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- Question
What will be the output of the program?
#include<stdio.h> int main() { int arr[1]={10}; printf("%d\n", 0[arr]); return 0; }
Options- A. 1
- B. 10
- C. 0
- D. 6
- Correct Answer
- 10
ExplanationStep 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.
Arrays problems
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- 1. What will be the output of the program?
#include<stdio.h> int main() { int a[5] = {5, 1, 15, 20, 25}; int i, j, m; i = ++a[1]; j = a[1]++; m = a[i++]; printf("%d, %d, %d", i, j, m); return 0; }
Options- A. 2, 1, 15
- B. 1, 2, 5
- C. 3, 2, 15
- D. 2, 3, 20 Discuss
Correct Answer: 3, 2, 15
Explanation:
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 toa[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 incremented 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
- 2. 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
- 3. 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
- 4. 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'
- 5. 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.
- 6. What will be the output of the program in 16-bit platform (under DOS)?
#include<stdio.h> int main() { struct node { int data; struct node *link; }; struct node *p, *q; p = (struct node *) malloc(sizeof(struct node)); q = (struct node *) malloc(sizeof(struct node)); printf("%d, %d\n", sizeof(p), sizeof(q)); return 0; }
Options- A. 2, 2
- B. 8, 8
- C. 5, 5
- D. 4, 4 Discuss
Correct Answer: 2, 2
- 7. What will be the output of the program?
#include<stdio.h> int main() { union var { int a, b; }; union var v; v.a=10; v.b=20; printf("%d\n", v.a); return 0; }
Options- A. 10
- B. 20
- C. 30
- D. 0 Discuss
Correct Answer: 20
- 8. What will be the output of the program in 16 bit platform (Turbo C under DOS)?
#include<stdio.h> int main() { struct value { int bit1:1; int bit3:4; int bit4:4; }bit; printf("%d\n", sizeof(bit)); return 0; }
Options- A. 1
- B. 2
- C. 4
- D. 9 Discuss
Correct Answer: 2
Explanation:
Since C is a compiler dependent language, in Turbo C (DOS) the output will be 2, but in GCC (Linux) the output will be 4.- 9. What will be the output of the program?
#include<stdio.h> int main() { enum days {MON=-1, TUE, WED=6, THU, FRI, SAT}; printf("%d, %d, %d, %d, %d, %d\n", ++MON, TUE, WED, THU, FRI, SAT); return 0; }
Options- A. -1, 0, 1, 2, 3, 4
- B. Error
- C. 0, 1, 6, 3, 4, 5
- D. 0, 0, 6, 7, 8, 9 Discuss
Correct Answer: Error
Explanation:
Because ++ or -- cannot be done on enum value.- 10. What will be the output of the program?
#include<stdio.h> struct course { int courseno; char coursename[25]; }; int main() { struct course c[] = { {102, "Java"}, {103, "PHP"}, {104, "DotNet"} }; printf("%d ", c[1].courseno); printf("%s\n", (*(c+2)).coursename); return 0; }
Options- A. 103 DotNet
- B. 102 Java
- C. 103 PHP
- D. 104 DotNet Discuss
Correct Answer: 103 DotNet
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- 1. What will be the output of the program?