How did its output come "A"?-CodePudding

#include<stdio.h>
int main()
{
  int i = 577;
  printf("%c",i);
  return 0;
}

After compiling, its giving output "A". Can anyone explain how i'm getting this?

CodePudding user response：

%c will only accept values up to 255 included, then it will start from 0 again !

577 % 256 = 65; // (char code for 'A')

CodePudding user response：

577 in hex is 0x241. The ASCII representation of 'A' is 0x41. You're passing an int to printf but then telling printf to treat it as a char (because of %c). A char is one-byte wide and so printf looks at the first argument you gave it and reads the first byte. Since your computer's architecture is little-Endian, that first byte is the least significant byte which is 0x41.

To print an integer, you need to use %d or %i.

CodePudding user response：

The program has undefined behavior and doesn't have to output A. It happens to do so in your case since %c makes it take one byte (from the stack most likely) and it found that part of an int which happend to represent an A. The leftovers from the int you placed there corrupts the stack.

CodePudding user response：

Just output the value of the variable i in the hexadecimal representation

#include <stdio.h>

int main( void )
{
    int i = 577;

    printf( "i = %#x\n", i );
}

The program output will be

i = 0x241

So the least significant byte contains the hexadecimal value 0x41 that represents the ASCII code of the letter 'A'.