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- Question
What will be the output of the program?
#include<stdio.h> int main() { float arr[] = {12.4, 2.3, 4.5, 6.7}; printf("%d\n", sizeof(arr)/sizeof(arr[0])); return 0; }
Options- A. 5
- B. 4
- C. 6
- D. 7
- Correct Answer
- 4
ExplanationThe sizeof function return the given variable. Example: float a=10; sizeof(a) is 4 bytesTags: Bank ExamsStep 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'.
Arrays problems
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- 1. What will be the output of the program?
#include<stdio.h> int main() { static int a[2][2] = {1, 2, 3, 4}; int i, j; static int *p[] = {(int*)a, (int*)a+1, (int*)a+2}; for(i=0; i<2; i++) { for(j=0; j<2; j++) { printf("%d, %d, %d, %d\n", *(*(p+i)+j), *(*(j+p)+i), *(*(i+p)+j), *(*(p+j)+i)); } } return 0; }
Options- A. 1, 1, 1, 1
2, 3, 2, 3
3, 2, 3, 2
4, 4, 4, 4 - B. 1, 2, 1, 2
2, 3, 2, 3
3, 4, 3, 4
4, 2, 4, 2 - C. 1, 1, 1, 1
2, 2, 2, 2
2, 2, 2, 2
3, 3, 3, 3 - D. 1, 2, 3, 4
2, 3, 4, 1
3, 4, 1, 2
4, 1, 2, 3
Discuss
Correct Answer: 1, 1, 1, 1
2, 2, 2, 2
2, 2, 2, 2
3, 3, 3, 3
- 2. What will be the output of the program if the array begins at address 65486?
#include<stdio.h> int main() { int arr[] = {12, 14, 15, 23, 45}; printf("%u, %u\n", arr, &arr); return 0; }
Options- A. 65486, 65488
- B. 65486, 65486
- C. 65486, 65490
- D. 65486, 65487 Discuss
Correct Answer: 65486, 65486
Explanation:
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
- 3. What will be the output of the program?
#include<stdio.h> int main() { static int arr[] = {0, 1, 2, 3, 4}; int *p[] = {arr, arr+1, arr+2, arr+3, arr+4}; int **ptr=p; ptr++; printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr); *ptr++; printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr); *++ptr; printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr); ++*ptr; printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr); return 0; }
Options- A. 0, 0, 0
1, 1, 1
2, 2, 2
3, 3, 3 - B. 1, 1, 2
2, 2, 3
3, 3, 4
4, 4, 1 - C. 1, 1, 1
2, 2, 2
3, 3, 3
3, 4, 4 - D. 0, 1, 2
1, 2, 3
2, 3, 4
3, 4, 5
Discuss
Correct Answer: 1, 1, 1
2, 2, 2
3, 3, 3
3, 4, 4
- 4. Will it result in to an error if a header file is included twice?
Options- A. Yes
- B. No
- C. It is compiler dependent. Discuss
Correct Answer: It is compiler dependent.
Explanation:
Unless the header file has taken care to ensure that if already included it doesn't get included again.
Turbo C, GCC compilers would take care of these problems, generate no error.- 5. Will the program compile successfully?
#include<stdio.h> int main() { printf("India" "CURIOUSTAB\n"); return 0; }
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
Explanation:
Yes, It prints "CuriousTab"- 6. What will be the output of the program?
#include<stdio.h> int main() { void fun(int, int[]); int arr[] = {1, 2, 3, 4}; int i; fun(4, arr); for(i=0; i<4; i++) printf("%d,", arr[i]); return 0; } void fun(int n, int arr[]) { int *p=0; int i=0; while(i++ < n) p = &arr[i]; *p=0; }
Options- A. 2, 3, 4, 5
- B. 1, 2, 3, 4
- C. 0, 1, 2, 3
- D. 3, 2, 1 0 Discuss
Correct Answer: 1, 2, 3, 4
Explanation:
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
- 7. What will be the output of the program in Turb C (under DOS)?
#include<stdio.h> int main() { int arr[5], i=0; while(i<5) arr[i]=++i; for(i=0; i<5; i++) printf("%d, ", arr[i]); return 0; }
Options- A. 1, 2, 3, 4, 5,
- B. Garbage value, 1, 2, 3, 4,
- C. 0, 1, 2, 3, 4,
- D. 2, 3, 4, 5, 6, Discuss
Correct Answer: Garbage value, 1, 2, 3, 4,
Explanation:
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.
- 8. What will be the output of the program?
#include<stdio.h> void fun(int **p); int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; int *ptr; ptr = &a[0][0]; fun(&ptr); return 0; } void fun(int **p) { printf("%d\n", **p); }
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
Correct Answer: 1
Explanation:
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'
- 9. What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?
#include<stdio.h> int main() { int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; printf("%u, %u\n", a+1, &a+1); return 0; }
Options- A. 65474, 65476
- B. 65480, 65496
- C. 65480, 65488
- D. 65474, 65488 Discuss
Correct Answer: 65480, 65496
Explanation:
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
- 10. What will be the output of the program if the array begins 1200 in memory?
#include<stdio.h> int main() { int arr[]={2, 3, 4, 1, 6}; printf("%u, %u, %u\n", arr, &arr[0], &arr); return 0; }
Options- A. 1200, 1202, 1204
- B. 1200, 1200, 1200
- C. 1200, 1204, 1208
- D. 1200, 1202, 1200 Discuss
Correct Answer: 1200, 1200, 1200
Explanation:
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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- 1. What will be the output of the program?