Why does this C code not print 'a' in output?

ghz 8months ago ⋅ 73 views 
#include<stdio.h>
#include<string.h>
void main(){
   char x[9]="helloabc";
   x[8]='\0';
   int a=2,b=3;
   strcpy(&x[2],&x[4]);
   for(int i=0;i<8;i++){
       printf("%c",x[i]);
   }
}

Almost all compilers output "heocbc\0c". Why does this C code not print 'a' in output?

Answers

The output you are observing, "heocbc\0c", is due to the following sequence of actions in your code:

  1. You define a character array x with a size of 9 and initialize it with the string "helloabc".
  2. You explicitly set the last character of x to null terminator ('\0') with x[8]='\0', effectively terminating the string.
  3. You use strcpy(&x[2], &x[4]) to copy the substring starting from index 4 to index 2 in x.
  4. After the strcpy operation, the resulting string in x is "heocbc\0c".
  5. Finally, you print the contents of x character by character, resulting in the output "heocbc\0c".

The reason 'a' is not present in the output is because it gets overwritten during the strcpy operation. Here's what happens in more detail:

  • After strcpy(&x[2], &x[4]), the string in x becomes "heocbcabc\0", with 'a' being copied over the characters starting from index 2.
  • However, the original string "helloabc" only had 'a' as the last character. Since the null terminator ('\0') is placed at index 8, there is no 'a' in the output after the null terminator.

To better understand, let's break down the string manipulation step by step:

  1. Before strcpy operation: "helloabc\0"
  2. After strcpy operation: "heocbcabc\0"
  3. Printing characters of x: "heocbc\0c"

As a result, 'a' is overwritten by the characters copied by strcpy, and it does not appear in the output.

#c