Boost lambda 困惑
为什么回调只调用一次？
bool callback()
{
static bool res = false;
res = !res;
return res;
}
int main(int argc, char* argv[])
{
vector<int> x(10);
bool result=false;
for_each(x.begin(),x.end(),var(result)=var(result)||bind(callback));
return 0;
}
Why is callback called once only?
bool callback()
{
static bool res = false;
res = !res;
return res;
}
int main(int argc, char* argv[])
{
vector<int> x(10);
bool result=false;
for_each(x.begin(),x.end(),var(result)=var(result)||bind(callback));
return 0;
}
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评论（2）
||表达式短路 第一次bind返回true后。第一次评估
bind时会被调用，因为这是确定false || 值的唯一方法。 绑定（...）。 由于bind(...)返回true，因此result设置为true。每当你说
...
bind(...)表达式不会被计算，因为它返回什么并不重要; 表达式true || Anything始终为true，并且||表达式 短路。确保始终调用
bind的一种方法是将其移动到||的左侧，或者将||更改为&&，具体取决于您想要通过result实现的目标。The
||expression short circuits after the first timebindreturnstrue.The first time you evaluate
bindis called, because that's the only way to determine the value offalse || bind(...). Becausebind(...)returnstrue,resultis set totrue.Every other time you say
... the
bind(...)expression isn't evaluated, because it doesn't matter what it returns; the expressiontrue || anythingis alwaystrue, and the||expression short circuits.One way to ensure that
bindis always called would be to move it to the left side of the||, or change the||to an&&, depending on what you are trying to accomplish withresult.在您的特定示例中，Boost.Lambda 并没有真正为您带来任何好处。 去掉 lambda 部分，也许你会更清楚地看到发生了什么：
这仍然依赖于你知道
||运算符是短路的，如 丹尼尔解释了。In your particular example, Boost.Lambda doesn't really gain you anything. Get rid of the lambda parts, and maybe you'll see more clearly what's going on:
This still relies on you to know that the
||operator is short-circuited, as Daniel explained.