我正在編寫一個程序來重置密碼,並且在最終(接受新密碼)條件產生之前,我有一些轉義程序的問題。我正在使用conio.h的getch()函數來接受將其寫入字符串的輸入(使用string.h)getch()等於回車到esc鍵
我使用Borland TurboC++ V4.0來保持兼容性,因爲這是附加代碼一箇舊的MS DOS程序。如果可以的話,我會使用更新的ide /編譯器。
這是程序的一小部分;我在他們的atm函數中省略了標題和新密碼/檢查密碼行。我對C編程非常陌生,我知道這可能不是最簡單的編寫方式,而且有很多冗餘。我曾嘗試創建一個退出該程序的標誌以及一箇中斷函數,並且它們都遇到了同樣的問題;他們將'\ r'視爲esc,反之亦然。
main()
{
char password[] = "0000000000000000000";
char newpassword[] = "0000000000000000000";
char checkpassword[] = "0000000000000000000";
char answer[] = "hrigsetup";
initgraph(&gd,&gm,"C://TC//BGI");
//draw screen
rectangle(0,0,screen_x,screen_y);
rectangle(screen_x/4,screen_y/3,(3*(screen_x/4)),(2*(screen_y/3)));
rectangle(((screen_x/4)+50),((screen_y/3)+45),(3*(screen_x/4)-50),((screen_y/3)+20));
rectangle(((screen_x/4)+50),((screen_y/3)+90),(3*(screen_x/4)-50),((screen_y/3)+65));
rectangle(((screen_x/4)+50),((screen_y/3)+135),(3*(screen_x/4)-50),((screen_y/3)+110));
//write text
setcolor(headercolor);
settextstyle(DEFAULT_FONT,HORIZ_DIR,2);
settextjustify(CENTER_TEXT,TOP_TEXT);
outtextxy(screen_x/2,((screen_y/4)),"Change Password");
setcolor(drawcolor);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
settextjustify(LEFT_TEXT,TOP_TEXT);
outtextxy((screen_x/2)-108,((screen_y/2)-70),"Enter Old Password");
outtextxy((screen_x/2)-108,((screen_y/2)-25),"Enter New Password");
outtextxy((screen_x/2)-108,((screen_y/2)+20),"Re-Enter New Password");
//enter old password
//password entry
while(unlock != 1 && esc_key != 1)
{
while((password[p] = getch()) != '\r') //mask text logic
{
if(password[p] == '\b') //tolerate a backspace
{
if(p == 17 && endchar == 0) //Logic password field
{ //bounds/overflow protection
password[p] = ' ';
setcolor(clear);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
settextjustify(LEFT_TEXT,TOP_TEXT);
outtextxy((screen_x/2)-(100-(p*8)),(screen_y/2)-50,"Û");
endchar = 1;
}
else
{
password[p] = ' ';
setcolor(clear);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
settextjustify(LEFT_TEXT,TOP_TEXT);
outtextxy((screen_x/2)-(100-((p-1)*8)),(screen_y/2)-50,"Û");
p--;
if(p <= -1)
p = 0;
}
}
else if(password[p] == 27) //esc logic
{ //have tried 27, 0x1b, '\027
break;
}
else
{
if(clear_bar == 1) //erase contents of password
{ //box
setcolor(clear);
settextstyle(DEFAULT_FONT,HORIZ_DIR,3);
outtextxy((screen_x/2)-2,(screen_y/2)-59,"ÛÛÛÛÛÛÛÛÛ");
setcolor(drawcolor);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
settextjustify(LEFT_TEXT,TOP_TEXT);
clear_bar = 0;
}
setcolor(drawcolor);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
settextjustify(LEFT_TEXT,TOP_TEXT);
outtextxy((screen_x/2)-(100-(p*8)),(screen_y/2)-50,"*");
if(p == 17)
{
p = 17;
endchar = 0;
}
else
p++;
}
}
if(strcmp(password, answer) !=0 && esc_key != 1) //check password = wrong,
{ //clear password box, fill
setcolor(headercolor); //read and write "incorrect"
settextjustify(CENTER_TEXT,TOP_TEXT); //in the box, write "Try Again"
settextstyle(DEFAULT_FONT,HORIZ_DIR,3); //below input box
outtextxy((screen_x/2)-2,(screen_y/2)-59,"ÛÛÛÛÛÛÛÛÛ");
setcolor(drawcolor);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
outtextxy(screen_x/2,(screen_y/2)-50,"Incorrect");
outtextxy(screen_x/2,(screen_y/2)+100,"Try Again");
unlock = 0;
clear_bar = 1;
newp_ent = 0;
password[p] = 0;
p = 0;
}
if(strcmp(password, answer) == 0 && esc_key != 1) //check password = correct
{ // clear and fill input box
setcolor(unlock_color); //green and display "Correct"
settextjustify(CENTER_TEXT,TOP_TEXT); //write "Press Any Key" below
settextstyle(DEFAULT_FONT,HORIZ_DIR,3); //the input box
outtextxy((screen_x/2)-2,(screen_y/2)-59,"ÛÛÛÛÛÛÛÛÛ");
setcolor(clear);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
outtextxy(screen_x/2,(screen_y/2)+100,"ÛÛÛÛÛÛÛÛÛ");
setcolor(drawcolor);
settextstyle(DEFAULT_FONT,HORIZ_DIR,1);
outtextxy(screen_x/2,(screen_y/2)-50,"Correct");
unlock = 1;
newp_ent = 0;
}
}
這條線,在貼代碼的行87:'p值= 17;'絕對沒有如現有'if'已經證明,「p」爲已經等於17 – user3629249
有很多的'魔術「號碼。強烈建議(當你執行維護時,你會理解這一點):所有這些數字都用有意義的名稱#定義,然後在整個代碼中使用那些有意義的名稱 – user3629249
爲了代碼的可讀性和清晰性,請始終縮進代碼。在每個大括號之後建議4個空格'{',並在每個大括號之前取消縮進'}'切勿使用製表符來縮進,因爲每個文字處理器/編輯器的製表符/製表符寬度設置不同。推薦4個空格作爲是可見的,即使在該代碼塊可變寬度的字體 – user3629249