A question about memory allocation

I've just got a quick question about how memory is stored on the computer..
Say you had 16bytes of memory, had 3 arrays of characters, one 4 bytes, one 2 bytes and one 6 bytes long:

cstr1 = (char*)malloc(sizeof(char)*4)
cstr2 = (char*)malloc(sizeof(char)*2)
cstr3 = (char*)malloc(sizeof(char)*6)
cstr1="test";
cstr2="ok";
cstr3="testing";

0001 t
0010 e
0011 s
0100 t
0101 o
0110 k
0111 t
1000 e
1001 s
1010 t
1011 i
1100 n
1101 g
1110
1111

Now if you were to free up cstr2, there would be a gap of 2 bytes between cstr1 and cstr2. If you were to attempt to create another array of characters, 4 bytes long, would the compiler shift the array holding 'testing' up two bytes, to fit in the new array, or would it simply generate an error?

well, first off the memory wouldn't look like that.

you're allocation memory, then assigning a pointer to a constant string, which is automatically stored elsewhere in memory.. not to mention the absence of null terminators.

i think i can answer the question though. i'm not 100% on what you intend to do, but if there are 2 bytes of free memory and you attempt to allocate 4, the program will look for the first space where there's 4 available bytes in a row. if you attempted to copy the contents of the array into that space, then yeah you would have a problem.

It depends on the malloc implementation.
In general, malloc uses a chain of pointers to available memory blocks. This is called the free chain. When you allocate memory, malloc walks the free chain looking for a chunk big enough. This is quite a primitive system.

So in principle, if you have at least 4 bytes free, you will be able to allocate a 4-byte block of memory. However, depending on how scattered the allocated blocks are, malloc may have to first clean up the free chain by re-arranging the currently allocated blocks of memory. This can take quite a long time, but you will get your 4 bytes.

An error will occur if you try to allocate more bytes than are available anywhere in the free chain.

ok, thanks for the answers guys =)

Quote:

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Originally Posted by penagate

malloc may have to first clean up the free chain by re-arranging the currently allocated blocks of memory.

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Um, it can't do that in C or C++, not on common implementations where pointers are direct memory addresses.

It can (and does) do it on hosted environments like the Java or .Net runtime.

Sorry, I meant the free blocks.