Can someone please explain to me the difference between
int *x = malloc(sizeof(int));
&&
int *x = (int*)malloc(sizeof(int));
Thanks!
Asked
Active
Viewed 90 times
1
Haris
- 12,120
- 6
- 43
- 70
Oliver Atienza
- 631
- 4
- 8
-
`malloc` returns a void pointer. As you are assigning it to `int` pointer, it's always good to type cast. It won't make any difference in code but you will avoid one warning. – Swanand Nov 19 '15 at 08:28
-
@Swanand, you will not get any warning if you don't type caste the return of `malloc()`. – Haris Nov 19 '15 at 08:29
-
1Its compiler dependent. eg, If you are writing `C` program in a `C++` compiler, then you need to cast. – SKD Nov 19 '15 at 08:37
-
@Swanand `malloc`'s return is `void*`, this means that in `C` there is no need to `cast` `malloc`, but if you need to do that probably is because you are using `a wrong Compiler` or maybe the `wrong Language`. Now, could you please explain us why **it's always good to type cast.**? Trust me, there will be a big difference at some point by doing that. Please explain. – Michi Nov 19 '15 at 09:43
-
@Haris & @ Milchi Yes... You both are correct. My compiler is wrong. I tried with GCC and it was without any warning. I am using an eclipse based compiler with lot of saftey plugins (MISRA, Lint etc) that's why I got warning. – Swanand Nov 19 '15 at 09:46
-
@Swanand [Please take a look here](http://stackoverflow.com/questions/7545365/why-does-this-code-segfault-on-64-bit-architecture-but-work-fine-on-32-bit). – Michi Nov 19 '15 at 09:46
1 Answers
2
The difference is that you are casting the return of malloc() in the second example. malloc() returns a void* pointer, which is automatically and safely promoted to any other pointer type in this case.
Therefore casting in this case is not required and should not be done. Check here.
