How to store int in char * in iphone

Question

Can anyone help converting the Int to char array as i have buffer as

char *buffer = NULL;  
int lengthOfComponent = -1;   
char *obj;
buffer[index]= (char *)&lengthOfComponent;

if i do this it is thorwing EXCESS BAD ACCESS after the execution how to store the value of the obj to buffer using memcpy

Use sprintf or printf, or NSLog or NSString stringWithFormat. — nhahtdh, Jul 23 '12 at 12:42

score 2 · Accepted Answer · answered Jul 23 '12 at 12:54

2

Of course you cannot write in buffer[index], it is not allocated!

buffer = malloc(sizeof(char) * lengthOfBuffer);

should do it. After that you can write the buffer with memcpy or with an assignation, like you are doing.

answered Jul 23 '12 at 12:54

Jorge Aguirre

Mahesh · Answer 2 · 2012-07-23T12:53:35.267

0

buffer[index] = (char *)&lengthOfComponent;

buffer[index] is like dereferencing the pointer. But buffer is not pointing to any valid location. Hence the runtime error.

The C solution is using snprintf. Try -

int i = 11;
char buffer[10];

snprintf(buffer, sizeof(buffer), "%d", i);

edited Jul 23 '12 at 12:53

answered Jul 23 '12 at 12:43

Mahesh

`Incompatible pointer to integer conversion assigning to 'char' from 'char *'; dereference with *` it is throwing this warning – 012346 Jul 23 '12 at 12:47
It is exactly what is mentioned in the answer. `buffer[index]` gives the character at the index location but not the pointer. The question is why do you want a character pointer to point to an integer location. – Mahesh Jul 23 '12 at 12:49
basically i want the length of component value to store in the buffer[index] please help me out doing this – 012346 Jul 23 '12 at 12:53

2 Answers2