MCQs
The sizeof function return the given variable. Example: float a=10; sizeof(a) is 4 bytes
Step 1: float arr[] = {12.4, 2.3, 4.5, 6.7}; The variable arr is declared as an floating point array
and it is initialized with the values.
Step 2: printf("%dn", sizeof(arr)/sizeof(arr[0]));
The variable arr has 4 elements. The size of the float variable is 4 bytes.
Hence 4 elements x 4 bytes = 16 bytes
sizeof(arr[0]) is 4 bytes
Hence 16/4 is 4 bytes
Hence the output of the program is '4'.
Since C is a compiler dependent language, it may give different outputs at different platforms.
We have given the TurboC Compiler (Windows) output.
Please try the above programs in Windows (Turbo-C Compiler) and Linux (GCC Compiler),
you will understand the difference better.
Step 1: int arr[] = {12, 14, 15, 23, 45}; The variable arr is declared as an integer
array and initialized.
Step 2: printf("%u, %u`setminus`n", arr, &arr); Here,
The base address of the array is 65486.
=> arr, &arr is pointing to the base address of the array arr.
Hence the output of the program is 65486, 65486
Step 1: int arr[1]={10}; The variable arr[1] is declared as an integer array with size '2'
and it's first element is initialized to value '10'(means arr[0]=10)
Step 2: printf("%d`setminus`n", 0[arr]); It prints the first element value of the variable arr.
Hence the output of the program is 10.
Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer
array having the 3 rows and 4 colums dimensions.
Step 2: printf("%u, %u`setminus`n", a+1, &a+1);
The base address(also the address of the first element) of array is 65472.
For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore,
a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 +
(4 ints * 2 bytes) = 65480
Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes
"12 ints * 2 bytes * 1 = 24 bytes".
Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496
Hence the output of the program is 65480, 65496