Linked lists in C. Segmentation fault

Question

#include <stdio.h>


struct mychar{
    char value;
    struct mychar *nextPtr;
};

typedef struct mychar Mychar;
typedef Mychar *MycharPtr;


void insert(MycharPtr *, char );
void printlist(MycharPtr);


int main(){
    MycharPtr startPtr = NULL;

    char b = 'b';

    insert(&startPtr, b);

    printlist(startPtr);
}

void insert(MycharPtr *sPtr, char newvalue){

    MycharPtr newlinkPtr;

    if (*sPtr == NULL){
        newlinkPtr->value = newvalue;
        newlinkPtr->nextPtr = NULL;
    }

    *sPtr = newlinkPtr;
}


void printlist(MycharPtr currentPtr){
    printf("%c", currentPtr->value);
}

I'm just starting by only adding one char. If I can't even do that, I can't go on doing else.

It gives me segmentation fault, but I don't really know why.

Also, I still don't get the reason why in insert call I should write & but in printlist call I shouldn't write &.

P.P · Accepted Answer · 2020-05-23 07:18:59Z

1

You haven't actually allocated memory for newlinkPtr. So you're just dereferencing and attempting to write to an uninitialized pointer, resulting in undefined behaviour.

In insert, you can modify it to:

MycharPtr newlinkPtr = malloc(sizeof *newlinkPtr);
if (!newlinkPtr) {
    perror("malloc");
    return;
}

...

This also, sort of, illustrates why typedef'ing a struct pointer could mislead and considered a bad practice. I'd suggest avoiding it.

answered May 23, 2020 at 7:18

P.P

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6 Comments

Mnkisd Over a year ago

Thanks a lot. I hate this typedef of ptr too... it was in my book so I thought that for good practice of programming I'd better use it, but it actually confuses me. One question: I don't really get why initializing with malloc is needed... Shouldn't creating a new pointer without malloc actually allocates memory? Why it doesn't and I need malloc?

P.P Over a year ago

To be clear, typedef'ing pointer is not always bad (opaque types are created that way). But in this case, it's unnecessary and misleading. re. need for malloc: the definition of mylinkPtr just creates a pointer (for the pointer itself memory is allocated). But it doesn't point to anywhere valid. So you need to make it point to a valid block of memory in order to write.

Mnkisd Over a year ago

So when I do MycharPtr newlinkPtr = malloc(sizeof *newlinkPtr); I am actually creating a new struct node and a pointer that points to that struct, right? So this is the reason why before it didn't work, because I was only creating a pointer with MycharPtr newlinkPtr; but not creating the new struct.

P.P Over a year ago

That's correct. Without the typedef, it'd look like: struct mychar *newlinkPtr = malloc(sizeof *newlinkPtr);. Does it make it easier to understand?

Mnkisd Over a year ago

Yes, thanks you! Wouldn't be the same if I do: Mychar *newlinkPtr = malloc(sizeof(Mychar)); ?

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