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).