C ++中的变量初始化 [英] Variable initialization in C++

问题描述

我的理解是 int 变量会自动初始化为 0 然而，它不是。下面的代码打印一个随机值;任何人都可以告诉我什么规则，如果有的话，适用于初始化？具体来说，在什么条件下自动初始化变量？

My understanding is that the int variable will be initialized to 0 automatically; however, it's not. The code below prints a random value; can anybody tell me what rules, if any, apply to initialization? Specifically, under what conditions are variable initialized automatically?

int main () 
{   
    int a[10];
    int i;
    cout << i << endl;
    for(int i = 0; i< 10; i++)
        cout << a[i] << " ";
    return 0;
}

推荐答案

• 它是一个类/结构实例，默认构造函数初始化所有基本类型; MyClass实例;


• 您使用数组初始化语法，例如。 int a [10] = {} （全零）或 int a [10] = {1,2}; （ a [0] == 1 和 a [1] == 2 ）


• 同样适用于非聚合类/结构体，例如MyClass instance = {}; （有关此问题的详情，请访问这里）


• 它是一个全局/外部变量


• 定义变量 static



• it's a class/struct instance in which the default constructor initializes all primitive types; like MyClass instance;
• you use array initializer syntax, e.g. int a[10] = {} (all zeroed) or int a[10] = {1,2}; (all zeroed except the first two items: a[0] == 1 and a[1] == 2)
• same applies to non-aggregate classes/structs, e.g. MyClass instance = {}; (more information on this can be found here)
• it's a global/extern variable
• the variable is defined static (no matter if inside a function or in global/namespace scope) - thanks Jerry

不要信任一个简单类型（int，long，...）的变量被自动初始化！它可能发生在像C＃的语言，但不是在C& C ++。

Never trust on a variable of a plain type (int, long, ...) being automatically initialized! It might happen in languages like C#, but not in C & C++.

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