Question

Is there a leak in this C code?
Although the answer is coming correct, but just wanted to know if it is Ok to write like this:

// _mm_adds_epi16 : Adds the 8 signed 16-bit integers in a to the 8 signed 
//16-bit integers in b and saturates
__m128i t7=_mm_adds_epi16(  t5 ,t6  );
unsigned short *p= (unsigned short *)malloc(8);
p=(unsigned short *)&t7;
for(int i=0;i<8;i++)
{
    printf("%d\n", p[i]);
}

Updated

So now I am updating it as follows:

// _mm_adds_epi16 : Adds the 8 signed 16-bit integers in a to the 8 signed 
//16-bit integers in b and saturates
__m128i t7=_mm_adds_epi16(  t5 ,t6  );
unsigned short *p= (unsigned short *)malloc(8);
p=(unsigned short *)&t7;
for(int i=0;i<8;i++)
{
    printf("%d\n", p[i]);
}

 free(p);

Do I still have leak?

What is the correct way to print t7

Answer 1

Every malloc in C needs to have some corresponding free which is reachable before termination. Any exception to this constitutes a memory leak.

You store the return of malloc in p and lose the pointer by overwriting it. As such there is no chance for the pointer to be freed. In your particular case, the malloc itself is redundant since you are not using the return in any way.

---

In response to your question of whether the new code has a memory leak, yes it does. malloc allocates memory and returns a pointer to the allocated memory. You are losing the pointer by overwriting its value then never using the allocated memory. Even worse you are now calling free on what appears to be an automatic variable which is undefined behaviour.

If the printing is working correctly, you can just do this:

__m128i t7 = _mm_adds_epi16(t5, t6);
unsigned short *p = (unsigned short *)&t7;
for(int i=0;i<8;i++) {
    printf("%d\n", p[i]);
}

The malloc is redundant because you are not even using the memory it allocates.