如何解析根据C语言环境设置格式化的双精度

std::replace(sSource.begin(),sSource.end(),getDecimalSeparator(),'.');

但是,我没有发现另一个问题,也没有找到解决方法.如果值为负且千位分隔符是一个点(“.”),则strtod返回0并且_EndPtr指向字符串的开头：

// PRECONDITIONS:
//  * digit grouping symbol (thousands separator) = "."
//  * decimal symbol                              = ","
//  * digital grouping                            = "123.456.789"
//  * negative sign symbol                        = "-"
//  * negative number format                      = "- 1,1"
//  * OS WinXP SP2
//  * the rest doesn't matter

double parse(const char* cszValue,char** szStop)
{
    // NOTE: error handling code has been removed to simplify the sample
    return strtod(sSource.c_str(),szStop);
}

//...
char* szStop = NULL;
double dVal = 0.0;
dVal = parse("123.45",&szStop); // works!

// All the next DON'T WORK!
dVal = parse("123,45",&szStop); // dVal == 123.0 szStop == ",45"
dVal = parse("- 123.45",&szStop); // dVal == 0.0   szStop == "- 123.45"
// the same for "- 123,45"
dVal = parse("1.123.45",&szStop); // dVal == 1.123 szStop == ".45"
// the same for "1.123,45"
dVal = parse("1 123.45",&szStop); // dVal == 1     szStop == " 123.45"
// the same for "1 123,45"
dVal = parse("- 1 123,&szStop); // dVal == 0     szStop == start of the string
// the same for "- 1 123.45","- 1.123,"- 1.123.45"

有问题：

>我做错了什么？ (回答)
>如果根据本地设置格式化小数分隔符,为什么strtod会失败？ (回答)
>即使我用一个点(“.”)替换当前的小数点分隔符并删除所有千位分隔符,如何解析一个负值？检测否定唱,删除它,将值解析为正数并在此之后反转负号？