Why do i get 2 different output from when printing out the value in the same address?

the pointer ptr is pointing at the index 0 of the accessed Element (bar).

yet is showing me different results?

unsigned int bar[5];


int main(){

unsigned int * ptr = &bar[0];

printf("%lu\n",sizeof(ptr)); // Console output : 8 (bytes)
printf("%lu\n",sizeof(bar[0])); //Console output : 4 (bytes)
  return 0;
}

CodePudding user response：

Why do i get 2 different output from when printing out the value in the same address?

These two statements

printf("%lu\n",sizeof(ptr)); // Console output : 8 (bytes)
printf("%lu\n",sizeof(bar[0])); //Console output : 4 (bytes)

do not output "values in the same address".

The first statement outputs the size of the pointer ptr that has the type unsigned int *. This statement is equivalent to

printf("%zu\n",sizeof( unsigned int * )); // Console output : 8 (bytes)

The second call of printf outputs the size of an object of the type unsigned int. This call is equivalent to

printf("%zu\n",sizeof( unsigned int ) ); //Console output : 4 (bytes)

As you can see the arguments of the expressions with the operator sizeof in these two calls of printf are different

printf("%zu\n",sizeof( unsigned int * )); // Console output : 8 (bytes)
printf("%zu\n",sizeof( unsigned int ) ); //Console output : 4 (bytes)

If you will rewrite the second call of printf for example the following way

printf("%zu\n",sizeof( bar   0 ) ); //Console output : 8 (bytes)

then you will get the same value as the value produced by the firs call because the expression bar 0 has the type unsigned int * due to the implicit conversion of the array designator to a pointer to its first element in this expression.

CodePudding user response：

ptr is a unsigned int *. The size of this kind of pointer is 8 bytes in that environment.

bar[0] is a unsigned int. The size of this is 4 bytes in that environment.

Maybe you thought you were using *ptr?