Step 1: int a[5] = {5, 1, 15, 20, 25}; The variable arr is declared as an integer array with a size of 5 and it is initialized to a[0] = 5, a[1] = 1, a[2] = 15, a[3] = 20, a[4] = 25 . Step 2: int i, j, m; The variable i,j,m are declared as an integer type. Step 3: i = ++a[1]; becomes i = ++1; Hence i = 2 and a[1] = 2 Step 4: j = a[1]++; becomes j = 2++; Hence j = 2 and a[1] = 3. Step 5: m = a[i++]; becomes m = a[2]; Hence m = 15 and i is increemented by 1(i++ means 2++ so i=3) Step 6: printf("%d, %d, %d", i, j, m); It prints the value of the variables i, j, m Hence the output of the program is 3, 2, 15
Since C is a compiler dependent language, it may give different outputs at different platforms. We have given the Turbo-C 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\n", arr+1, &arr+1); Here, the base address(also the address of first element) of the array is 65486. => Here, arr is reference to arr has type "pointer to int". Therefore, arr+1 is pointing to second element of the array arr memory location. Hence 65486 + 2 bytes = 65488 => Then, &arr is "pointer to array of 5 ints". Therefore, &arr+1 denotes "5 ints * 2 bytes * 1 = 10 bytes". Hence, begining address 65486 + 10 = 65496. So, &arr+1 = 65496 Hence the output of the program is 65486, 65496
Step 1: void fun(int, int[]); This prototype tells the compiler that the function fun() accepts one integer value and one array as an arguments and does not return anything. Step 2: int arr[] = {1, 2, 3, 4}; The variable a is declared as an integer array and it is initialized to a[0] = 1, a[1] = 2, a[2] = 3, a[3] = 4 Step 3: int i; The variable i is declared as an integer type. Step 4: fun(4, arr); This function does not affect the output of the program. Let's skip this function. Step 5: for(i=0; i<4; i++) { printf("%d,", arr[i]); } The for loop runs untill the variable i is less than '4' and it prints the each value of array a. Hence the output of the program is 1,2,3,4
Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; The variable a is declared as an multidimensional integer array with size of 3 rows 4 columns. Step 2: int *ptr; The *ptr is a integer pointer variable. Step 3: ptr = &a[0][0]; Here we are assigning the base address of the array a to the pointer variable *ptr. Step 4: fun(&ptr); Now, the &ptr contains the base address of array a. Step 4: Inside the function fun(&ptr); The printf("%d\n", **p); prints the value '1'. because the *p contains the base address or the first element memory address of the array a (ie. a[0]) **p contains the value of *p memory location (ie. a[0]=1). Hence the output of the program is '1'
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\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
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[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\n", 0[arr]); It prints the first element value of the variable arr. Hence the output of the program is 10.
11.
What will be the output of the program if the array begins at address 65486?
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\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
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("%d\n", 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'.
Step 1: int arr[]={2, 3, 4, 1, 6}; The variable arr is declared as an integer array and initialized. Step 2: printf("%u, %u, %u\n", arr, &arr[0], &arr); Here, The base address of the array is 1200. => arr, &arr is pointing to the base address of the array arr. => &arr[0] is pointing to the address of the first element array arr. (ie. base address) Hence the output of the program is 1200, 1200, 1200
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