Should I cast the result of malloc (in C)?

In modern C programming, you generally should not cast the result of malloc (or related memory allocation functions such as calloc or realloc). Here’s why:

No Cast Required in C
- In C, malloc returns a void*. An implicit conversion to any other pointer type (e.g., int*, char*, a user-defined structure pointer, etc.) is automatically performed by the compiler. Therefore, explicitly writing something like (int*) malloc(...) is not necessary.
Potentially Masks Missing Includes
- If you forget to include <stdlib.h> (where malloc is declared), the compiler will assume malloc returns an int by default (in pre-C99 compilers or without proper warnings). Casting that assumed int to some pointer type can hide the warning or error you’d otherwise get—potentially causing undefined behavior.
Consistency with C++
- Although C++ code typically uses new instead of malloc, if you do use malloc in C++ for some reason, then you do need a cast (C++ does not allow an implicit conversion from void*). But for C codebases, this is irrelevant.
Historical Reason
- Older K&R compilers sometimes required casts, but modern conforming C compilers do not. This practice remains in some codebases primarily for historical reasons or cross-language code that might be compiled as C++.

Example

#include <stdlib.h>  // for malloc

int main() {
    int *arr = malloc(5 * sizeof(int)); // No cast needed in C
    if (!arr) {
        // Handle allocation failure
    }

    // Use arr...
    free(arr);
    return 0;
}

Correct (no cast):
```
int *arr = malloc(5 * sizeof(int));
```

Works but discouraged (casting in C):

int *arr = (int *)malloc(5 * sizeof(int));

Recommended Resources

Conclusion

In C, don’t cast the result of malloc. It’s unnecessary and can mask compiler warnings about missing function prototypes. Just be sure to include the relevant header (<stdlib.h>) so that your compiler recognizes malloc properly.

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