#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> void fun(void *p); int i; int main() { void *vptr; vptr = &i; fun(vptr); return 0; } void fun(void *p) { int **q; q = (int**)&p; printf("%d\n", **q); }
#include<stdio.h> #include<string.h> int main() { static char s[] = "Hello!"; printf("%d\n", *(s+strlen(s))); return 0; }
#include<stdio.h> int main() { int i=4, j=8; printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j); return 0; }
#include<stdio.h> int main() { int i; printf("%d\n", scanf("%d", &i)); return 0; }
printf("%d\n", scanf("%d", &i)); The scanf function returns the value 1(one).
Therefore, the output of the program is '1'.
int *ptr[30];
#include<stdio.h> int main() { char huge *near *far *ptr1; char near *far *huge *ptr2; char far *huge *near *ptr3; printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3)); return 0; }
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.