Exercise: Python

Questions for: Arrays

Which of the following statements are correct about the program below? 
#include<stdio.h>

int main()
{
    int size, i;
    scanf("%d", &size);
    int arr[size];
    for(i=1; i<=size; i++)
    {
        scanf("%d", arr[i]);
        printf("%d", arr[i]);
    }
    return 0;
}
A:
The code is erroneous since the subscript for array used in for loop is in the range 1 to size.
B:
The code is erroneous since the values of array are getting scanned through the loop.
C:
The code is erroneous since the statement declaring array is invalid.
D:
The code is correct and runs successfully.
Answer: C

The statement int arr[size]; produces an error, because we cannot initialize the size of array dynamically. Constant expression is required here.

Example: int arr[10];

One more point is there, that is, usually declaration is not allowed after calling any function in a current block of code. In the given program the declaration int arr[10]; is placed after a function call scanf().

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Which of the following statements mentioning the name of the array begins DOES NOT yield the base address?
1: When array name is used with the sizeof operator.
2: When array name is operand of the & operator.
3: When array name is passed to scanf() function.
4: When array name is passed to printf() function.
A:
A
B:
A, B
C:
B
D:
B, D
Answer: B

The statement 1 and 2 does not yield the base address of the array. While the scanf() and printf() yields the base address of the array.

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Which of the following is correct way to define the function fun() in the below program? 
#include<stdio.h>

int main()
{
    int a[3][4];
    fun(a);
    return 0;
}
A: 
void fun(int p[][4])
{
}
B: 
void fun(int *p[4])
{
}
C: 
void fun(int *p[][4])
{
}
D: 
void fun(int *p[3][4])
{
}
Answer: A

void fun(int p[][4]){ } is the correct way to write the function fun(). while the others are considered only the function fun() is called by using call by reference.

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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: B

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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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: B

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'.

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