Assign the values of an array to a pointer

Question

I am first concatenating a series of elements in an auxiliary char array to then assign the concatenated array to the pointer. The problem comes when assigning this char array to the pointer, where it produces a segmentation fault.

My approach was the following one:

char aux_name [12];
char * name = (char *) malloc(sizeof(char)*13);
int i;

for(i = 0; i < 5; i++){
    sprintf(aux_name, "id_%i", i);
    *name = (void *) (intptr_t) aux_name; //Conflict line

   //Do something with variable name (it is required a pointer)

}

Standard Warning : Please do not cast the return value of malloc() and family in C. — Sourav Ghosh
– Sourav Ghosh, Commented Apr 27, 2015 at 11:16
Side note: what's the point in performing the exact same assignment 5 times??? — barak manos
– barak manos, Commented Apr 27, 2015 at 11:46

Community · Accepted Answer · 2017-05-23 12:14:23Z

1

You don't assign a pointer value to an already malloc()-ed pointer, you'll be facing memory-leak there. You have to use strcpy() to achieve what you want.

OTOH, if you don't allocate memory dynamically, then you can assign the pointer like

name = aux_name;

That said,

I am first concatenating a series of elements in an auxiliary char array

Well, you're not. you're simply overwriting the array every time in every iteration. What you need to do is

Collect the return value of sprintf() every time.
next iteration, advance the pointer to buffer by that many locations to concatinate the new input after the previous one.

Note / Suggestion:

do not cast the return value of malloc() and family in C.
sizeof(char) is guranteed to be 1 in c standard. You don't need to use that, simply drop that part.

edited May 23, 2017 at 12:14

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answered Apr 27, 2015 at 11:17

Sourav Ghosh

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2 Comments

Lundin Over a year ago

The remark about sizeof(char) is subjective. Some believe it should be there to make all malloc calls consistent. To use it or not is personal preference, there's no right or wrong (unlike casting the result of malloc, which is always bad/superfluous).

Sourav Ghosh Over a year ago

@Lundin sir, you're very right. But just as per one of the logics behind not casting malloc() result [the It makes you repeat yourself, which is generally bad. part], we can suggest not to use sizeof(char), isn't it? However, I updated my remark. Hope it is ok now. :-)

Iharob Al Asimi · Accepted Answer · 2015-04-27 11:18:38Z

0

You can't do that, and you don't really need to, this would work

size_t nonConstanSizeInBytes = 14;
char *name = malloc(nonConstanSizeInBytes);

snprintf(name, 13, "id_%i", i);

answered Apr 27, 2015 at 11:18

Iharob Al Asimi

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