在opencv中未声明“ true”(此功能首次使用)

shr*_*eya 2 c opencv

可能重复:
在C中使用布尔值

我是C的新手,想编写一个程序来检测网络摄像头中的人脸,我上线了,我在Eclipse CDT上使用opencv-2.4.3,我在网上搜索了解决方法,但没有找到合适的解决方案解决我的问题的方法,因此将其发布为新问题。以下是代码:

 // Include header files
 #include "/home/OpenCV-2.4.3/include/opencv/cv.h"
 #include "/home/OpenCV-2.4.3/include/opencv/highgui.h"
 #include "stdafx.h"

 int main(){

//initialize to load the video stream, find the device
 CvCapture *capture = cvCaptureFromCAM( 0 );
if (!capture) return 1;

//create a window
cvNamedWindow("BLINK",1);

 while (true){
    //grab each frame sequentially
    IplImage* frame = cvQueryFrame( capture );
    if (!frame) break;

    //show the retrived frame in the window
    cvShowImage("BLINK", frame);

    //wait for 20 ms
    int c = cvWaitKey(20);

    //exit the loop if user press "Esc" key
    if((char)c == 27 )break;
}
 //destroy the opened window
cvDestroyWindow("BLINK");

//release memory
cvReleaseCapture(&capture);
return 0;
 }
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而且我得到的错误是“未声明为真”(此函数的首次使用),它在while循环中引起问题,我读到使用while(true)并不是一种好习惯,但我应该怎么做。谁能帮我。

LSe*_*rni 5

用例如替换

while(1)
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要么

for(;;)
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或者你可以做(c在循环之前定义):

while (c != 27)
{
    //grab each frame sequentially
    IplImage* frame = cvQueryFrame( capture );
    if (!frame)
        break;
    //show the retrieved frame in the window
    cvShowImage("BLINK", frame);
    //wait for 20 ms
    c = cvWaitKey(20);
    //exit the loop if user press "Esc" key
}
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c根本没有,但这将以20毫秒的等待时间开始循环:

while (cvWaitKey(20) != 27)
{
    //grab each frame sequentially
    IplImage* frame = cvQueryFrame( capture );
    if (!frame)
        break;
    //show the retrieved frame in the window
    cvShowImage("BLINK", frame);
}
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第三种可能性:

for(;;)
{
    //grab each frame sequentially
    IplImage* frame = cvQueryFrame( capture );
    if (!frame)
        break;
    //show the retrieved frame in the window
    cvShowImage("BLINK", frame);
    if (cvWaitKey(20) == 27)
        break;
}
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更新:同时想知道定义是否更正确

#define true  1
#define false 0
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要么

#define true 1
#define false (!true)
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或再次

#define false 0
#define true  (!false)
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因为如果我说:

int a = 5;
if (a == true) { // This is false. a is 5 and not 1. So a is not true }
if (a == false){ // This too is false. So a is not false              }
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我会得出一个非常奇怪的结果,我发现此链接与一个稍微奇怪的结果有关。

我怀疑以安全的方式解决此问题将需要一些宏,例如

#define IS_FALSE(a)  (0 == (a))
#define IS_TRUE(a)   (!IS_FALSE(a))
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