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मैं x86_64 असेंबली सीखने की कोशिश कर रहा हूं और मैं आज मानक इनपुट आउटपुट का प्रयास कर रहा था और इस पोस्ट पर ठोकर खाई Learning assembly - echo program name एसटीडीआईएन (SYSCALL निर्देश का उपयोग करके) से इनपुट पढ़ने के लिए मैं वही कैसे करूं? विशेष रूप से अगर मुझे पता है कि इनपुट हमेशा एक पूर्णांक होगा और मैं इसे एक रजिस्टर में पढ़ना चाहता हूं?x86_64 असेंबली में STDIN से इनपुट कैसे पढ़ा जाए?
संपादित करें: नीचे @Daniel Kozar का जवाब मुझे लिनक्स पर syscall अनुदेश के साथ कैसे STDIN और STDOUT सामान काम समझने में मदद की। मैंने एक छोटा प्रोग्राम लिखने का प्रयास किया, जो कंसोल इनपुट से एक संख्या पढ़ता है और उस संख्या से संबंधित एसीआईआई चरित्र को प्रिंट करता है। कहें कि क्या आप इनपुट के रूप में 65 टाइप करते हैं, आपको आउटपुट के रूप में ए प्राप्त करना चाहिए। और एक नया लाइन चरित्र भी। अगर सब पर है, यह किसी और किसी भी एक :-)
section .text
global _start
_start:
mov rdi, 0x0 ; file descriptor = stdin = 0
lea rsi, [rsp+8] ; buffer = address to store the bytes read
mov rdx, 0x2 ; number of bytes to read
mov rax, 0x0 ; SYSCALL number for reading from STDIN
syscall ; make the syscall
xor rax, rax ; clear off rax
mov rbx, [rsp+8] ; read the first byte read into rsp+8 by STDIN call to rbp
sub rbx, 0x30 ; Since this is read as a character, it is obtained as ASCII value, so subtract by 0x30 to get the number
and rbx, 0xff ; This ensures that everything other than the last byte is set to 0 while the last byte is as is
mov rax, rbx ; move this value to rax since we want to store the final result in rax
shl rbx, 0x1 ; We need to multiply this by 10 so that we can add up all the digits read so multiplying the number by 2 and then by 8 and adding them up, so multiply by 2 here
shl rax, 0x3 ; multiply by 8 here
add rax, rbx ; add 8 times multiplied value with 2 times multiplied value to get 10 times multiplied value
mov rbx, [rsp+9] ; now read the next byte (or digit)
sub rbx, 0x30 ; Again get the digit value from ASCII value of that digit's character
and rbx, 0xff ; clear higher bytes
add rax, rbx ; Add this to rax as unit's place value
mov [rsp+8], rax ; Move the entire byte to rax
mov rdi, 0x1 ; file descriptor = stdout
lea rsi, [rsp+8] ; buffer = address to write to console
mov rdx, 0x1 ; number of bytes to write
mov rax, 0x1 ; SYSCALL number for writing to STDOUT
syscall ; make the syscall
xor rax, rax ; clear off rax
mov rax, 0xa ; move the new line character to rax
mov [rsp+8], rax ; put this on the stack
mov rdi, 0x1 ; file descriptor = stdout
lea rsi, [rsp+8] ; buffer = address to write to console
mov rdx, 0x1 ; number of bytes to write
mov rax, 0x1 ; SYSCALL number for writing to STDOUT
syscall ; make the syscall
mov rdi, 0 ; set exit status = 0
mov rax, 60 ; SYSCALL number for EXIT
syscall ; make the syscall
संपादित 2 में मदद करता है: यहाँ है मेरी, मानक इनपुट से एक अहस्ताक्षरित 32-बिट दशमलव पूर्णांक पढ़ संगणना के लिए पूर्णांक के रूप में संग्रहीत और फिर लिखने के लिए प्रयास वह वापस बाहर निकलने के लिए। सब से
section .text
global _start
_start:
;Read from STDIN
mov rdi, 0x0 ; file descriptor = stdin = 0
lea rsi, [rsp+8] ; buffer = address to store the bytes read
mov rdx, 0xa ; number of bytes to read
mov rax, 0x0 ; SYSCALL number for reading from STDIN
syscall ; make the syscall
; Ascii to decimal conversion
xor rax, rax ; clear off rax
mov rbx, 0x0 ; initialize the counter which stores the number of bytes in the string representation of the integer
lea rsi, [rsp+8] ; Get the address on the stack where the first ASCII byte of the integer is stored.
