Declaration: char *strstr(const char *s1, const char *s2);
Return Value:
On success, strstr returns a pointer to the element in s1 where s2 begins (points to s2 in s1).
On error (if s2 does not occur in s1), strstr returns null.
Example:
#include <stdio.h>
#include <string.h>
int main(void)
{
char *str1 = "CuriousTab", *str2 = "ia", *ptr;
ptr = strstr(str1, str2);
printf("The substring is: %s\n", ptr);
return 0;
}
Output: The substring is: iaCURIOUSTAB
Example: #define CUBE(X)(X*X*X)
#include<stdio.h> int fun(int *f) { *f = 10; return 0; } int main() { const int arr[5] = {1, 2, 3, 4, 5}; printf("Before modification arr[3] = %d", arr[3]); fun(&arr[3]); printf("\nAfter modification arr[3] = %d", arr[3]); return 0; }
arr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5
Step 2: printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4).
Step 3: fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10.
A const variable can be indirectly modified by a pointer.
Step 4: printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10).
Hence the output of the program is
Before modification arr[3] = 4
After modification arr[3] = 10
#include<stdio.h> int main() { const int x=5; const int *ptrx; ptrx = &x; *ptrx = 10; printf("%d\n", x); return 0; }
Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer.
Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx.
Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error.
To change the value of const variable x we have to use *(int *)&x = 10;
#include<stdio.h> int fun(int **ptr); int main() { int i=10; const int *ptr = &i; fun(&ptr); return 0; } int fun(int **ptr) { int j = 223; int *temp = &j; printf("Before changing ptr = %5x\n", *ptr); const *ptr = temp; printf("After changing ptr = %5x\n", *ptr); return 0; }
#include<stdio.h> int main() { const int x; x=128; printf("%d\n", x); return 0; }
Hence Option B is correct
#include<stdio.h> #define MAX 128 int main() { const int max=128; char array[max]; char string[MAX]; array[0] = string[0] = 'A'; printf("%c %c\n", array[0], string[0]); return 0; }
Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.
Step 3: char array[max]; This statement reports an error "constant expression required". Because, we cannot use variable to define the size of array.
To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];
Note: The above program will print A A as output in Unix platform.
#include<stdio.h> int fun(int **ptr); int main() { int i=10, j=20; const int *ptr = &i; printf(" i = %5X", ptr); printf(" ptr = %d", *ptr); ptr = &j; printf(" j = %5X", ptr); printf(" ptr = %d", *ptr); return 0; }
#include<stdio.h> int main() { const int i=0; printf("%d\n", i++); return 0; }
Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero).
Step 2: printf("%d\n", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object".
Because, we cannot modify a const variable.
#include<stdio.h> #include<stdlib.h> union employee { char name[15]; int age; float salary; }; const union employee e1; int main() { strcpy(e1.name, "K"); printf("%s", e1.name); e1.age=85; printf("%d", e1.age); printf("%f", e1.salary); return 0; }
#include<stdio.h> #define MAX 128 int main() { char mybuf[] = "India"; char yourbuf[] = "CURIOUSTAB"; char *const ptr = mybuf; *ptr = 'a'; ptr = yourbuf; return 0; }
Step 2: char yourbuf[] = "CURIOUSTAB"; The variable yourbuf is declared as an array of characters and initialized with string "CURIOUSTAB".
Step 3: char *const ptr = mybuf; Here, ptr is a constant pointer, which points at a char.
The value at which ptr it points is not a constant; it will not be an error to modify the pointed character; There will be an error only to modify the pointer itself.
Step 4: *ptr = 'a'; The value of ptr is assigned to 'a'.
Step 5: ptr = yourbuf; Here, we are changing the pointer itself, this will result in the error "cannot modify a const object".
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