int d;
scanf("%c", &d);
printf("%d", d);
here giving a alpha value as input and am trying to get ascii value as output why is not working instead it shows address of the char all other ways it works, like getting decimal in char and convert into char.
You should have your code the other way - scan an integer, print a character.Originally Posted by zero_code
since the %c in scanf expects pointer to char your d var should be declared as char d
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
LIke this?
Code:#include<stdio.h> main() { char d; scanf(" %c", &d); printf("%d\n", d); return 0; }
Looks better and even works...
You may benefit from reading FAQ > main() / void main() / int main() / int main(void) / int main(int argc, char *argv[]) - Cprogramming.com
And I will add loop so several conversions could be made in the same run...
Code:#include<stdio.h> int main(void) { char d; while(scanf(" %c", &d) == 1) { printf("%d 0x%x\n", d, d); } return 0; }
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
Looks better and even works...
You may benefit from reading FAQ > main() / void main() / int main() / int main(void) / int main(int argc, char *argv[]) - Cprogramming.com
And I would add loop so several conversions could be made in the same run...
Code:#include<stdio.h> int main(void) { char d; while(scanf(" %c", &d) == 1) { printf("%d 0x%x\n", d, d); } return 0; }
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
relating to the same question suppose if the variable is declared as one type and read as other type in printf is the behavior not undefined. For example
Is this behavior undefined or it is ok? Similarly for char and int types. I am getting confused.Code:float f; printf("%d",f);
The variable must be read as the same type float "%f" therefore your code is incorrect.
You are correct in that the behavior is undefined.
