#include<stdio.h>
int main()
{
int a;
char c;
int *A=&a;
char *C=&c;
printf("Enter the value of a,c\n");
scanf("%d,%d",&a,&c);
printf ("Adress of a,c= %d,%d\n",A,C);
printf("value of a,c= %d %d\n",a,c);
return 0;
}
output is:
c:\Users\Avinash\Desktop>a.exe
Enter the value of a,c
12,40
Adress of a,c= 6356740,6356739
value of a,c= 0, 40
As others said, the error is in the scanf() format expression '%d' used for reading the character c. What is happening is this: Note that the address of c is one lower than the address of a (this is expected for variables on the stack). Assume a character occupies one byte and an integer occupies 4 bytes. By using '%d' you are telling scanf() that the second pointer points to an integer; therefore it reads the second value as an integer, --padding out the high-order bits with zeros--, and stores the value --in the four bytes-- based at the address of c. The padding overwrites the low-order bits of a, explaining why the printf() shows its value as 0.
%c for scanning char. Change
scanf("%d,%d",&a,&c);
to
scanf("%d,%c",&a,&c); //make sure there is only single comma in between the inputs. Nothing else, nothing more.
you have some problem with the scanf format and the printf format
int a;
char c;
int *A = &a;
char *C = &c;
printf("Enter the value of a,c\n");
scanf("%d, %c", &a, &c); //%c for reading chars
printf("Adress of a= %p, c= %p\n", A, C); // printing the address %p
printf("value of a= %d c= %c\n", a, c); //%c for printing chars
scanf("%d,%d",&a,&c);-->scanf("%d, %c",&a,&c);orchar c;-->int c;aandcvariables, so they could have any value when your program starts. A good coding standard is to initialize variables with a value when they are defined.