C dynamic allocated array using memset for initialisation

Question

So i've make a program about an char array that is dinamically allocated, the program is not yet finished but i've got some different return values after i run the program, no warnings, no errors in compiler so i don't know why i get this also sometimes the program crashes..

#include <stdlib.h>
#include <stdio.h>
#include <malloc.h>
#include <string.h>
int main(){
int n,i;
char *tab=NULL;
printf("New size of array: ");
scanf("%d",&n);
tab=(char*)malloc(n*sizeof(char));
memset(tab,'X',n*sizeof(tab));
for(i=0;i<n;i++)
    printf("%c",tab[i]);
free(tab);
return 0;
}

memset(tab,'X',n*sizeof(tab)); --> memset(tab,'X',n*sizeof(*tab)); — BLUEPIXY
– BLUEPIXY, Commented Jan 17, 2016 at 23:17

Mysterion231 · Accepted Answer · 2016-01-17 23:14:14Z

2

In your memset you write n * sizeof(tab), I think you wanted to write : n * sizeof(char)

You can also add a +1 add the end of your malloc and check the return value, just for security.

answered Jan 17, 2016 at 23:14

Mysterion231

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3 Comments

user3134909 Over a year ago

thanks that did the trick :D but it's weird because I take the sintax for memset from a book and it is like this: memset(target, NULL, sizeof(target));

Mysterion231 Over a year ago

. Your book is not really clear. Instead of sizeof(target) think length. I invite you to read the man page about that it is clean. In your case n * sizeof(char) because your tab contain a certain number of char

Claudio Cortese Over a year ago

In this case you can also use sizeof *tab

Tony Tannous · Accepted Answer · 2016-01-17 23:12:28Z

2

The C library function void *memset(void *str, int c, size_t n) copies the character c (an unsigned char) to the first n characters of the string pointed to, by the argument str.

memset(tab,'X',n*sizeof(char));

You've written n*sizeof(tab), you are copying the character X to unknown memory which might cause crash during runtime.

answered Jan 17, 2016 at 23:12

Tony Tannous

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Comments

Mike -- No longer here · Accepted Answer · 2016-01-17 23:14:42Z

2

Look into this line:

memset(tab,'X',n*sizeof(tab));

You declared tab as a pointer, and a storage size of a pointer is more than one byte. Multiplying that by n which is the size of the array you want will cause you to copy data to memory space you're not allowed to access which is why you get a segmentation fault.

If you must use sizeof, then do this:

memset(tab,'X',n*sizeof(char));

or better yet, just do this:

memset(tab,'X',n);

because a char is one byte and n times 1 = n.

answered Jan 17, 2016 at 23:14

Mike -- No longer here

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