Assembly x86 , greeting program

Question

I tried to write simple greeting program using assembly x86 that takes user name and print out "Hello [userName]" the problem is the first char of user name is doubled when printing the greeting msg , for example :

input :

Black Knight 

output :

Hello BBlack Knight 

and here's my code

global _start 

section .data 
    msg1        db "Hello "
    user_input  times 20 db 0 


section .bss 

section .text 

_start : 

; read 
mov eax , 3 
mov ebx , 0
mov ecx , user_input
mov edx , 20
int 0x80

; write 
mov eax , 4
mov ebx , 1
mov ecx , msg1
mov edx , 7
int 0x80

mov eax , 4 
mov ebx , 1
mov ecx , user_input
mov edx , 20
int 0x80

; exit 
mov eax , 1
mov ebx , 0
int 0x80

Answer 1

This is happening because of this code:

; write 
mov eax , 4
mov ebx , 1
mov ecx , msg1
mov edx , 7
int 0x80

You are here telling the write instruction that the string to print is 7 bytes in length when it is, in fact, 6.

Why the doubled B? Because in memory, the inputted name appears beginning in the byte immediately following msg1.

You could solve this by printing only 6 characters (Hello + space) or by adding a null terminator to the end of your msg value.