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when using malloc, it returns void type. we need to type cast. And the acutall memory allocated is larger than you applied, the first part stores the number of bytes allocated (offset to teh next memory from the start of this memroy). For example, if you do int *i = (int*) malloc(12), it allocates 16 bytes. The first 4 bytes for actuall size, from 5th byte is return to i, now i can acess 12 bytes from this byte.

node points to first byte of address.
but shouldn't this be
List* node;
--- node=malloc(sizeof(node));
+++ node=malloc(sizeof(LIST));

because when you allocate using sizeof(node) it will returns size of address and not the complete struct size. and when we try to write we might get into segmentation faults.

malloc allocates the memory for the requested amount of bytes and returns the base address of the memory location that was allocated.

Node points to the first byte of the address that was allocated by malloc

List* node;
node=malloc(sizeof(node));
.........
above statement will give error:
cannot convert from 'void *' to List*


malloc return void* pointer to location where memory is allocated

correction:

List* node;
node=(node*)malloc(sizeof(node));

when using malloc, it returns void type. we need to type cast. And the acutall memory allocated is larger than you applied, the first part stores the number of bytes allocated (offset to teh next memory from the start of this memroy). For example, if you do int *i = (int*) malloc(12), it allocates 16 bytes. The first 4 bytes for actuall size, from 5th byte is return to i, now i can acess 12 bytes from this byte.

malloc returns void*.
malloc(sizeof(node)); will allocate memory of the size of node.
node points to the first byte of the address that was allocated by malloc.

but malloc(sizeof(node)) might not allocate the required memory for the data type List.

Not needed in c, i will not recomend to use in C, List* node;
node=malloc(sizeof(node)); will work fine in c But casting needed in c++.