How do I properly 'printf' an integer and

问题:

I have the following code:

char *s1, *s2;
char str[10];

printf("Type a string: ");
scanf("%s", str);

s1 = &str[0];
s2 = &str[2];

printf("%s\n", s1);
printf("%s\n", s2);

When I run the code, and enter the input "A 1" as follow:

Type a string: A 1

I got the following result:

A
�<�

I'm trying to read the first character as a string and the third character as an integer, and then print those out on the screen. The first character always works, but the screen would just display random stuffs after that.... How should I fix it?

回答1:

You're on the right track. Here's a corrected version:

char str[10];
int n;

printf("type a string: ");
scanf("%s %d", str, &n);

printf("%s\n", str);
printf("%d\n", n);

Let's talk through the changes:

1. allocate an int (n) to store your number in
2. tell scanf to read in first a string and then a number (%d means number, as you already knew from your printf

That's pretty much all there is to it. Your code is a little bit dangerous, still, because any user input that's longer than 9 characters will overflow str and start trampling your stack.

回答2:

scanf("%s",str) scans only until it finds a whitespace character. With the input "A 1", it will scan only the first character, hence s2 points at the garbage that happened to be in str, since that array wasn't initialised.

回答3:

Try this code my friend...

#include<stdio.h>
int main(){
   char *s1, *s2;
   char str[10];

   printf("type a string: ");
   scanf("%s", str);

   s1 = &str[0];
   s2 = &str[2];

   printf("%c\n", *s1);   //use %c instead of %s and *s1 which is the content of position 1
   printf("%c\n", *s2);   //use %c instead of %s and *s3 which is the content of position 1

   return 0;
}