Store int inside char array in C

Question

I am trying to store int value in char array, and access it again to retrieve value later. here is what i tried, but it is not working as i am looking for.

char mem[50];
int* size = (int*)&mem[0];

char* isFree = (char*)&mem[4];

char *t = (char *)&mem[4];

Now I want to store 1234, and 'f' in the char array according to the order. (within first 5 bytes)

I am not allowed to use other variables, have to use char array.

int a = 1234;
int *size = &a;

above one is not allowed.

You need to read the basics of *pointers*. To store a value you have to do `*size = 1234;` — Haris, Nov 12 '15 at 16:08
i am not allowed to use other variables than using array space. so this won't work. I edited question. so it will be more clear. int a = 1234; size = (int *)&a; — dhanushkac, Nov 12 '15 at 16:11
@Haris That also after allocating memory or pointing to another variable. — ameyCU, Nov 12 '15 at 16:12
actually i am implementing own malloc. so I am trying this for that work. — dhanushkac, Nov 12 '15 at 16:13
@ameyCU, `size` is pointing to the first element of `mem[]` array. Why would he need to allocate memory? — Haris, Nov 12 '15 at 16:14
@ameyCU, I meant instead of `size = (int *)1234;`, which he had before. Its now edited. — Haris, Nov 12 '15 at 16:16
@dhanush_c You might want to make it more clear what you want to store where (and possibly also *why*); I read "Now I want to store 1234, and 'f' in size and isFree" but I don't understand it. — anatolyg, Nov 12 '15 at 16:29
@anatolyg sorry, I edited question, my mistake. sorry about that. — dhanushkac, Nov 12 '15 at 16:32

score 5 · Answer 1 · answered Nov 12 '15 at 16:13

5

You have to use the type unsigned char to copy the memory or memcpy which does that for you. Aliasing a char array as an int is undefined in C.

store the value:

int value = 1234;
memcpy( mem, &value, sizeof(value) );

retrieve the value:

memcpy( &value, mem, sizeof(value) );

(The array mem must be large enough to hold the type int.)

answered Nov 12 '15 at 16:13

this

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thank you for answering my question. but this won't work. i am not allowed to use other variables than using array space. if not i can do this by using int a = 1234; size = (int *)&a; that's the issue – dhanushkac Nov 12 '15 at 16:16
Why not `char`? I see both mentioned in the description of strict aliasing [here](http://stackoverflow.com/a/99010/509868). Is there some additional rule I am not aware of? – anatolyg Nov 12 '15 at 16:16
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@anatolyg Signed char could potentially have a value with trap representation if using one's complement or sign+magnitude, if I read the C11 Standard correctly: 6.2.6.2,p2. – this Nov 12 '15 at 16:25

score 2 · Answer 2 · answered Nov 12 '15 at 16:20

2

You can use sprintf -

int size=1234;
char mem[50];
sprintf(mem,"%d",size);

Note- This will append null terminator.

answered Nov 12 '15 at 16:20

ameyCU

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score 1 · Accepted Answer · answered Nov 12 '15 at 16:27

1

I have tried another way and i got a solution. I am posting it as a help for others.

char mem[50];
int* size = (int*)&mem[0];
*size = 1234;

char* isFree = (char*)&mem[4];
*isFree = 'f';

char *t = (char *)&mem[4];

printf("%d\n",*(int*)&mem[0]);
printf("%c\n",*(char*)&mem[4]);

I assigned a pointer to a pointer. I worked. Thank you all for answering.

answered Nov 12 '15 at 16:27

dhanushkac

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On select systems, `mem[]` maybe not be properly aligned for an `int` array, so `int* size = (int*)&mem[0];` is UB. – chux - Reinstate Monica Nov 12 '15 at 18:47

Store int inside char array in C

3 Answers3