ending with ' _in ' and ' _t ' ! could you explain them to me. — hunterr986
– hunterr986, Commented Jun 15, 2012 at 12:44
Names that start with two underscores (or an underscore and a capital letter) are 'reserved for the implementation' for any use. Do not create such names yourself, and only use those documented, such as __FILE__ or __func__. Other names that start with an underscore are also mostly reserved to the implementation for a more restricted set of uses; again, stay clear. (C Standard, §7.1.3 Reserved Identifiers.) Types ending in _t are reserved to the implementation by POSIX. Again, tread with caution. I've not seen _in as a pervasive suffix; I'm not aware of any proscription on them. — Jonathan Leffler
– Jonathan Leffler, Commented Jun 15, 2012 at 14:13
As for C style guidelines, a Google search will uncover many. There are a few major styles. My observation about tabs and spaces is pragmatic rather than dogmatic. SO more or less treats tabs as 4 spaces, but code that was formatted under other settings doesn't look good. Indentation is crucial to easy understanding of code. Vim has a quick, effective system for indenting automatically, using =% when you are on a curly bracket { or }. Note that when editing a question or answer, the {} button indents or outdents the selection by 4 spaces. A character in columns 1-4 forces an indens. — Jonathan Leffler
– Jonathan Leffler, Commented Jun 15, 2012 at 14:19

wallyk · Accepted Answer · 2012-06-15 00:16:24Z

5

The problem is, as you said, right after printf():

*k=*(ptr+j)

I didn't get as far as looking at the right side. The left side definitely has a problem: the pointer is not initialized, so writing to that address will almost certainly cause trouble.

The right side has a memory access too, but, after inspection it looks like it might be okay.

answered Jun 15, 2012 at 0:16

wallyk

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2 Comments

user166390 Over a year ago

I wonder what the point of *k is ... looks like k alone would suffice? (As all access to indirect or "dereferenced" or whatever the correct term is ^^)

hunterr986 Over a year ago

*k is the rvalue of the pointer and if use only k then the rvalue of the pointer (ptr+j) will get placed into the pointer but not the address.

Kevin · Accepted Answer · 2012-06-15 00:24:00Z

2

You have declared k as a pointer but haven't given it any memory to point to, so there's no telling what will happen when you write to it. Give it some memory to write to with k = malloc(sizeof array).

answered Jun 15, 2012 at 0:24

Kevin

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4 Comments

hunterr986 Over a year ago

But why does it have to be initialized ? It can have any garbage value right like integers and characters etc. ? all i am doing is assigning it to an existing value inside a variable.

Jonathan Leffler Over a year ago

@hunterr986: Because if a pointer isn't initialized, you are accessing a quasi-random memory location, which typically is not a valid memory location, which is what leads to a segmentation fault. Or it might be a null pointer; that too is not valid and typically leads to a segmentation fault. If you don't know that a pointer is valid, it is not safe to use it! And No, you are not assigning to the pointer; you are assigning to what the (uninitialized) pointer points at. To assign to the pointer use: k = ptr + j; which gives you a valid pointer, assuming ptr and j are under control.

mlp Over a year ago

@hunterr986 you are partly correct, in that the thing-you-are-assigning-to can have any garbage value before the assignment, but the thing-you-are-assigning-to here is *k, and if k has a garbage value then *k is a meaningless (and invalid) concept, rather than a place you can usefully store something.

Kevin Over a year ago

@hunterr986 If you just used k = x, k wouldn't have to be initialized because, as I think you're trying to say, it's just going to be overwritten with the right data anyway. But in the code you have, you're not assigning the value to k, you're assigning the value to a point in memory, using k as the address at which the memory is located. Since you haven't told k the address it should point to, it could point you anywhere, which most likely won't be where you actually want to write the value.

SylvainL · Accepted Answer · 2012-06-15 10:36:09Z

1

As stated by the others, part of your problem is that you are using an unitialized pointer to store a value. After reading your code, it appears that you are using *k simply to store an integer value and nothing else; therefore you don't need to have a pointer and using an ordinary int value should suffice:

int i, j, k, x,array[]={5,3,4,1,8,9,2,7,6,0}; 
int *ptr=array; 

for(j=1;j<10;j++) { 
    printf("---------iteration %d--------------\n",j); 
    k=*(ptr+j);   // the segmentation error is occurring here at this line 
    printf("key=%d\n",k); 
    i=j-1; 

    while( i>=0 && k < *(ptr+i)) { 
        *(ptr+i+1)=*(ptr+i); 
        i--; 
    } 

    *(ptr+i+1) = k;
}

Furthermore, while *(ptr+i) represents the same thing as ptr[i], the usual C/C++ convention would be to use the later form.

answered Jun 15, 2012 at 10:36

SylvainL

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2 Comments

hunterr986 Over a year ago

i am trying to learn pointers so i am programming mostly using pointers.

SylvainL Over a year ago

Then you should learn how to initialise pointer before accesssing its memory location: [code] int *k, km; k = &km; [/code] With km, you reserve on the stack the memory for an int and after that, you make k to point toward the address of this reserved memory. With this, you'll be able to use *k without any corruption problem.

Chilledrat · Accepted Answer · 2012-06-15 12:15:23Z

The code below is close to your original, with the bug fixed, and a couple of tweaks for good measure. The main fix is replacing *k with key (which is pretty much what @pst and @SylvainL have already said.)

The reason, from the algorithm standpoint, is you need somewhere outside of array to hold the value being moved while the other array elements are shifted. Otherwise you end up overwriting array elements, which I suspect is what you found judging from the "...and when I remove the * from the both the sides it works but the answer is wrong." comment.

The Wikipedia entry on insertion sort has a nice animation that illustrates the point nicely:
enter image description here (image by Swfung8 CC BY-SA License page)

See code comments for other tweaks, as they aren't relevant to the question, just helpful pointers that you can take or leave ;-)

#include<stdio.h>

int main() 
{
    int array[]={5,3,4,1,8,9,2,7,6,0};

    int elm, key, i;
    int *ptr=array;

    // Calculate the number of array elements so the code will still 
    // work if the data set changes.  A good habit, rather than a necessity 
    // in this code, but it is used in two places... ;-) .

    int array_len = sizeof(array) / sizeof(int);

    for(elm=1; elm < array_len; elm++) {
        printf("---------iteration %d--------------\n",elm);
        key=(ptr+elm);
        printf("key=%d\n", key);

        // The while loop here was a for loop in disgise.

        for ( i=elm-1; i >= 0 && key < *(ptr+i); i-- ) {
            *(ptr+i+1) = *(ptr+i);
        }

        *(ptr+i+1) = key;
        printf("%d\n",*(ptr+i+1));

        // x declaration moved to here as it is only used as a loop counter

        for(int x=0; x < array_len; x++)
            printf("%d,",*(ptr+x));

        printf("\n");
    }

    for( i=0; i < 10; i++)
        printf("%d,",*ptr++);

    printf("\n");
}

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