What is the the output of the following C programme? - c

Question

I don't know what the compilar is doing with ++*p;
Can anyone explain me pictorically what is going on inside the memory in this code?
int main()
{
int arr[]={1,2,3,4};
int *p;
p=arr;
++*p;
printf("%d",*p);
}

Answer 1

The answer should be 2
The reason is ++*p is is actually incrementing the first member in the array by 1.

Answer 2

You are incrementing the first element in that array by creating another int pointer p that points to the element. The line
++*p
increments the value of the object pointed to by p - in this case it is the first element in the array.

Answer 3

Making up the actual memory addresses and using "ma" for memory address
at memory address starting at 1000 we have 4 continuous 4-byte (sizeof(int) = 4) slots.
each slot contains the integer value given in the array initializer:
arr
ma1000 ... ma1015
_____________________
| 1| 2| 3| 4|
_____________________
arr gives the starting address of the 4 int slots and how many there are.
p holds the address of an integer and refers to one 8-byte slot in memory (assuming we are on a 64-bit system where pointers are 8 bytes - 64 address bits/8bits-per-byte) at location 2000.
After the statement p = arr, p holds the address 1000
p *p or arr[0]
ma2000 .. ma2007 ma1000 .. ma1003
__________ ________
| 1000| | 1 |
__________ ________
*p gives the value at the memory address pointed to by p. p holds memory address 1000 and memory address 1000 contains 1, thus *p results in 1.
++*p says to increment the value of the int "pointed to" by p. p holds memory address 1000 which holds the value 1. The value at address 1000 then goes from 1 to 2
arr
ma1000 ... ma1015
_____________________
| 2| 2| 3| 4|
_____________________
printf then prints the int value at the address "pointed to" by p, which is 2.
p *p or arr[0]
ma2000 .. ma2007 ma1000 .. ma1003
___________ ___________
| 1000| | 2|
___________ ___________

Answer 4

Let's take this line :
++*p
This will first dereference the p pointer, so access arr[0], then increment it.
if you print arr[0] now, it will be 2.
Then you print *p that is the same as printing arr[0], it equals 2.
Try replacing the first element by 41, your code will print 42.

Answer 5

When you have a pointer int *p, then p means "the memory address p points to and *p means "the contents of the memory address where p points to". ++ means pre-increment, which means increase the value by 1. Since the associativity of unary operators such as * and ++(prefix operators) is right-to-left, and * is closer to p than ++ when traversed right-to-left, * operates before ++. Therefore ++*p means "increment the value pointed to by p", not "increment p then get the value". It is clearer to write ++(*p), but it means the same thing.
The thing to understand is the difference between "increment the pointer", which means point to another value (e.g. ++p) and "increment the value that the pointer points to" (e.g. ++*p).
You can work this out, but that is all the information you need to understand it.

Answer 6

This is a very basic concept in understanding of pointers.
p = arr ;
The above code will make the pointer " p " to point towards what the " arr " is pointing to. " arr " itself is pointing to the 1st element.
*arr = *(arr + 0) = arr[0] = 1
*(arr + 1) = arr[1] = 2
*(arr + 2) = arr[2] = 3
and so on...
So now when you do ++(*arr) that means ++(1) = 2
As p = arr, you can do the remaining replacement and math.