我应该使用一个程序(一次一个char)读取GPS日志文件并解析它。当我使用一些.log文件时,我得到正常的响应,而不是与其他人(特别是不长的)文件。 抓取信息并存储它的代码如下:
class ReadInput{
public:
string massiveString;
};
void ReadInput::assign(string inMassiveString){ //assign the string passed to it to the object
this->massiveString.empty(); //clear whatever might be in the string
this->massiveString=inMassiveString; //copy the string into where it belongs.
}
int main(int argc, const char * argv[]) {
ifstream inputStream;
inputStream.open("gps3.log");
int i = 1;
char temp;
string tempString;
ReadInput *sentence[258];
//this is the part that grabs the characters
while(1){ //
if (!inputStream.eof()){ //
inputStream.get(temp);
if (temp!=' ') {
tempString.append(&temp); //this appends the characters to a temporary string
// cout << i << "\t" << tempString << endl; //testing the input
}
if (temp == '\n') { //if this is the end of the line, pass it on.
sentence[i]=new ReadInput;
sentence[i]->assign(tempString); //the part that copies the temporary string to the object
cout << i << "\t" << sentence[i]->massiveString << endl; //another testing line
i++;
tempString.clear();
}
}
else{
break;
}
}
}
我逐字逐句地阅读它是非常重要的。 应该阅读的内容如下所示: $ GPGGA,155008.000,3732.7239,N,07726.9956,W,1,05,1.7,89.0,男,-33.6,男,, 0000 * 53 这是一个有问题的阅读的例子。格式为 entry#,stringStoredInEntry。
1 $GPGGA,155002.000,3732.7239,N,07726.9956,W,1,05,1.7,89.1,M,-33.6,M,,0000*58
2 $GPGSA,A,3,17,06,28,02,24,,,,,,,,3.2,1.7,2.7*3E
3 $GPRMC,155002.000,A,3732.7239,N,07726.9956,W,0.00,80.30,141014,,,A*48
4 $GPGGA,155003.000,3732.7239,N,07726.9956,W,1,05,1.7,89.1,M,-33.6,M,,0000*59
5 $GPGSA,A,3,17,06,28,02,24,,,,,,,,3.2,1.7,2.7*3E
6 $GPGSV,3,1,12,17,64,038,42,06,63,250,43,28,44,150,45,02,25,239,38*7B
7 $GPGSV,3,2,12,24,17,294,22,10,21,176,22,12,25,316,23,51,36,223,37*71
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29 $GPGSA,A,3,17,06,28,02,24,,,,,,,,3.2,1.7,2.7*3E
30 $GPRMC,155009.000,A,3732.7239,N,07726.9956,W,0.00,80.30,141014,,,A*43
31 $GPGGA,155010.000,3732.7239,N,07726.9956,W,1,05,1.7,89.0,M,-33.6,M,,0000*5A
2 $ G P G S A , A , 3 , 1 7 , 0 6 , 2 8 , 0 2 , 2 4 , , , , , , , , 3 . 2 , 1 . 7 , 2 . 7 * 3 E
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"34 $"G"P"G"G"A","1"5"5"0"1"1"."0"0"0","3"7"3"2"."7"2"3"9","N","0"7"7"2"6"."9"9"5"6","W","1","0"5","1"."7","8"9"."0","M","-"3"3"."6","M",","0"0"0"0"*"5"B"
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)41 $)G)P)G)S)A),)A),)3),)1)7),)0)6),)2)8),)0)2),)2)4),),),),),),),),)3).)2),)1).)7),)2).)7)*)3)E)
*42 $*G*P*G*S*V*,*3*,*1*,*1*2*,*1*7*,*6*4*,*0*3*8*,*4*1*,*0*6*,*6*3*,*2*5*0*,*4*2*,*2*8*,*4*4*,*1*5*0*,*4*3*,*0*2*,*2*5*,*2*3*9*,*3*8***7*F*
+43 $+G+P+G+S+V+,+3+,+2+,+1+2+,+2+4+,+1+7+,+2+9+4+,+2+3+,+1+0+,+2+1+,+1+7+6+,+2+2+,+1+2+,+2+5+,+3+1+6+,+2+4+,+5+1+,+3+6+,+2+2+3+,+3+5+*+7+5+
,44 $,G,P,G,S,V,,,3,,,3,,,1,2,,,2,0,,,3,4,,,0,6,2,,,,,0,1,,,0,7,,,0,5,1,,,,,0,4,,,0,6,,,0,5,2,,,,,3,2,,,0,2,,,0,3,3,,,*,7,F,
-45 $-G-P-R-M-C-,-1-5-5-0-1-3-.-0-0-0-,-A-,-3-7-3-2-.-7-2-3-9-,-N-,-0-7-7-2-6-.-9-9-5-6-,-W-,-0-.-0-0-,-8-0-.-3-0-,-1-4-1-0-1-4-,-,-,-A-*-4-8-
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A65 $AGAPAGASAAA,AAA,A3A,A1A7A,A0A6A,A2A8A,A0A2A,A,A,A,A,A,A,A,A,A4A.A8A,A2A.A3A,A4A.A3A*A3A0A
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B67 $GPGGA,155007.000,3732.7239,N,07726.9956,W,1,05,1.7,89.0,M,-33.6,M,,0000*5C
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21 $GPRMC,155007.000,A,3732.7239,N,07726.9956,W,0.00,80.30,141014,,,A*4D
22 $GPGSA,A,3,17,06,28,02,24,,,,,,,,3.2,1.7,2.7*3E,1.7,89.0,M,-33.6,M,,0000*53
23 $GPGSA,A,3,17,06,28,02,24,,,,,,,,3.2,1.7,2.7*3E
34 $GPRMC,155002.000,A,3732.7239,N,07726.9956,W,0.00,80.30,141014,,,A*48
25 $GPGSV,3,2,12,24,17,294,22,10,21,176,22,12,25,316,24,51,36,223,35*74
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1-7是完美的。大。
然后8变怪了。
它永远不会正常。
这是从哪里来的?这是.log文件的问题吗?这是我的代码吗?这个废话在哪里蔓延?更重要的是,它如何搞乱左边的整数?如何能够始终如一地制作一个无意义的整数?
答案 0 :(得分:0)
使用tempString.push_back(temp); 而不是tempString.append(&amp; temp); 修复一切。