I am trying to print the out put of name and phone. Keeps showing this two errors in terminal
format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’
and
format ‘%s’ expects argument of type ‘char *’, but argument 3 has type ‘int’
my code:
int main(void)
{
char name,phone;
int age;
printf("Kindly input your name, age and phone number");
scanf("%s %s %d", &name, &phone, &age);
printf("%s %s %d\n", name, phone, age);
return 0;
}
The error message means that in this call of printf
printf("%s %s %d\n", name, phone, age);
argument expressions name and phone having the type char are promoted to the type int and there is used the conversion specification %s designed to output strings instead of integers.
Instead of declaring these variables as having the scalar type char you need to declare them as having a character array type to be able to store strings.
And you should change the prompt
printf("Kindly input your name, age and phone number");
like
printf("Kindly input your name, phone number and age: ");
For example
char name[20], phone[20];
int age;
printf("Kindly input your name, phone number and age: ");
scanf("%19s %19s %d", name, phone, &age);
printf("%s %s %d\n", name, phone, age);
Your sample code is likely to fault .. Actually, you are storing an array of characters (string) into a single char memory location. You should rather do this:
/* Note the array definition `[]`.
* Note that 32 is an arbitrary length, change it as per your needs.
*/
char name[32], phone[32];
Then, as an array is already a pointer to memory location, you can use it without the &.
scanf("%31s %31s %d", name, phone, &age);
EDIT: as suggested in comment, you can restrict the maximum input size. Knowing a string is NULL-terminated, in the example the maximum total string length becomes 31.
