Question about malloc()

[ALa Rihane]

Hello.
I'm new in this forum and it looks a nice site for me to get helped and maybe help others .


I'm new on the C laguage and i want to know if i can do this with malloc:


Allocating a memory for an int array of n*sizeof(int) and and changing the array size just with changing the value of n, just like that example:

void main(){
int *array
int n=5; //n is the size of the array.

//allocate the memorie for the array
array = (int *) malloc(n*sizeof(int));

n++; //when i change n value does the size of the array will change also ?

array[5]=7; //giving a value to the new array square without realloc() or other thing...

}

The compiler don't show any error and it works, but it seems to be not logic...

Thank you for helping.
Regards.

[laserlight]

when i change n value does the size of the array will change also ?

No, not unless you also do a realloc or something similiar to "expand" the dynamic array.

The compiler don't show any error and it works, but it seems to be not logic...

You are accessing the array out of bounds, resulting in undefined behaviour. It seems to work, but it might not.

Other things to note:

• #include <stdlib.h> for malloc.
• void main should be int main
• Remember your semi-colons.
• Do not cast the return value of malloc unless you need the code to be compilable as C++.
• free what you malloc.
• Indent your code properly.

[ALa Rihane]

thank you for reply, but works because of compiler do some changes on the compilation ? or what ? because also when i do this code it work also:

#include <stdio.h>#include <stdlib.h>
#include<conio.h>


void main(){
    int tab[0];
    printf("sizeof tab %d\n",sizeof(tab));
    tab[5] = 6;
    tab[7] = 8;
    tab[6] = 7;
    printf("%d, %d, %d,",tab[5], tab[7], tab[6]);
    printf("\nsizeof tab %d\n",sizeof(tab));
    getch();
}

[laserlight]

thank you for reply, but works because of compiler do some changes on the compilation ? or what ?

Because you were "lucky". In real life, your program could appear to work until your boss demonstrates it to the client, shortly after which you will find yourself looking for a new job. Don't write code that accesses arrays out of bounds. Just don't.

because also when i do this code it work also:

Zero sized arrays are not allowed in standard C. I compiled and ran your program, and the process crashed. Even if there was no crash, it is still wrong.

[grumpy]

Zero sized arrays are not allowed in standard C.

I don't have my copy of the C standard handy, but don't recall anything in the standard that actually disallows zero-sized arrays.

Except the fact that accessing array elements out of bounds is always undefined. Even if creation of a zero-sized array is valid, any operation that accesses its elements will yield undefined behaviour. But "undefined behaviour" is a different concept than "not allowed".

I also seem to recall that zero sized arrays are allowed by most implementations (albeit with a warning, if enabled) because some automatic code generators routinely produce such constructs (but do not subsequently access elements in an invalid way).

[cas]

C99 6.7.5.1, under "Constraints": "If the expression is a constant expression, it shall have a value greater than zero." So it's a constraint violation. The same holds for C89, with different wording (because C89 allowed only constant expressions).

For VLAs, a size of 0 or smaller is undefined, because of course it cannot (generally) be detected at compile time if an object has the value 0.

[iMalc]

Zero-sized arrays may be allowed on some compilers as a language extension, but it's certainly not standard or portable.