Consider the following C program segment. #include <stdio.h> int main() p = s1 + 2; printf("%s", s1); return 0;
{
char s1[7] = "1234";
char *p;
*p = '0';
}
1204
Analyzing the C Program Segment and Output
The provided C program segment demonstrates how pointers can be used to access and modify elements within a character array, which is commonly used to represent strings in C. Let's break down the program step by step to understand its execution and determine the final output.
Step-by-Step Program Execution Analysis
- Declaration and Initialization:
char s1[7] = "1234";This line declares a character array named
s1with a size of 7. It is initialized with the string literal "1234". When a character array is initialized with a string literal, the characters of the literal are placed in the array, followed by a null terminator character'\0'. Since the size of the array (7) is larger than the length of the string "1234" (4 characters + 1 null terminator = 5), the array will contain:Index 0 1 2 3 4 5 6 Content '1' '2' '3' '4' '\0' ? ? The '?' denotes uninitialized or garbage values in the remaining memory locations within the array
s1. - Pointer Declaration:
char *p;This declares a character pointer named
p. A pointer variable stores the memory address of another variable, in this case, the address of a character. - Pointer Assignment using Pointer Arithmetic:
p = s1 + 2;This is where pointer arithmetic comes into play.
s1represents the base address of the arrays1(which is the address ofs1[0]). Adding an integernto an array name (or a pointer) results in a pointer that pointsnelements away from the original address, considering the size of the data type. Sinces1is achararray,s1 + 2calculates the address that is 2 character sizes away from the start ofs1. This address corresponds to the memory location ofs1[2]. So, pointerpnow points tos1[2], which currently holds the character '3'. - Modifying Array Content using Pointer:
*p = '0';The expression
*prefers to the value at the memory address pointed to byp. Sinceppoints tos1[2], this line changes the value stored ats1[2]from '3' to '0'.The array
s1now becomes:Index 0 1 2 3 4 5 6 Content '1' '2' '0' '4' '\0' ? ? - Printing the String:
printf("%s", s1);The
printffunction with the%sformat specifier is used to print a string. It starts printing characters from the memory address provided (which is the beginning of the arrays1) and continues until it encounters a null terminator character ('\0').The function will print:
s1[0]which is '1's1[1]which is '2's1[2]which is '0' (the modified value)s1[3]which is '4'- It encounters
s1[4]which is '\0', so printing stops.
Therefore, the output printed by the program will be "1204".
Revision Table: Key C Concepts
| Concept | Description |
|---|---|
| Character Array (String) | An array of characters used to store sequences of characters. Strings in C are null-terminated, meaning they end with a '\0' character. |
| String Initialization | Initializing a char array with a string literal automatically includes the null terminator. The array size should be large enough to hold all characters plus the null terminator. |
| Pointer | A variable that stores the memory address of another variable. A char * stores the address of a character. |
| Pointer Arithmetic | Operations like addition or subtraction on pointers. Adding an integer n to a pointer moves the pointer n times the size of the pointed-to data type. arrayName + n is equivalent to &arrayName[n]. |
| Dereferencing Pointer | Using the * operator before a pointer variable (e.g., *p) accesses the value stored at the memory address the pointer holds. |
printf("%s", ...) |
Prints a string starting from the given address until a null terminator ('\0') is found. |
Additional Information on C Strings and Pointers
Strings in C are fundamentally arrays of characters. The null terminator '\0' is crucial because it marks the end of the string, allowing functions like printf("%s", ...) and strlen() to know where the string ends. Without the null terminator, these functions would continue reading memory until they accidentally find a zero byte, potentially leading to errors or crashes.
Pointers are extremely powerful in C for manipulating strings and arrays efficiently. Pointer arithmetic provides a concise way to move through arrays. For example, iterating through a string can be done by incrementing a character pointer: p++ moves the pointer to the next character.
Modifying a string using a pointer, as shown in the example, directly changes the content of the underlying character array. This is a common technique in C string manipulation. However, it's important to be careful not to write beyond the bounds of the array, which can cause buffer overflows and security vulnerabilities.
In this specific program, s1 + 2 correctly calculates the address of the third element (index 2) because s1 decays to a pointer to its first element in this context, and adding 2 performs pointer arithmetic based on the size of a char.
Important Questions from Array
Which of the following initialization statement store six integer values in array?
Which of the following expression will delete the entire array pointed to by q?
#include <stdio.h> void foo(int *p, int x){ *p = x; } int main(){ int *z; int a = 20, b = 25; z = &a; foo(z, b); printf("%d", a); return 0; }The output of the given C program is __________. (Answer in integer)- Consider an array $A$ of integers of size $n$. The indices of $A$ run from 1 to $n$. An algorithm is to be designed to check whether $A$ satisfies the condition given below.
$\forall i,j \in \{1, \dots, n-1\}$ such that $i > j$, $(A[i + 1] - A[i]) > (A[j + 1] - A[j])$
Which one of the following gives the worst case time complexity of the fastest algorithm that can be designed for the problem? What is the output the given C language code snippet?
#include<stdio.h>
int main()
{
int x[] = {10,20,30};
int *p = x;
p++;
printf("%d",*p);
return 0;
}
