[英]C# - Index was out of range
我试图将C ++类转换为C#,并在此过程中学习一些C ++。 我以前从未遇到过vector <>,而我的理解是,这就像C#中的List <>函数一样。 在类转换期间,我使用List futures_price = New List(Convert.ToInt32(no_steps)+ 1);重新编写了代码。 一旦运行代码,就会出现“索引超出范围”错误。
看了SOF之后,我相信问题在于与此相关的参数不在索引范围内,但是我看不到使用以下代码来解决此问题的简单解决方案。
特别是,这是触发错误的行:futures_prices [0] = spot_price * Math.Pow(d,no_steps);
下面是完整的代码:
public double futures_option_price_call_american_binomial(double spot_price, double option_strike, double r, double sigma, double time, double no_steps)
{
//double spot_price, // price futures contract
//double option_strike, // exercise price
//double r, // interest rate
//double sigma, // volatility
//double time, // time to maturity
//int no_steps
List<double> futures_prices = new List<double>(Convert.ToInt32(no_steps) + 1);
//(no_steps+1);
//double call_values = (no_steps+1);
List<double> call_values = new List<double>(Convert.ToInt32(no_steps) + 1);
double t_delta = time/no_steps;
double Rinv = Math.Exp(-r*(t_delta));
double u = Math.Exp(sigma * Math.Sqrt(t_delta));
double d = 1.0/u;
double uu= u*u;
double pUp = (1-d)/(u-d); // note how probability is calculated
double pDown = 1.0 - pUp;
futures_prices[0] = spot_price * Math.Pow(d, no_steps);
int i;
for (i=1; i<=no_steps; ++i) futures_prices[i] = uu*futures_prices[i-1]; // terminal tree nodes
for (i=0; i<=no_steps; ++i) call_values[i] = Math.Max(0.0, (futures_prices[i]-option_strike));
for (int step = Convert.ToInt32(no_steps) - 1; step >= 0; --step)
{
for (i = 0; i <= step; ++i)
{
futures_prices[i] = d * futures_prices[i + 1];
call_values[i] = (pDown * call_values[i] + pUp * call_values[i + 1]) * Rinv;
call_values[i] = Math.Max(call_values[i], futures_prices[i] - option_strike); // check for exercise
};
};
return call_values[0];
}
这是C ++的原始源代码:
double futures_option_price_call_american_binomial(const double& F, // price futures contract
const double& K, // exercise price
const double& r, // interest rate
const double& sigma, // volatility
const double& time, // time to maturity
const int& no_steps) { // number of steps
vector<double> futures_prices(no_steps+1);
vector<double> call_values (no_steps+1);
double t_delta= time/no_steps;
double Rinv = exp(-r*(t_delta));
double u = exp(sigma*sqrt(t_delta));
double d = 1.0/u;
double uu= u*u;
double pUp = (1-d)/(u-d); // note how probability is calculated
double pDown = 1.0 - pUp;
futures_prices[0] = F*pow(d, no_steps);
int i;
for (i=1; i<=no_steps; ++i) futures_prices[i] = uu*futures_prices[i-1]; // terminal tree nodes
for (i=0; i<=no_steps; ++i) call_values[i] = max(0.0, (futures_prices[i]-K));
for (int step=no_steps-1; step>=0; --step) {
for (i=0; i<=step; ++i) {
futures_prices[i] = d*futures_prices[i+1];
call_values[i] = (pDown*call_values[i]+pUp*call_values[i+1])*Rinv;
call_values[i] = max(call_values[i], futures_prices[i]-K); // check for exercise
};
};
return call_values[0];
};
List<double>
开始为空,直到您向其中添加项目。 (传递构造函数参数只是设置容量,以防止昂贵的调整大小)
在您Add()
之前,您无法访问[0]
。
要按原样使用它,请改用数组。
正如SLaks所说,在这种情况下最好使用数组。 C#列表填充有Add方法,而值则通过Remove方法删除...这将更加复杂,内存/性能昂贵,因为您还要替换值。
public Double FuturesOptionPriceCallAmericanBinomial(Double spotPrice, Double optionStrike, Double r, Double sigma, Double time, Double steps)
{
// Avoid calling Convert multiple times as it can be quite performance expensive.
Int32 stepsInteger = Convert.ToInt32(steps);
Double[] futurePrices = new Double[(stepsInteger + 1)];
Double[] callValues = new Double[(stepsInteger + 1)];
Double tDelta = time / steps;
Double rInv = Math.Exp(-r * (tDelta));
Double u = Math.Exp(sigma * Math.Sqrt(tDelta));
Double d = 1.0 / u;
Double uu = u * u;
Double pUp = (1 - d) / (u - d);
Double pDown = 1.0 - pUp;
futurePrices[0] = spotPrice * Math.Pow(d, steps);
for (Int32 i = 1; i <= steps; ++i)
futurePrices[i] = uu * futurePrices[(i - 1)];
for (Int32 i = 0; i <= steps; ++i)
callValues[i] = Math.Max(0.0, (futurePrices[i] - optionStrike));
for (Int32 step = stepsInteger - 1; step >= 0; --step)
{
for (Int32 i = 0; i <= step; ++i)
{
futurePrices[i] = d * futurePrices[(i + 1)];
callValues[i] = ((pDown * callValues[i]) + (pUp * callValues[i + 1])) * rInv;
callValues[i] = Math.Max(callValues[i], (futurePrices[i] - option_strike));
}
}
return callValues[0];
}
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