#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 %d %f", e1.name, e1.age, e1.salary); return 0; }
The output will be (in 16-bit platform DOS):
K 75 0.000000#include<stdio.h> int main() { const char *s = ""; char str[] = "Hello"; s = str; while(*s) printf("%c", *s++); return 0; }
Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".
Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".
Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.
Hence the output of the program is "Hello".
#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> int main() { int y=128; const int x=y; printf("%d\n", x); return 0; }
Step 2: const int x=y; The constant variable 'x' is declared as an integer and it is initialized with the variable 'y' value.
Step 3: printf("%d\n", x); It prints the value of variable 'x'.
Hence the output of the program is "128"
#include<stdio.h> int get(); int main() { const int x = get(); printf("%d", x); return 0; } int get() { return 20; }
Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".
The function get() returns the value "20".
Step 3: printf("%d", x); It prints the value of the variable x.
Hence the output of the program is "20".
#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 main() { const c = -11; const int d = 34; printf("%d, %d\n", c, d); return 0; }
Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.
Step 3: printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed.
Hence the output of the program is -11, 34
Example:
#include <stdio.h>
#include <string.h>
int main()
{
char str[30] = "12345678910111213";
printf("The last position of '2' is %d.\n",
strrchr(str, '2') - str);
return 0;
}
Output: The last position of '2' is 14.
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