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标题:二次的密码验证机制
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ianlin1024
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已结贴  问题点数:18 回复次数:1 
二次的密码验证机制
又来发第二次帖了,这次遇到的问题是临时想到的
我想在原有程式的基础下再添加一个二次输入验证的机制(就有点像在注册帐号时的"请再输入一次密码"的感觉)
但因为我在先前有写一段纪录按键延迟的阵列,想把延迟也当作验证的一部份

大致流程
1.照"节奏"第一次输入密码时:显示输入的数(矩阵记录下密码与间隔时间),按下回车跳至第二次的验证
2.第二次验证密码:第二次验证需重复第一次的密码与节奏(考虑到人的节奏感,希望在第一次的节奏基础上有+-的容错时间差)
3.如果符合则跳至 Keys pressed/ intervals between key presses那一段,如不符合需再重复输入(希望至多5次)

麻烦各位老师了!

程式碼如下:
程序代码:
#include <stdio.h>
#include <windows.h>

#define MAX_KEYS 50
#define MAX_INTERVALS (MAX_KEYS - 1) 

int main()
{
    INPUT_RECORD input_record;
    HANDLE console = GetStdHandle(STD_INPUT_HANDLE);
    DWORD events_read; 
    LARGE_INTEGER start_time, end_time, frequency;
    BOOL last_event_was_release = FALSE;
    LARGE_INTEGER last_release_time;
    int key_count = 0;
    char keys[MAX_KEYS] = {0};
    char password[50];
    char verify_password[50];
    double intervals[MAX_INTERVALS - 1] = {0}; 
    
    QueryPerformanceFrequency(&frequency);

    
    printf("請設定節拍密碼\n\n");
    while (1) {
        
    
        if (ReadConsoleInput(console, &input_record, 1, &events_read)) {
            if (input_record.EventType == KEY_EVENT) {
                if (input_record.Event.KeyEvent.bKeyDown) {
                    if (last_event_was_release && input_record.Event.KeyEvent.wVirtualKeyCode != VK_RETURN) {
                        LARGE_INTEGER press_time;
                        QueryPerformanceCounter(&press_time);
                        double interval = (double)(press_time.QuadPart - last_release_time.QuadPart) / frequency.QuadPart * 1.0;
                        intervals[key_count - 1] = interval;
                        printf("Time since last release: %.3fs\n", (double)(press_time.QuadPart - last_release_time.QuadPart) / frequency.QuadPart * 1.0);
                    }
                    QueryPerformanceCounter(&start_time);
                    last_event_was_release = FALSE;
                     
                    if (key_count < MAX_KEYS) {
                        keys[key_count++] = input_record.Event.KeyEvent.uChar.AsciiChar;
                    }
                } else {
                    LARGE_INTEGER release_time;
                    QueryPerformanceCounter(&release_time);
                    last_release_time = release_time;
                    last_event_was_release = TRUE;
                    if (input_record.Event.KeyEvent.uChar.AsciiChar != '\0') {
                        printf("'%c' key pressed for %.3f ms\n", input_record.Event.KeyEvent.uChar.AsciiChar,
                        (double)(release_time.QuadPart - start_time.QuadPart) / frequency.QuadPart * 1.0);
                    } else {
                        printf("Key pressed for %.3f s\n", (release_time.QuadPart - start_time.QuadPart) / frequency.QuadPart * 1.0);
                    }
                }
            } 
        }

        if (GetAsyncKeyState(VK_RETURN) & 1) {
            
            printf("\n==============================================================================\n");
            printf("\n Keys pressed ");
            for (int i = 0; i < key_count; i++) {
                printf("%c ", keys[i]);
            }
            printf("\n");
            
            // Output the intervals between key presses
            printf(" intervals between key presses: ");
            for (int i = 0; i < key_count - 2; i++) {
                printf("%.3fs ", intervals[i]);
            }
            printf("\n");
            
            break;
        }
    }

    return 0;
} 
搜索更多相关主题的帖子: printf 密码 验证 Event key 
2023-03-26 15:56
rjsp
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「容错时间差」当然是你自己定义了

程序代码:
#include <stdio.h>
#include <math.h>
#include <windows.h>

void InputPassword( size_t maxcount, char keys[], double timepoints[] )
{
    LARGE_INTEGER frequency, start_time;
    QueryPerformanceFrequency( &frequency );
    start_time.QuadPart = 0;

    bool keys_down[128] = { false }; // 记录 char 0-127 的按键状态
    HANDLE console = GetStdHandle(STD_INPUT_HANDLE);
    for( size_t i=0; i!=maxcount; )
    {
        INPUT_RECORD input_record;
        DWORD events_read;
        if( !ReadConsoleInput(console,&input_record,1,&events_read) || events_read!=1 || input_record.EventType!=KEY_EVENT ) // 忽略非按键信息
            continue;

