以下用C语言编写的程序的输出是什么？

Question

I am a novice for C language.but I could understand why this following code is giving output as 'A' .

one thing that is bothering me is the array name p in the printf statement.how this p is being treated by the compiler?

How can the p is replaced by the character array "%c\\n" after line no 5 ?

I know that this is a silly question so sorry to post this hare.

Can anyone will help me to understand the concept behind this?

 line1:      #include<stdio.h>
 line2:      int main()
 line3:      {
 line4:          char p[]="%d\n";
 line5:          p[1]='c';
 line6:          printf(p,65);
 line7:          return 0;
             }

Answer 1

The first argument to printf() is a const char* that contains the format specifiers. It is more common to see it as a string literal:

printf("%c\n", 65);

but it is legal to use a variable containing a string.

The assignment of p[1] = 'c' changes the d to c in the buffer p , resulting in the character A (as 65 is decimal value for A ) being written to standard output (as %c instructs printf() to print the character, rather than %d which will print the numeric value).

Answer 2

You are not replacing the whole array, just the character at array's offset #1 (second character). you are replacing it with 'c' making the content to be "%c\\n" which, when used as a formatting string, formats the integer 65 as an upper-case Latin A

Answer 3

In line 6:
printf(p,65);
will be changed to
printf("%c\\n",65); Ascii Value of 'A' is 65.
http://www.asciitable.com/

Answer 4

Explanation below:

char p[]="%d\n";

After the above executes, P will contain -> "%d\\n"

 line5:          p[1]='c';

Here, P will now be "%c\\n", as you are changing the 1th character of a zero based indexing.

line6:          printf(p,65);

This is equivalent to:

printf("%c\n",65)

or

printf("%c\n",'A')

Hence you get the output of A

Answer 5

     #include<stdio.h>
       int main()
       {
          char p[]="%d\n";   #This is stored at p[1]
          p[1]='c'           # d is replaced by c
          printf(p,65);      # p is taken as p[1] and is replaced by "%c\n"
           return 0;
         }

output: A # ASCII value 65

If you give the value as 66 output will be 'B" and so on.