Suppose i creating a linked list, the data structure is...

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Answer by Vikas
Submitted on 5/12/2006
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Malloc/Calloc work on their own way. They are provided to support non-contiguous dynamic allocation, whenever static allocation is not efficient (or even possible). They search for the available memory area for the specified number of bytes in one chunk. For the next allocation request , the amount of memory may not be available in the RAM contigously where it ended earlier. Hence you will get the memory allocation non-contigously. However if you still want to have the memory allocation contigous only and you can't use array for this purpose then use realloc it will allocate a bigger chunk of memory at some other place of RAM by copying the data of previous location. However, be cautious in using it because it may harm your program's performance greatly. Do it only when nothing else can be done. My suggestion is, use array because it is most efficient in terms of time. Memory is not a big deal these days but saving time by even 10 - 20 % is considered very good.

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Question by randal
Submitted on 4/7/2004
Related FAQ: comp.lang.c Answers to Frequently Asked Questions (FAQ List)
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Suppose i creating a linked list, the data structure is given below. struct _list { struct _list next; int data; }; The number of bytes used by this struct is 8 (4 for the next ptr, 4 for int). When i allocate objects of this type using malloc/calloc the memory addresses of the objects are not properly aligned. For eg the address of the first object (returned by malloc/calloc) is 10 then the address of the next node (using malloc/calloc) comes to 20 instead of 18. How can i force the allocation of the next node to start at address 18 rather than 20 ??

Answer by faisal
Submitted on 7/5/2004
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phobic

Answer by skishore
Submitted on 3/8/2006
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The question must be corrected as the structure can't have the structure of its own type as one of its member. so the structure must be struct _list { struct _list *next; int data; }; instead of struct _list { struct _list next; int data; }; ::SOLUTION:: just put the #pragma pack(1) preprocessor directive before this declaration.

Answer by Xzone
Submitted on 7/25/2006
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struct list { struct list * next; int data; } abc; for above struct, byte allignment can be forced by: a) Using compiler setting to change padding pattern. b) Using union Union list { char byte_allign[8]; struct list abc; } alligned_list;

Answer by Siva
Submitted on 10/16/2006
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i think , it is nothing but a structure padding and also the problem is u are creating a non-pointer self-refferential structure.

Answer by Chandra Sekhar
Submitted on 2/22/2007
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1.Pass the compiler option not to align structures at word boundaries. this option depends on compiler. 2.Get enough memory to store all the structures. and use placement new operator to place the structure at th required place

Answer by melwyn
Submitted on 3/20/2007
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Use gcc attribute __aligned__(32) which make the structure to be aligned in multiple of 4 bytes in memory

Answer by kezar
Submitted on 5/16/2007
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You can assign using calloc 'n' number of nodes which will be having sequential address in that block. For example, if your block size is for ten nodes, then those ten nodes will be in sequential address. But for the next block the address will change. that is First block is from 10 - 90 then the next block might start from 120 - 200 (ten nodes in each block).

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