### Find Output Of Program

Question 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;
}

A). 2, 1, 15
B). 1, 2, 5
C). 3, 2, 15
D). 2, 3, 20
Answer : Option C

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 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 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

Question 2 :

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;
}

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
Answer : Option C

Question 3 :

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;
}

A). 2, 3, 4, 5
B). 1, 2, 3, 4
C). 0, 1, 2, 3
D). 3, 2, 1 0
Answer : Option B

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

Question 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);
}

A). 1
B). 2
C). 3
D). 4
Answer : Option A

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 5: 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'

Question 5 :

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;
}

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
Answer : Option C

Question 6 :

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;
}

A). 65474, 65476
B). 65480, 65496
C). 65480, 65488
D). 65474, 65488
Answer : Option B

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

Question 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;
}

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,
Answer : Option B

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.

Question 8 :

What will be the output of the program ?

#include < stdio.h >
int main()
{
int arr[1]={10};
printf("%d\n", 0[arr]);
return 0;
}

A). 1
B). 10
C). 0
D). 6
Answer : Option B

Explanation :

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.

Question 9 :

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;
}

A). 65486, 65488
B). 65486, 65486
C). 65486, 65490
D). 65486, 65487
Answer : Option B

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

Question 10 :

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;
}

A). 5
B). 4
C). 6
D). 7
Answer : Option B

Explanation :

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'.

Question 11 :

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;
}

A). 1200, 1202, 1204
B). 1200, 1200, 1200
C). 1200, 1204, 1208
D). 1200, 1202, 1200
Answer : Option B

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