如何使用x86程序集创建一个循环来打印输入值的星形行？ [英] How do I create a loop using x86 assembly to print rows of stars from inputted values?
问题描述
我设法打印了一排星星但是当我尝试从新值打印另一行时，它会创建一个无限循环。它停在0.我想使用x86程序集，我正在使用visual studio。
示例
输入：
1
5
8
0
输出：< br $>
*
*****
********
我的尝试：
#includestdafx.h
#include< stdlib.h>
void printChar（char c）;
void printStr（char * strAddr）;
void printInt（int someInt）;
int numItems; //项目编号
int * items; //指向项目的指针
int anItem; //将对应于所看到的星号数
int wmain（int argc，_TCHAR * argv []）
{
items =（int *）malloc（1000）;
numItems = 0;
do
{
printf（输入项目％d（0表示数据结束）：，numItems + 1）;
scanf（％d，& anItem）;
items [numItems] = anItem;
numItems ++;
} while（anItem！= 0）; //用于输入和读取的数据
_asm
{
调用printNewLine
mov esi，[items] // esi指向'numItems'的开头（'items'指向'numItems'的地址）
mainLoop：mov eax，[esi] // eax的内存地址为'numItems'
mov [anItem]，eax //内存地址'anItem'的内存地址为'numItems'
mov eax，0
starLoop：push eax // eax first in stack
mov al，'*'
push al
call printChar //打印一张*
pop eax // eax out of stack
inc eax // eax增加1
cmp eax，[anItem ] //与'anItem'比较的eax值
jne starLoop //如果eax不等于'anItem'的值，starLoop再次启动
jmp完成
printNewLine：
push'\ r'//用于打印字符的两行。
调用printChar
push'\ n'//两行打印另一个字符。
调用printChar
ret //回到<无论我们来自哪里>
完成：//做什么都不用
}
printf（按enter键退出quant \ n）;
char dummy [10]; //以防万一缓冲区中的几个键
scanf（％c，dummy）; //暂停。
scanf（％c，虚拟）; //再次暂停
}
void printChar（char c）
{
printf（ ％c，c）; //％c表示作为字符
}
void printStr（char * strAddr）
{
printf（％s，strAddr）;
}
void printInt（int someInt）
{
printf（％d，someInt）;
}
I have managed to print a row of stars but when I try to print another row from a new value, it creates an infinite loop. It stops at 0. I want to use x86 assembly and I am using visual studio.
example
input:
1
5
8
0
output:
*
*****
********
What I have tried:
#include "stdafx.h"
#include <stdlib.h>
void printChar(char c);
void printStr(char *strAddr);
void printInt(int someInt);
int numItems; //item number
int *items; // Pointer to the items
int anItem; //will correspond to number of asterisks seen
int wmain(int argc, _TCHAR* argv[])
{
items = (int *)malloc(1000);
numItems = 0;
do
{
printf("Enter item %d (0 means end of data): ", numItems + 1);
scanf("%d", &anItem);
items[numItems] = anItem;
numItems++;
} while (anItem != 0); //for data to be input and read
_asm
{
call printNewLine
mov esi, [items] //esi points to start of 'numItems' ('items' points to address of 'numItems')
mainLoop: mov eax, [esi] //eax has memory address of 'numItems'
mov [anItem], eax //memory address of 'anItem' has the memory address of 'numItems'
mov eax, 0
starLoop: push eax //eax first in stack
mov al, '*'
push al
call printChar //Prints a single *
pop eax //eax out of stack
inc eax //eax increases by 1
cmp eax, [anItem] //value of eax compared to 'anItem'
jne starLoop //if eax doesn't equal value of 'anItem', starLoop starts again
jmp finish
printNewLine:
push '\r' // Two lines to print a character.
call printChar
push '\n' // Two lines to print another character.
call printChar
ret // And back to <wherever we came from>
finish : // Do nothing
}
printf("press enter to quit\n");
char dummy[10]; //Just in case several keys in buffer
scanf("%c", dummy); //pause.
scanf("%c", dummy); //pause again.
}
void printChar(char c)
{
printf("%c", c); //%c means as a char
}
void printStr(char *strAddr)
{
printf("%s", strAddr);
}
void printInt(int someInt)
{
printf("%d", someInt);
}
推荐答案
mov esi, [items] //esi points to start of 'numItems' ('items' points to address of 'numItems')从什么时候开始？
看看你的代码。 项目在哪里设置为 numitmes ？
严肃地说，如果你想学习汇编程序，那么我强烈建议你完全抛弃c函数并使用纯汇编程序解决方案。我还建议您在考虑学习汇编程序之前需要更好的编程背景，因为您的问题历史记录并未显示您正在考虑自己的代码：您提出的问题是任何具有合理编码背景的人都应该能够解决，即使他们不知道汇编程序！
Since when?
Look at your code. Where does items get set to numitmes?
In all seriousness, if you want to learn assembler, then I'd strongly suggest you dump the c stuff completely and use a "pure" assembler solution. I'd also suggest that you need a much better programming background before you even think of learning assembler as your question history does not show that you are thinking about your own code: you are asking questions that anyone with a reasonable background in coding should be able to work out, even if they don't know assembler!
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