Question: c programming int count = 56; int *int_ptr = &count; count++; int **int_ptr2 = &int_ptr; **int_ptr2 += 2; *int_ptr = 104; printf(%i, *int_ptr2+2); Given the

c programming

c programming int count = 56; int *int_ptr = &count; count++;

int count = 56; int *int_ptr = &count; count++; int **int_ptr2 = &int_ptr; **int_ptr2 += 2; *int_ptr = 104; printf("%i", *int_ptr2+2); Given the code above run in x86 32-bit Linux Machine. Assuming the memory address of variable count is 3509, the memory address of variable int_ptr is 8338, and the memory address of variable int_ptr2 is 17867. What's the output

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