c++中基本语法问题

的输出是？

    答案：构造函数的初始化列表

字符串转化为整形的代码：

enum Status{ kValid = 0,kInvalid };
int g_nStatus = kValid;
int StrToInt(const char* str)
{
g_nStatus = kInvalid;
long long num = 0;
if (str != NULL&&*str != ‘\0‘)
{
bool minus = false;
if (*str == ‘+‘)
str++;
else if (*str == ‘-‘)
{
str++;
minus = true;
}
if (*str != ‘\0‘)
{
num = StrToIntCore(str, minus);
}
}
return (int)num;
}
long long StrToIntCore(const char* digit, bool minus)
{
long long num = 0;
while (*digit！ = ‘\0‘)
{
if (*digit >= ‘0‘&&*digit <= ‘9‘)
{
int flag = minus ? -1 : 1;
num = num * 10 + flag*(*digit - ‘0‘);
if ((!minus&&num > 0x7FFFFFFF) || (minus&&num < (signed int)0x80000000))
{
num = 0;
break;
}
digit++;
}
else
{
num = 0;
break;
}
}
if (*digit == ‘\0‘)
{
g_nStatus = kValid;
}
return num;
}

c++中基本语法问题

原文：http://blog.csdn.net/wangfengfan1/article/details/46848751