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cpointersprintfsizeofmemory-address

Why do i get 2 different output from when printing out the sizeof() pointer vs variable

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;
}
Solution

  • 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.