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我必须将数组从一个汇编程序传递给一个C++函数。我最终想出了如何让两个文件交谈，现在我无法弄清楚我应该如何将MASM数组的地址传递给我的C++函数。我尝试在masm中使用ptr和addr来调用递归DFS。不知道我在做什么错误，因为我只在程序集和C++中进行了超过2个月的编程。 printSomething被调用的函数会显示出来，所以我知道这两个程序正在进行通信，但是，当DFS被调用时，我得到了一个deque迭代器而不可解引用，所以我不确定发生了什么，但它与使用我的堆栈的DFS相关DFS算法中的“路径”。我尝试使用int * hex_array []作为DFS函数中的参数，但它不是那样的。 DFS在数组中搜索1（红色搜索）的值，向每个访问的红色十六进制数加3。如果找不到路径，它会通过从每个“visited hex”中减去3来重置“visited”的hexes，并返回-1，如果找到了有效的路径，它将返回1，当整个程序使用C++时，DFS工作，所以它不是函数本身或堆栈的问题，尽管当我在VS2012中使用我的调试器时，我注意到每次堆栈= 1时都会失败，并且我将array_index设置为路径顶端，弹出堆栈并调用DFS与array_index。在那一点上，堆栈是空的，但我不明白为什么它在这里失败时，当它完美的工作，当整个程序在C++文件中，我猜测它与MASM有关阵列没有得到由++的运作方式与c访问应该???如何将MASM数组传递给C++函数
的我的汇编代码部分：
INCLUDE Irvine32.inc
printSomething PROTO C ;displays "Goobers"
DFS PROTO C, color:BYTE, bptr:PTR DWORD, arrayIndex:SDWORD
PDWORD TYPEDEF PTR DWORD
.data
ALIGN SDWORD
bptr PDWORD board
board SDWORD 121 DUP (0) ;array to hold the hex board
.code
main PROC
INVOKE printSomething ;test to see if MASM talking to C++ program
Start:
CALL PlaceRed ;prompt user to place a red stone
CALL ShowBoard ;redraw board to show update
;check if there is a valid path using C++ DFS
;What needs to be saved...? not aX since DFS will return -1 (no path) or 1(path) in aX from C++ function
PUSH EDX
PUSH EBX
PUSH ECX
;INVOKE DFS, 1, ADDR board, 0 ; color red, board pointer, arrayIndex 0
INVOKE DFS, 1, bptr, 0 ; color red, board pointer, arrayIndex 0
POP ECX
POP EBX
POP EDX
CMP AX,1 ;if Ax == 1 winning path found
JNE Continue ;Ax != 1 no valid path...continue game
MOV EDX, OFFSET redWins ; move "Red wins..." to eDx and display
CALL WriteString
JMP END_GAME
Continue:
CALL PlaceBlue ;place a blue stone
CALL ShowBoard ;redraw the board
;check if there is a valid path using C++ DFS
PUSH EDX
PUSH EBX
PUSH ECX
;INVOKE DFS, 2, ADDR board, 0; color blue (2), pointer, arrayIndex 0
INVOKE DFS, 2, bptr, 0; color blue (2), pointer, arrayIndex 0
POP ECX
POP EBX
POP EDX
CMP AX,1 ;if Ax == 1 winning path found
JNE Start ;Ax != 1 no valid path...continue game
MOV EDX, OFFSET blueWins ; move "Blue wins..." to eDx and display
CALL WriteString
END_GAME:
Retn
main ENDP
END main
和部分
#include "stdafx.h"
#include<iostream>
#include<stack>
#include "DFSAlgorithm.h"//include definition of class DFSAlgorithm
using namespace std;
//int *board[121];
int adjacency[6];
stack<int> path; //stack to hold the last hex visited
//test printsomething
extern "C" void printSomething(){
cout<<"goobers";
}
//First call of DFS always starts with array_index == 0
int DFS(int color, int hex_array[], int array_index){
//DFS code here...blah blah...
