如果我给我的程序提供txt文件:
BB
MB
150 570 2
240 570 3
360 570 0
FB
E
T
它读取错误,而是将其读作
BB
150 0 0
240 570 2
360 570 3
0 570 0
MB
FB
E
T
以下是我用来阅读此内容的简化版本:
string one,two,three,four;
ifstream file;
filename+=".txt";//filename is a string
file.open(filename.c_str());
while (file >> one >> two>>three&&one!="MB")
{
//do stuff with it here
}
等等。 有人可以解释为什么两个和三个最初被设置为0?
完整版代码:
阅读:
void load(string filename)
{
string one,two,three,four;
ifstream file;
filename+=".txt";
file.open(filename.c_str());
//blocks
//backblock list
while (file >> one >> two>>three&&one!="MB")
{
backBlockList.push_back(
Block(atoi(two.c_str()),atoi(three.c_str()),atoi(one.c_str())));
}
while (file >> one >> two>>three&&one!="FB")
{
midBlockList.push_back(
Block(atoi(two.c_str()),atoi(three.c_str()),atoi(one.c_str())));
}
while (file >> one >> two>>three&&one!="E")
{
foreBlockList.push_back(
Block(atoi(two.c_str()),atoi(three.c_str()),atoi(one.c_str())));
}
while (file >> one &&one!="T")
{
enemyList.push_back(Enemy(atoi(one.c_str())));
//loads waypoints
while (file >> one>>two )
{
enemyList.at(enemyList.size()-1).addWaypoint(
atoi(one.c_str()),atoi(two.c_str()));
}
while(file>>one>>two>>three>>four)
{
textBlockList.push_back(
TextBlock(atoi(one.c_str()),atoi(two.c_str())));
textBlockList.at(
textBlockList.size()-1).setText(three);
textBlockList.at(
textBlockList.size()-1).setRange(atoi(four.c_str()));
}
}
}
写:
void printOut(string filename )
{
cout<<"printing "<<endl;
ofstream myfile;
filename+=".txt";
myfile.open (filename.c_str());
myfile << "BB\n";
//prints out blocks
cout<<"printing backblocks";
unsigned int i = 0;
for( i = 0; i<backBlockList.size(); i++)
{
backBlockList.at(i).print(myfile);
}
cout<<" printed "<<i<<endl;
cout<<"printing midblocks";
myfile << "MB\n";
for( i = 0; i<midBlockList.size(); i++)
{
midBlockList.at(i).print(myfile);
}
cout<<" printed "<<i<<endl;
cout<<"printing foreblocks";
myfile << "FB\n";
for( i = 0; i<foreBlockList.size(); i++)
{
foreBlockList.at(i).print(myfile);
}
cout<<" printed "<<i<<endl;
cout<<"printing enemies "<<endl;
myfile<<"E\n";
for( i =0; i<enemyList.size(); i++)
{
enemyList.at(i).print(myfile);
}
cout<<"printing text";
myfile<<"T\n";
for( i =0; i<textBlockList.size(); i++)
{
if(textBlockList.at(i).complete())
textBlockList.at(i).print(myfile);
}
cout<<" printed "<<i<<endl;
cout<<"printing other"<<endl;
//Additional stuff goes here EX BACKGROUND
myfile.close();
cout<<"printing done";
}
阻止写:
void Block::print(ofstream & file)
{
file << x;
file << " ";
file<< y;
file<< " ";
file<< Type;
file<< " \n";
}
TextBlock写道:
void TextBlock::print(ofstream & file)
{
file<< x;
file<<" ";
file<< y;
file<<" ";
file<< text;
file<<" ";
file<<range;
file<<" \n";
}
敌人写道:
void Enemy::print(ofstream & file)
{
file<<type;
for(unsigned int i =0; i<X.size()-1; i++)
{
file<<" ";
file<< X.at(i);
file<<" ";
file<< Y.at(i);
}
file<<"\n";
}
答案 0 :(得分:2)
我原以为它会把文件读成:
BB MB 150
570 2 240
570 3 360
570 0 FB
E T
因为它总是一直读三个字符串。如果您想要总是读取三个字符串,您可能希望用虚拟0
s来填充MB和BB指示符(例如MB 0 0
)。
可能有助于实现
在处理换行符时, cin >> a >> b >> c;
与cin >> a; cin >> b; cin >> c;
没什么区别。
答案 1 :(得分:1)
您获得的数字三倍的原因:
150 0 0
240 570 2
360 570 3
0 570 0
如下:您的输入仅从第一个循环中读取:
while (file >> one >> two>>three&&one!="MB")
如下:
Loop | one | two | three | atoi | atoi | atoi
| | | | one | two | three
----------------------------------------------
1 | BB | MB | 150 | 0 | 0 | 150
2 | 570 | 2 | 240 | 570 | 2 | 240
3 | 570 | 3 | 360 | 570 | 3 | 360
4 | 570 | 0 | FB | 570 | 0 | 0
5 | breaks the loop because three can't be read
表格中的最后三列是观察到的三元组数。
答案 2 :(得分:0)
该行
file >> one >> two>>three&&one!="FB"
从文件中读取三个字符串而不是一个。