C Error: comparison between pointer and integer [enabled by default]

Question

I was trying to learn some basic c code, to bolster my general skills. I wrote a basic script that tells me if a number is greater or less than 10:

#include <stdio.h>
#include <unistd.h>
#include <string.h>


int main()
{
   int number;

   printf("Enter an integer\n");

   scanf("%d",&number);

   printf("Integer entered by you is %d\n", number);

   if ( "%d" > 10)
      printf("%d is greater than 10.\n", number);
   else
      printf("%d is less than 10.\n", number);

   return 0;
}

And yet when I compile it, I get an error stating that I'm trying to compare a pointer and an integer:

dave@dave-[laptop]:~/Code/C/Examples$ gcc 004----takeandif.c -o 004----takeandif
004----takeandif.c: In function ‘main’:
004----takeandif.c:16:14: warning: comparison between pointer and integer [enabled by default]
    if ( "%d" > 10)
              ^

And when I run it, it says that all numbers are less than 10:

dave@dave-[laptop]:~/Code/C/Examples$ ./004----takeandif
Enter an integer
2
Integer entered by you is 2
2 is greater than 10.

None of the other answers apply to my situation. What should I change?

Answer 1

Replace

if ( "%d" > 10)

With

if( number > 10 )

Answer 2

我非常确定％d只是一个占位符，并且由于您想将“ number”的值与10进行比较，因此应写成“ if（number> 10）”而不是“ if（”％d“> 10 ）”