Need help in understanding left shift operator in C

Question

I have a few issues understanding the left shift operator. I know that if shifts by x amount of bits to the left, therefore pretty much multiplying the number to 2 to the power of the number of bits shifted.

But consider this program:

int a=4,*pA=&a,r3;
printf("%d\n",pA);
printf("%d\n",(*pA<<2));
r3=pA-(*pA<<2);
printf("%d",r3);

It prints out pA, and (*pA<<2) separately, and if you were to subtract these two, it won't equal to the r3 written in the expression, being off by 4 times, which is the size of int. But where do you need to take the size of int into consideration, since you have both pA and (*pA<<2), and their subtraction doesn't equal to what it's supposed to.

Any help is appreciated...

For the record, I am not interested in printing out the actual value of the pointer, but actually its address being shifted by 2 bits. And I don't understand the exact process that is happening.

For a=4, pA comes up as 2686724, and *pA<<2 is 80. But r3 comes up as 2686404, instead of 2686644. — , Jan 20 '17 at 19:53
`pA` is a pointer.. You're printing the address of the pointer. To print the actual value, you'd need to dereference it. — tambre, Jan 20 '17 at 19:53
Possible duplicate of [What does "dereferencing" a pointer mean?](http://stackoverflow.com/questions/4955198/what-does-dereferencing-a-pointer-mean) — tambre, Jan 20 '17 at 19:54
No, I don't want to use the value that pA is pointing at, I want to use the address. — , Jan 20 '17 at 19:56
arithmetic on a pointer scales with the size of what it points to. So when you do `pA - 80` your code actually does `pA - 80*sizeof(*pA)`. This has nothing to do with `<<`. — Kevin, Jan 20 '17 at 20:07
Instead of `r3=pA-(*pA<<2);` did you mean `r3=*pA-(*pA<<2);`? — Weather Vane, Jan 20 '17 at 20:07

score 5 · Accepted Answer · answered Jan 20 '17 at 20:08

5

Arithmetic on a pointer scales with the size of what it points to. So when you do pA - 80 your code actually does pA - 80*sizeof(*pA). This has nothing to do with <<.

Also you should print addresses using the %p format specifier. And to store pointer values in an integer type use intptr_t or uintptr_t (from stdint.h). Enable warnings on your compiler and it should complain if you don't do this.

answered Jan 20 '17 at 20:08

Kevin

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Makes the most sense. Thank you. – Jan 20 '17 at 20:16

score 1 · Answer 2 · answered Jan 20 '17 at 20:11

The dereference operator * has higher precedence than the left-shift operator <<. So the expression *pA<<2 actually evaluates to (*pA)<<2. This give you the value pointed to by pA, i.e. a left shifted by 2.

It seems like what you want is pA<<2, however a pointer is not a valid operand to the << operator. Multiplying a pointer value is not defined, nor does it makes sense. You can however add to or subtract from a pointer.

The index operator [] is a syntactic sugar around adding to a pointer and dereferencing, so you could use that. However, pA points to a single int, not an array, so doing so is undefined behavior.

@Radu That's what you're doing now, i.e. shifting the value not the pointer. Shifting a pointer makes no sense. — dbush, Jan 20 '17 at 20:17

Need help in understanding left shift operator in C

2 Answers2