        bool bKeyDown = input_record.Event.KeyEvent.bKeyDown!=0;
        char AsciiChar = input_record.Event.KeyEvent.uChar.AsciiChar;
        if( bKeyDown && AsciiChar=='\r' ) // 回车后不再记录
        {
            keys[i++] = 0;
            break;
        }
        if( AsciiChar<' ' || AsciiChar>=127 ) // 忽略无法映射的字符(对应着locale("C")下的isprint)
            continue;
        AsciiChar = (AsciiChar>='a' && AsciiChar<'z') ? (AsciiChar-'a'+'A') : AsciiChar; // 忽略大小写(否则要监控Shift键和Caps键,过于复杂)

        if( bKeyDown && !keys_down[AsciiChar] )
        {
            LARGE_INTEGER timepoint;
            QueryPerformanceCounter( &timepoint );
            if( i == 0 )
                start_time = timepoint;

            keys_down[AsciiChar] = true;
            keys[i] = AsciiChar;
            timepoints[i++] = (timepoint.QuadPart-start_time.QuadPart)/(frequency.QuadPart+0.0);
            putchar( '*' );
        }
        else if( !bKeyDown && keys_down[AsciiChar] )
        {
            LARGE_INTEGER timepoint;
            QueryPerformanceCounter( &timepoint );

            keys_down[AsciiChar] = false;
            keys[i] = AsciiChar;
            timepoints[i++] = (timepoint.QuadPart-start_time.QuadPart)/(frequency.QuadPart+0.0);
        }
    }
    putchar( '\n' );
}

void ShowPassword( size_t maxcount, char keys[], double timepoints[] )
{
    printf( "{\n" );
    bool keys_down[128] = { false };
    for( size_t i=0; i!=maxcount && keys[i]!=0; ++i )
    {
        keys_down[keys[i]] = !keys_down[keys[i]];
        printf( "    '%c' %-4s at %10.6fs\n", keys[i], (keys_down[keys[i]]?"Down":"Up"), timepoints[i] );
    }
    printf( "}\n" );
}

bool ShowDifference( size_t maxcount, char a_keys[], double a_timepoints[], char b_keys[], double b_timepoints[] )
{
    for( size_t i=0; i!=maxcount && a_keys[i]!=0; ++i )
    {
        if( a_keys[i] != b_keys[i] )
        {
            puts( "{ 按键顺序不一致 }" );
            return false;
        }
    }

    printf( "{\n" );
    double a_keys_down[128] = { 0 };
    double b_keys_down[128] = { 0 };
    for( size_t i=0; i!=maxcount && a_keys[i]!=0; ++i )
    {
        if( a_keys_down[a_keys[i]] == 0 )
        {
            a_keys_down[a_keys[i]] = a_timepoints[i] + 1.0;
            b_keys_down[b_keys[i]] = b_timepoints[i] + 1.0;
            if( i == 0 )
                printf( "    '%c' Down:\n", a_keys[i] );
            else
                printf( "    '%c' Down: 按键时刻相差 %10.6fs (%4.2f%%)\n", a_keys[i], fabs(a_timepoints[i]-b_timepoints[i]), fabs(a_timepoints[i]-b_timepoints[i])*100/a_timepoints[i] );
        }
        else
        {
            double d1 = fabs( a_timepoints[i]+1.0 - a_keys_down[a_keys[i]] );
            double d2 = fabs( b_timepoints[i]+1.0 - b_keys_down[b_keys[i]] );
            printf( "    '%c' Up  : 按键时长相差 %10.6fs (%4.2f%%)\n", a_keys[i], fabs(d1-d2), fabs(d1-d2)*100/d1 );
            a_keys_down[a_keys[i]] = 0;
            b_keys_down[b_keys[i]] = 0;
        }
    }
    printf( "}\n" );

    puts( "你认为这个偏差可以接受,就返回true;否则,返回false" );
    return true;
}

int main( void )
{
    char a_keys[50];
    double a_intervals[50];
    InputPassword( 50, a_keys, a_intervals );
    ShowPassword( 50, a_keys, a_intervals );

    char b_keys[50];
    double b_intervals[50];
    InputPassword( 50, b_keys, b_intervals );
    ShowPassword( 50, b_keys, b_intervals );

    ShowDifference( 50, a_keys, a_intervals, b_keys, b_intervals );
}


测试时,两次输入都是「按下A,按下B,按下C,松开B,松开A,按下回车键」,然后程序输出
***
{
    'A' Down at   0.000000s
    'B' Down at   0.503053s
    'C' Down at   1.165817s
    'B' Up   at   1.990602s
    'A' Up   at   2.470596s
}

***
{
    'A' Down at   0.000000s
    'B' Down at   0.529204s
    'C' Down at   0.952299s
    'B' Up   at   1.359857s
    'A' Up   at   1.832545s
}

{
    'A' Down:
    'B' Down: 按键时刻相差   0.026151s (5.20%)
    'C' Down: 按键时刻相差   0.213518s (18.31%)
    'B' Up  : 按键时长相差   0.656896s (44.16%)
    'A' Up  : 按键时长相差   0.638051s (25.83%)
}
你认为这个偏差可以接受,就返回true;否则,返回false
2023-03-27 12:33
快速回复:二次的密码验证机制
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