How to set a pointer to struct to NULL

Question

typedef struct _DocumentRow
{
    char * code /** The code */;
    char * designation /** The designation */;
    double quantity /** The quantity */;
    char * unity /** The unity */;
    double basePrice /** The base price */;
    double sellingPrice /** The selling price */;
    double discount /** The discount */;
    double rateOfVAT /** The rate of VAT */;
    struct _DocumentRow * next /** The pointer to the next row */;
} DocumentRow;

void DocumentRowList_init(DocumentRow ** list) {
    DocumentRow *L;
    list = ( DocumentRow ** ) malloc( sizeof( DocumentRow* ) );
    if ( list == NULL ) {
        fatalError( "memory is not enough" );
    }
    L = NULL;
    list = &L;
}

After using the function DocumentRowList_init, when I test if ( *list == NULL ), it evaluates to false, why ? I have already set list = &L and L = NULL.

don't follow you question... if *list==NULL returns 0, *list is not NULL. — tristan
– tristan, Commented Jan 3, 2014 at 8:06
To me, this code seems to do nothing, e.g. L = NULL; list = &L;?! — Sebastian
– Sebastian, Commented Jan 3, 2014 at 8:07
also L = malloc( sizeof( DocumentRow ) ); if(L == NULL) ... — BLUEPIXY
– BLUEPIXY, Commented Jan 3, 2014 at 8:08
What you wrote is called a memory leak. You allocate memory pointed by list and then you make list to point somewhere else. You will not be able to free(list). And *list is not NULL because *list is L which is a pointer to NULL. Outside this function, testing *list may result in a segmentation fault. — francis
– francis, Commented Jan 3, 2014 at 8:12

Devolus · Accepted Answer · 2014-01-03 08:31:30Z

2

You would have undefined behaviour here. L is a local variable, so when you return it's address via the pointer pointer, the variable no longer exists when DocumentRowList_init is returning.

So even though you assign NULL to it, it will point to invalid memory.

But list is local to DocumentRowList_init, so it will not return the value anyway, as you only assign it a value and then return.

If you want to return a structure of DocumentRow you'd have to use this

  *list = malloc( sizeof *L);

to allocate a structure and return the pointer to it.

edited Jan 3, 2014 at 8:31

answered Jan 3, 2014 at 8:26

Devolus

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Comments

Lee Duhem · Accepted Answer · 2014-01-03 08:36:23Z

2

Looks like you want to change (initialize) something pointed to by list, here is how it usually be done:

void DocumentRowList_init(DocumentRow ** list) {
    *list = ( DocumentRow * ) malloc( sizeof( DocumentRow ) );
    if ( *list == NULL ) {
        fatalError( "memory is not enough" );
    }
}

edited Jan 3, 2014 at 8:36

answered Jan 3, 2014 at 8:22

Lee Duhem

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1 Comment

Boris Brodski Over a year ago

You malloc only size(DocumentRow*) bytes for a pointer, not size(DocumentRow) for the struct. In this case the type cast to ( DocumentRow ** ) is correct.

Pandrei · Accepted Answer · 2014-01-03 08:40:29Z

0

void DocumentRowList_init(DocumentRow ** list) {
    DocumentRow *L;
    list = ( DocumentRow ** ) malloc( sizeof( DocumentRow* ) );
    if ( list == NULL ) {
        fatalError( "memory is not enough" );
    }
    L = NULL;
    list = &L;
}

Please mind that list=&L assigns list the address of a variable located on the stack. So once you exist the function the variable is out of scope and your list is left pointing to a some location (on the stack).

answered Jan 3, 2014 at 8:40

Pandrei

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2 Comments

Alex Over a year ago

So, what can I do ? I want to make *list = NULL, but it also can't work

Pandrei Over a year ago

you can make DocumentRow *L a global variable; or dynamically allocate memory for L: L= (DocumentRow *)malloc(sizeof(DocumentRow *))

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