Manipulating data in memory

**congi** · 02-01-2015

So, it seems I have a bug in my code:

Code:

char *mem = malloc(9), *bottom = malloc(3), *in = "abc";
	int position = 2, i;


	mem = "onetwo";


	printf("OLD mem = %s\n", mem);


	/* save the second half */
	for(i = position; i < (9 - position); i++)
	{
		bottom[i] = mem[i + position];
		position++;
	}


	/* insert the middle */
	for(i = position; i < (position + 3); i++)
	{
		mem[i] = in[i];
	}


	/* put it back together */
	for(i = (position + 3); i < 9; i++)
	{
		mem[i + position] = bottom[i];
	}


	printf("NEW mem = %s\n", mem);

What I'm trying to do is insert "abc" in the 3rd position of "onetwo". Results would be "oneabctwo", but all I get is crashing.

There's got to be a better way to do this, I'm not sure.

Thank you for your suggestions,

Congi

**Salem** · 02-01-2015

> mem = "onetwo";
You didn't copy the memory, you just made it point somewhere else.

This new place has the distinction of being read-only, so as soon as you try to write to it, the program crashes.

> bottom[i] = mem[i + position];
It seems unlikely that bottom has enough memory to contain whatever you're trying to write into it.
It also seems likely that mem isn't long enough either.

**congi** · 02-01-2015

I don't quite understand it.. is there a better way to do this or an example?

Goal is to insert (not overwrite) data ("abc") in a position of one piece ("onetwo") of memory from another one.

Currently I'm trying to do this by saving the data after the position in another buffer, overwriting with inserted data, and then putting back after the inserted data to make a new memory.

**Salem** · 02-02-2015

Code:

strncpy(&s2[to],&s1[from],n);

Copies n chars from a position in s1, to a position in s2.

You only need this, and something to track where the \0 goes when you're done.

**congi** · 02-02-2015

I see..

Just curious, would this also work if the data is not a string (which is ended with a \0)?

**Salem** · 02-02-2015

You should try it, and find out.

I know I can do it, but there are still plenty of ways to make a mess of it.

Post your attempt, then we can see whether you've understood everything.

**congi** · 02-02-2015

Ok!

Code:

char *mem = malloc(9), *bottom = malloc(3), *in = "abc";
	int position = 2, i;


	mem = "onetwo";


	printf("OLD mem = %s\n", mem);


	strncpy(&mem[position], in, 3);


	printf("NEW mem = %s\n", mem);

Getting a crash when running... Just prints "OLD mem" and then stops.

**laserlight** · 02-03-2015

This overwrites mem, which contained the return value of the malloc call:

Code:

mem = "onetwo";

**Salem** · 02-03-2015

> mem = "onetwo";

You missed this part of the information previously given
> This new place has the distinction of being read-only, so as soon as you try to write to it, the program crashes.

I suggest you forget about micro-managing malloc down to the last byte (you seem to be almost always off by 1 anyway), and start with

Code:

char mem[100] = "onetwo";
char in[100] = "abc";
int position = 2;
strncpy(&mem[position], in, 3);

At least until you understand exactly what is going on in memory, and the actual number of bytes needed to store a string.

Then make a go of the malloc version.

**congi** · 02-11-2015

I think I understand it now... thanks for your help!!

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