这是我的代码:
section .bss
bufflen equ 1024
buff: resb bufflen
whatread: resb 8
section .data
section .text
global main
main:
nop
read:
mov eax,3 ; Specify sys_read
mov ebx,0 ; Specify standard input
mov ecx,buff ; Where to read to...
mov edx,bufflen ; How long to read
int 80h ; Tell linux to do its magic
; Eax currently has the return value from linux system call..
add eax, 30h ; Convert number to ASCII digit
mov [whatread],eax ; Store how many byte reads info to memory at loc whatread
mov eax,4 ; Specify sys_write
mov ebx,1 ; Specify standart output
mov ecx,whatread ; Get the address of whatread in ecx
mov edx,4 ; number of bytes to be written
int 80h ; Tell linux to do its work
mov eax,3 ; Specify sys_read
mov ebx,0 ; Specify standard input
mov ecx,buff ; Where to read to...
mov edx,bufflen ; How long to read
int 80h ; Tell linux to do its magic
; Eax currently has the return value from linux system call..
add eax, 30h ; Convert number to ASCII digit
mov [whatread],eax ; Store how many byte reads info to memory at loc whatread
mov eax,4 ; Specify sys_write
mov ebx,1 ; Specify standart output
mov ecx,whatread ; Get the address of whatread in ecx
mov edx,4 ; number of bytes to be written
int 80h ; Tell linux to do its work
mov eax, 1;
mov ebx, 0;
int 80h
我有一个名为 all.txt 的文件,内容为:
61 62 63 0A 64 65 66 0A (abc - new line - def - new line)
以下是一个示例运行:
koray@koray-VirtualBox:~/asm/buffasm$ ./buff < all.txt
80koray@koray-VirtualBox:~/asm/buffasm$
因此,第一次读取代码会尝试从文件中读取1024个字节。文件本身有8个字节..然后它在控制台打印8,我猜是好的。但是,为什么它打印'0'表示文件结束?我希望它再次读取相同的8个字节?为什么要尝试读取下一个1024字节?
答案 0 :(得分:0)
程序没有逻辑再次读取相同的字节。它们已经被读取并且在缓冲区中。输入缓冲区(由重定向文件填充)为空,因此无法读取更多字节。这就是sys_read以0读取字节结束的原因。
如果它不是STDIN,那么同样的逻辑将是一个真实的文件(当然,之前由sys_open打开)。
顺便说一句,返回0的数字实际上就是所谓的EOF。