c++ - 隐式转换和调用函数

Question

当我调用一个将参数作为 int 的函数，同时传递一个 double 参数时，在 C++ 中隐式转换有什么用？为什么这不是编译时错误？允许这种情况发生的运行时堆栈中发生了什么？

#include <iostream>
void foo(int num);

int main() {
    double doub= 4.9;
    foo(doub);
    return 0;
}

void foo(int num){
    std::cout<<num<<std::endl;
}

Answer 1

这是所有 implicit conversions 的一长串 list C++ 做的。这个特定的包含在:

Floating–integral conversions A prvalue of floating-point type can be converted to a prvalue of any integer type. The fractional part is truncated, that is, the fractional part is discarded. If the value cannot fit into the destination type, the behavior is undefined (even when the destination type is unsigned, modulo arithmetic does not apply). If the destination type is bool, this is a boolean conversion (see below).