#include<stdio.h> const char *fun(); int main() { char *ptr = fun(); return 0; } const char *fun() { return "Hello"; }
#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() { 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> const char *fun(); int main() { *fun() = 'A'; return 0; } const char *fun() { return "Hello"; }
#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 a constant; it will be an error to modify the pointed character; There will not be any error to modify the pointer itself.
Step 4: *ptr = 'a'; Here, we are changing the value of ptr, this will result in the error "cannot modify a const object".
#include<stdio.h> int main() { const int x; x=128; printf("%d\n", x); return 0; }
Hence Option B is correct
#include<stdio.h> int main() { const int k=7; int *const q=&k; printf("%d", *q); return 0; }
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