在C ++中两个整数的乘法 [英] Multiplication of two integers in C++

问题描述

我有一个很基本的问题，但我不知道我是否理解这个概念。假设我们有：

I have a pretty basic question, but I am not sure if I understand the concept or not. Suppose we have:

int a = 1000000;
int b = 1000000;
long long c = a * b;

当我运行这个时，c显示负值，所以我改变了a和b为long long，然后一切都很好。所以为什么我必须改变a和b，当他们的值在int的范围，并且他们的产品被分配到c（长long）？

When I run this, c shows negative value, so I changed also a and b to long long and then everything was fine. So why do I have to change a and b, when their values are in range of int and their product is assigned to c (which is long long)?

使用C / C ++

I am using C/C++

推荐答案

int 不会提升为 long long 在乘法之前，它们仍然保留 int s和产品。然后产品投放到长期，但太晚了，溢出已经触发。

The ints are not promoted to long long before multiplication, they remain ints and the product as well. Then the product is cast to long long, but too late, overflow has struck.

c $ c> a 或 b long long 将被晋升。

Having one of a or b long long should work as well, as the other would be promoted.

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