To answer this question, let's go through the code step by step:
main() {
int a[5] = {10,20,30,40,50};
char *p;
p=(char *)a;
printf("%d\n",a);
}
In this code, an integer array a
is declared and initialized with values 10, 20, 30, 40, and 50.
The variable p
is declared as a character pointer.
The line p=(char *)a;
casts the integer array a
to a character pointer p
. This means that p
now points to the first element of the integer array a
.
The printf("%d\n",a);
statement prints the value of the integer array a
. However, the format specifier %d
is used to print integers, not arrays.
So, the correct format specifier to print the value of a
would be %p
, which is used to print pointers.
Now, let's go through the options:
A) Some value - This option is correct. When we use the incorrect format specifier %d
to print the array a
, it will print some unexpected value. The exact value printed will depend on the system and compiler being used.
B) Compilation error - This option is incorrect. The code will compile without any errors.
C) 10 - This option is incorrect. Since the incorrect format specifier %d
is used, it will not print the value 10.
D) 50 - This option is incorrect. Since the incorrect format specifier %d
is used, it will not print the value 50.
The correct answer is A) Some value, as explained above.