}
我的头文件
//DFSAlgorithm.h
//Definition of DFSAlgorithm class that does the DFS for the game of HEX
//Member functions are defined in DFSAlgorithm.ccp
#ifndef DFSAlgorithm_H
#define DFSAlgorithm_H
extern "C" void printSomething();
extern "C" int DFS(int color, int hex_array[], int array_index);
#endif
DFS代码是每费鲁乔要求全部
#include "stdafx.h"
#include<iostream>
#include<stack>
#include "DFSAlgorithm.h"//include definition of class DFSAlgorithm
using namespace std;
int adjacency[6];
//int hex_array[];
//int array_index;
extern stack<int> path; //stack to hold the last hex visited
//test printsomething
extern "C" void printSomething(){
cout<<"I'm not dead yet...";
}
//First call of DFS always starts with array_index == 0
extern "C" int DFS(int color, int hex_array[], int array_index){
if (hex_array[array_index] == color){ //if hex has an appropriately colored stone
hex_array[array_index] += 3; //mark the hex as visited
path.push(array_index); //push the hex onto the path stack
}
if ((color == 1 && array_index % 11 == 10 && hex_array[array_index] == 4) ||
(color == 2 && array_index/11 == 10 && hex_array[array_index] == 5)){
return 1; //winner base case==>reached the other side
}
//If a visited/unvisited hex has a stone of correct color==> search the adjacent hexes
if ((color == 1 && hex_array[array_index] == 4) ||
(color == 2 && hex_array[array_index] == 5)){
//get adjacencies
if(array_index == 0){//top left 2 corner
adjacency[ 0 ] = 1;
adjacency[ 1 ] = 11;
adjacency[ 2 ] = - 1;
adjacency[ 3 ] = - 1;
adjacency[ 4 ] = - 1;
adjacency[ 5 ] = - 1;
}
else if(array_index == 10){//top right three corner
adjacency[ 0 ] = 9;
adjacency[ 1 ] = 20;
adjacency[ 2 ] = 21;
adjacency[ 3 ] = - 1;
adjacency[ 4 ] = - 1;
adjacency[ 5 ] = - 1;
}
else if(array_index == 110){//bottom left corner
adjacency[ 0 ] = 99;
adjacency[ 1 ] = 100;
adjacency[ 2 ] = 111;
adjacency[ 3 ] = - 1;
adjacency[ 4 ] = - 1;
adjacency[ 5 ] = - 1;
}
else if(array_index==120){//bottom right corner
adjacency[ 0 ] = 109;
adjacency[ 1 ] = 119;
adjacency[ 2 ] = -1;
adjacency[ 3 ] = -1;
adjacency[ 4 ] = -1;
adjacency[ 5 ] = -1;
}
else if(array_index/11 == 0){//top row minus corners
adjacency[ 0 ] = array_index - 1;
adjacency[ 1 ] = array_index + 1;
adjacency[ 2 ] = array_index + 10;
adjacency[ 3 ] = array_index + 11;
adjacency[ 4 ] = - 1;
adjacency[ 5 ] = - 1;
}
else if(array_index % 11 == 0){//left column minus corners
adjacency[ 0 ] = array_index - 11;
adjacency[ 1 ] = array_index + 11;
adjacency[ 2 ] = array_index - 10;
adjacency[ 3 ] = array_index + 1;
adjacency[ 4 ] = - 1;
adjacency[ 5 ] = - 1;
}
else if (array_index/11 == 10){//row 10 minus corners
adjacency[ 0 ]= array_index - 1;
adjacency[ 1 ]= array_index + 1;
adjacency[ 2 ]= array_index - 11;
adjacency[ 3 ]= array_index - 10;
adjacency[ 4 ]= - 1;
adjacency[ 5 ]= - 1;
}
else if(array_index % 11 == 10){//right column minus corners
adjacency[ 0 ] = array_index - 11;
adjacency[ 1 ] = array_index + 11;
adjacency[ 2 ] = array_index - 1;
adjacency[ 3 ] = array_index + 10;
adjacency[ 4 ] = - 1;
adjacency[ 5 ] = - 1;
}
else{//all interior hexes
adjacency[ 0 ] = array_index - 11;
adjacency[ 1 ] = array_index + 11;
adjacency[ 2 ] = array_index - 10;
adjacency[ 3 ] = array_index + 10;
adjacency[ 4 ] = array_index - 1;
adjacency[ 5 ]= array_index + 1;
}
/*Initialize adjacentHexes count to zero: if == 0 after all 6 adjacencies are
checked it means it is a dead end as there are no unvisited adjacent hexes with
the correct color stone*/
int adjacentHexes = 0;
for(int b = 0; b < 6; b++){//traverse adjacency array of the passed in index
//if one of the adjacent hexes has a red/blue stone
if((color == 1 && hex_array[adjacency[b]] == color) ||
(color == 2 && hex_array[adjacency[b]] == color)){
adjacentHexes++; //increment the adjacentHexes count
hex_array[adjacency[b]] += 3; //mark the hex as visited
path.push(adjacency[b]); //push visited adjacent hex onto path
//recursively call DFS with that adjacent hex index
return DFS(color, hex_array,adjacency[b]);
}
}
//If adjacentHexes == 0 ==> dead-end
if(adjacentHexes == 0 && path.size() > 1){
path.pop();//pop the top hex from the stack if stack > 1
//recursive call of DFS with the new top red/blue hex
return DFS(color, hex_array,path.top());
}
if(adjacentHexes == 0 && path.size() == 1){//back to Row 0/Column 0
//make the array_index = the top of the path stack
array_index = path.top();//this is the line causing deque iterator not dereferenceable problems+++++++++++++++++++++++
//pop remaining element from the stack so path is now zero
path.pop();
}
}
//if checking for a red path and path is empty
if (color == 1){
//search remaining column 0 hexes for unvisited red hexes
for(array_index ; array_index <= 99;){
//recursively call DFS with next Column 0 hex
return DFS(color, hex_array, array_index + 11);
}
}
//if checking for a blue path and path is empty
if (color == 2){
//search remaining row 0 hexes for unvisted blue hexes
for(array_index ; array_index <= 9;){
//recursively call DFS with next Row 0 hex
return DFS(color, hex_array, array_index + 1);
}
}
//No path exists reset all visited hexes to unvisited
for(int a = 0; a < 121; a++){
if(hex_array[a] >= 4)//if hex has been visited
hex_array[a] -= 3;//remove visited designation
}
return -1;//return false as no path exists
}
嗯。不知道我会如何匹配 – TryChick
@TryChick - 它不应该是SDWORD吗？ – Ferruccio
好吧，你要说的是在颜色中使颜色参数成为SDWORD？ – TryChick