（x86汇编）对于stdin，缓冲区大小指定为1个字节，但如果在stdin上提供，为什么缓冲超过1个字节？

Question

我在buff部分中定义.bss只是1个字节的缓冲区，用于将一个字符作为输入并将其转换为大写，然后在stdout上显示它。但是当我在输入（字符串）中提供多个字符时，它仍然将整个字符串转换为大写！如果我指定缓冲一个字节，它如何缓冲多个字节。在stdin之后是否有另一个缓冲区，它一次为我的缓冲区提供1个字节？

我的代码：

SECTION     .bss
buff        resb        1
SECTION     .data
SECTION     .text
global      _start
_start:
        NOP                             
read:           
        MOV     EAX, 3              ; Read syscall
        MOV     EBX, 0              ; Standard input  
        MOV     ECX, buff           ; Specify the buffer for stdin
        MOV     EDX, 1              ; 1 char at a time
        INT     0x80                ; Interrupt for the syscal
        CMP     EAX, 0              ; Look at the return of read syscall, if 
                                    ; unsuccessfull, exit
        JE      exit                ; If the input was 0, jump to exit
        CMP     byte [buff], 0x61   ; Test for lowercase, lowercase are above
                                    ; 0x61 - 'a'
        JB      write               ; Jump if below 'a' ASCII 0x61
        CMP     byte [buff], 0x7A   ; Test for above 'z'- ASCII 0x7A
        JA      write               ; Jump to write if input above 'z' 
        SUB     byte [buff], 0x20   ; A lowercase input finally, convert it to
                                    ;uppercase by subtracting 0x20h
write:          
        MOV     EAX, 4          ; Write syscall
        MOV     EBX, 1          ; Write on stdout
        MOV     ECX, buff       ; take from this buffer
        MOV     EDX, 1          ; 1 charcter at a time
        INT     0x80            ; Interrupt for the syscall
        JMP     read            ; Jump for reading again
exit:           
        MOV     EAX, 1          ; Exit syscall
        MOV     EBX, 0          ; Errorcode in EBX
        INT     0x80            ; Interrupt for exit syscall