rnext:
mov rcx, [rsi] ; Read the byte on the stack at the address represented by rsi
cmp rcx, 0xa ; Check if it is a newline character
je return ; If so we are done
cmp rbx, 0xa ; OR check if we have read 10 bytes (the largest 32 bit number contains 10 digits, so we will have to process at most 10 bytes
jg return ; If so we are done
sub rcx, 0x30 ; For the byte read, subtract by 0x30/48 to get the value from the ASCII code. 0 == 0x30 in ASCII, 1 == 0x31 in ASCII and so on.
and rcx, 0xff ; Clear off the higher order bytes to ensure there is no interference
mov rdx, rax ; We need to multiple this by 10 to get the next byte which goes to the unit's place and this byte becomes the ten's value. So make a copy
shl rax, 0x3 ; Multiply the original by 8 (Shift left by 3 is multiply by 8)
shl rdx, 0x1 ; Multiply the copy by 2 (Shift left by 1 is multiply by 2)
add rax, rdx ; Add these a * 8 + a * 2 to get a * 10.
add rax, rcx ; Add the digit to be at the units place to the original number
add rsi, 1 ; Advance the memory address by 1 to read the next byte
inc rbx ; Increment the digit counter
jmp rnext ; Loop until we have read all the digits or max is reached.
return:
push rax ; Push the read number on to the stack
; write New Line
mov rax, 0xa ; move the new line character to rax
mov [rsp+8], rax ; put this on the stack
mov rdi, 0x1 ; file descriptor = stdout
lea rsi, [rsp+8] ; buffer = address to write to console
mov rdx, 0x1 ; number of bytes to write
mov rax, 0x1 ; SYSCALL number for writing to STDOUT
syscall ; make the syscall
; Convert from Decimal to bytes
xor rdx, rdx ; Clear rdx which stores obtains a single digit of the number to convert to ASCII bytes
mov r8, 0x0 ; Initialize the counter containing the number of digits
pop rax ; Pop the read number from the stack
mov rbx, 0xa ; We store the divisor which is 10 for decimals (base-10) in rbx. rbx will be the divisor.
wnext:
div rbx ; Divide the number in rdx:rax by rbx to get the remainder in rdx
add rdx, 0x30 ; Add 0x30 to get the ASCII byte equivalent of the remainder which is the digit in the number to be written to display.
push rdx ; Push this byte to the stack. We do this because, we get the individial digit bytes in reverse order. So to reverse the order we use the stack
xor rdx, rdx ; Clear rdx preparing it for next division
inc r8 ; Increment the digits counter
cmp rax, 0x0 ; Continue until the number becomes 0 when there are no more digits to write to the console.
jne wnext ; Loop until there aren't any more digits.
popnext:
cmp r8, 0x0 ; Check if the counter which contains the number of digits to write is 0
jle endw ; If so there are no more digits to write
mov rdx, 0x1 ; number of bytes to write
mov rsi, rsp ; buffer = address to write to console
mov rdi, 0x1 ; file descriptor = stdout
mov rax, 0x1 ; SYSCALL number for writing to STDOUT
syscall ; make the syscall
dec r8 ; Decrement the counter
pop rbx ; Pop the current digit that was already written to the display preparing the stack pointer for next digit.
jmp popnext ; Loop until the counter which contains the number of digits goes down to 0.
endw:
; write New Line
xor rax, rax ; clear off rax
mov rax, 0xa ; move the new line character to rax
mov [rsp+9], rax ; put this on the stack
mov rdi, 0x1 ; file descriptor = stdout
lea rsi, [rsp+9] ; buffer = address to write to console
mov rdx, 0x1 ; number of bytes to write
mov rax, 0x1 ; SYSCALL number for writing to STDOUT
syscall ; make the syscall
; Exit
mov rdi, 0 ; set exit status = 0
mov rax, 60 ; SYSCALL number for EXIT
syscall ; make the syscall
आप किस ओएस का उपयोग कर रहे हैं? विंडोज? डॉस? लिनक्स? – Gabe
'syscall' का उपयोग ऑपरेटिंग सिस्टम निर्भर है। – hirschhornsalz
मैं लिनक्स का उपयोग कर रहा हूं। वह सटीक कोड मेरे लिए काम करता है। –