6. Arrays and Strings

 

Arrays

• Arrays are used to store multiple values in a single variable instead of declaring separate variables for each value.
• All the values stored in an array must be of same data type.
• Internally array representation gets converted into pointer representation.

    int arr[5]; // Here arr is a constant integer pointer

    arr[idx] = *(arr + idx) = *(idx + arr) = idx[arr];

1. What is the output of the following C code?
#include <stdio.h>


int main()
{
    int arr[5] = {1, 2, 3, 4, 5};
    for (int i = 0; i < 5; i++)
        printf("Address of %d idx element: %p\n", i, arr + i);
}

Address of 0 idx element: 000000189e5ff650
Address of 1 idx element: 000000189e5ff654
Address of 2 idx element: 000000189e5ff658
Address of 3 idx element: 000000189e5ff65c
Address of 4 idx element: 000000189e5ff660

2. What is the output of the following C code?
#include <stdio.h>


int main()
{
    int arr[5] = {1, 2, 3, 4, 5};
    for (int i = 0; i < 5; i++)
        printf("Value of arr[%d]: %d = %d = %d = %d\n", i, arr[i], *(arr + i), *(i + arr), i[arr]);
}

Value of arr[0]: 1 = 1 = 1 = 1
Value of arr[1]: 2 = 2 = 2 = 2
Value of arr[2]: 3 = 3 = 3 = 3
Value of arr[3]: 4 = 4 = 4 = 4
Value of arr[4]: 5 = 5 = 5 = 5

    int arr[5];

    int *ptr = arr;

arr = Array name = constant pointer
ptr = Pointer name = variable pointer

Then we can't change the value of arr i.e., arr will point to a address which can't be changed, but we can change the value of ptr.

Valid statements	Invalid statements
ptr = #	arr = #
ptr++;	arr++;
ptr = ptr - 1;	arr = arr - 1;

Pointer to an Array

    int arr[5] = {1,2,3,4,5};

    int (*arrPtr)[5] = &arr;

Now, arrPtr will behave same as arr, but we can change it unlike arr.

1. What is the output of the following C code?
#include <stdio.h>


int main()
{
    int a[10];
    printf("%d", *a + 1 - *a + 3);
}

4

2. What is the output of the following C code?
#include <stdio.h>


char *getString()
{
    char str[] = "Deepak Chandra Sharma";
    return str;
}


int main()
{
    printf("%s", getString());
    return 0;
}

warning: function returns address of local variable
return str;
(null)

3. What is the output of the following C code?
#include <stdio.h>


int main()
{
    int arr[] = {1, 2, 3};
    int *p = arr;
    if (&p == arr)
        printf("Same");
    else
        printf("Not same");
}

warning: comparison of distinct pointer types lacks a cast
if (&p == arr)
Not same

4. What is the output of the following C code?
#include <stdio.h>


int main()
{
    int arr[] = {1, 2, 3};
    int *p = arr;
    if (p == arr)
        printf("Same");
    else
        printf("Not same");
}

Same

Strings

• a string is a sequence of characters that are stored in contiguous memory locations.
• Strings in C are represented as arrays of characters, terminated by a null character '\0'.
char str[] = "abc";
• str is a string.
char *strPtr = "abc";
• strPtr is a string constant or string literal.

1. What is the output of the following C code?
#include <stdio.h>


int main()
{
    char str[] = "abc";
    str[1] = '1';
    printf(str);
    char *ptr = "abc";
    ptr[1] = 'p';
    printf(ptr);
}

a1c
Explanation: ptr is a string constant, so we can't change it's value. So when we tried to do so, we get an segmentation fault exception.

2. What is the output of the following C code?
#include <stdio.h>


int main()
{
    char arr[] = "deep";
    char *ptr = arr;
    while (*ptr != '\0')
        ++*ptr++;
    printf("%s %s", arr, ptr);
}

effq
Explanation: ptr points to the same address to which arr points. Firstly, the pointer ptr will be dereferenced, then its value will be pre-incremented and then the pointer will be post-incremented.

3. What is the output of the following C code?
#include <stdio.h>


int main()
{
    char str[] = "abc" "pqr""abc";
    printf(str);
}

abcpqrabc
Explanation: If we try to assign two string in a char array like above, then they will concatenate to a single character array.

4. What is the output of the following C code?
#include <stdio.h>


int main()
{
    char str[] = "geeks\nknowledge";
    char *ptr1, *ptr2;
    ptr1 = &str[3];
    ptr2 = str + 5;
    printf("%c", ++*str - --*ptr1 + *ptr2 + 2);
    printf("%s", str);
}

'\n'
heejs
knowledge
Explanation: After evaluating the expression inside first printf, we will get '\n' i.e., next line. Because ptr1 and ptr2 points to the same string, so the original string also changes.

5. What is the output of the following C code?
#include <stdio.h>


int main()
{
    char str[] = "Deepak";
    char *s1 = str, *s2 = str;
    int i;
    for (i = 0; i < 7; i++)
    {
        printf("%c", *str);
        ++s1;
    }
    for (i = 0; i < 6; i++)
    {
        printf("%c", *s2);
        ++s2;
    }
}

DDDDDDDDeepak
Explanation: s1, s2 and str points to the same string but all three are different pointers. So pre-incrementing s1 doesn't mean that we are incrementing str pointer.

char str[] = {'a', 'b', 'c'};

If we don't add '\0' as the last element of character array like above, then our string will get appended by some garbage value. But if we also give the size while initializing, then if there are 3 elements (excluding '\0') and we give size 4 or more, then it will be fine, but if we give size less than 4, it will add up some garbage values.

But if we declare a string using double quotes, then there is no need to add '\0' at the end, compiler automatically adds that.