An exercise on polyglossy: the same problem solved on multiple languages
Implemented reading and writing routines in the x86 implementation.
Finished the read_uint subroutine and created the print_uint subroutine.
1 file changed, 103 insertions, 29 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | x86/PROGRAM.ASM | 132 | 103 | 29 |
diff --git a/x86/PROGRAM.ASM b/x86/PROGRAM.ASM @@ -3,10 +3,11 @@ .stack 100h .data - buffer db 20 DUP(30h) - invalid db " is not a natural number!", 0dh, 0ah, 0dh, 0ah, "$" + buffer db 22 DUP(0) + invalid db "' is not a natural number!", 0dh, 0ah, 0dh, 0ah, "$" - proof db "The conjecture is proved for all natural numbers smaller or equals to 10000!", 0dh, 0ah, "$" + proof db 0dh, 0ah, "The conjecture is proved for all natural numbers smaller or equals to $" + counter db 0dh, 0ah, "The conjecture is disproved! Here's a counterexample: ($" intro1 db 0dh, 0ah, "This program is a simple test for the following conjecture:", 0dh, 0ah, "$" intro2 db 0dh, 0ah, "Let S: N -> N be the sum of the digits of a positive integer.", 0dh, 0ah, "$" @@ -42,16 +43,24 @@ read_uint: ; Load the user input into the buffer mov ax, cs mov ds, ax - mov dx, 0 mov buffer[0], 20 ; Configure the buffer length - mov dl, offset buffer + mov dx, offset buffer mov ah, 0ah int 21h - mov ax, 0 ; Initialize the register were the result will be stored - mov cx, 2 ; Initialize a counter + ; Print '\n' + mov ah, 2h + mov dx, 0dh + int 21h + mov dx, 0ah + int 21h + + ; Move si to the to adress of the first character in the buffer mov si, offset buffer - add si, 3 + add si, 2 + + mov ax, 0 ; Initialize the register where the result will be stored + mov cx, 1 ; Initialize a counter read_loop: mov bx, 0 ; Clear bx @@ -76,35 +85,34 @@ read_loop: add ax, bx ; Jump if the counter is still smaller than the buffer length - cmp cl, buffer[0] - jb short read_loop cmp cl, buffer[1] - jb short read_loop + jbe short read_loop ret invalid_input: - mov ah, 2 + mov ah, 2h ; Print '\'' + mov dx, 0ah + int 21 mov dx, 27h - int 21h + int 21 mov si, offset buffer - mov cx, 0 + add si, 2 + mov cx, 1 invalid_input_loop: - mov dl, [si] + mov dx, [si] int 21h + inc cx inc si - cmp cl, 6 + cmp cl, buffer[1] jbe short invalid_input_loop invalid_input_end: - ; Print '\'' - mov dx, 27h - int 21h ; Print the rest of the message mov ah, 9h @@ -114,19 +122,22 @@ invalid_input_end: call quit ret -; Iterate ax from 0 to 10000 +; Iterate ax from 0 to maximum value, stored in bx iter: + push bx call test_num + pop bx inc ax - cmp ax, 10000 + cmp ax, bx jbe short iter - + call proved ; Iterate bx from 0 to ax test_num: mov bx, 0 + test_loop: push bx push ax @@ -147,7 +158,7 @@ test_pair: ; Calculate the 1st + 2nd into cx mov cx, ax add cx, bx - + push cx ; Push 1st + 2nd to the stack push bx ; Push 2nd to the stack @@ -156,7 +167,7 @@ test_pair: pop ax ; Store the value of 2nd in ax push bx ; Push S(1st) to the stack call sum_digits ; Calculate S(2nd) into bx - + pop ax ; Store S(1st) in ax add ax, bx ; Store S(1st) + S(2nd) in ax pop bx ; Store 1st + 2nd in bx @@ -168,7 +179,7 @@ test_pair: pop ax ; Store S(1st) + S(2nd) in ax sub bx, ax ; Calculate S(1st + 2nd) - (S(1st) + S(2nd)) into bx - + mov ax, bx ; Store S(1st + 2nd) - (S(1st) + S(2nd)) in ax mov cx, 9 ; Set the devident to 9 mov dx, 0 ; Clear the register where the rest will be stored @@ -178,7 +189,7 @@ test_pair: sum_digits: mov cx, 10 ; Store the devident in cx - mov bx, 0 ; Clear the register where the result will be stored + mov bx, 0 ; Clear the register where the result will be stored sum_loop: mov dx, 0 ; Clear the rest of the division @@ -195,16 +206,78 @@ proved: mov dx, offset proof int 21h + mov ax, bx + call print_uint + + ; Print '!\n' + mov ah, 2h + mov dx, 21h + int 21h + mov dx, 0dh + int 21h + mov dx, 0ah + int 21h + call quit ret disproved: + push bx + push ax + mov ah, 9h - mov dx, offset proof + mov dx, offset counter + int 21h + + pop ax + call print_uint + + ; Print ', ' + mov ah, 2h + mov dx, 2ch + int 21h + mov dx, 20h + int 21h + + pop ax + call print_uint + + ; Print ')\n' + mov ah, 2h + mov dx, 29h + int 21h + mov dx, 0ah int 21h call quit - ret + ret + +print_uint: + mov bx, 10 + mov cx, 0 + +print_uint_collect: + mov dx, 0 + div bx + add dx, 30h + + push dx + inc cx + + cmp ax, 0 + ja short print_uint_collect + + mov ah, 2h + +print_uint_loop: + pop dx + int 21h + dec cx + + cmp cx, 0 + ja short print_uint_loop + ret + quit: int 20h -end start +end start+ \ No newline at end of file