三元表达式语句不起作用 [英] C ternary expression-statement not working

问题描述

#include <stdio.h>
int main()
{
  int a=3,b=4,g;
  a > b ? g = a : g = b ;
  printf("%d",g);
  return 0;
}

为什么 g 无法打印？编译器说需要 lvalue 。是什么意思？

Why does the value of g not get printed? And compiler says lvalue required. What does it mean?

推荐答案

由于运算符的优先级更高？：在 = 上，表达式

Due to higher precedence of operator ?: over =, the expression

a > b ? g = a : g = b;   

将被解析为

(a > b ? g = a : g) = b;  

表达式（a> b？g = a：g）将给出一个右值。赋值操作符（ = ）的左操作数必须是左值 ¹ （可修改的 ² ）。

The expression (a > b ? g = a : g) will give an rvalue. The left operand of assignment (=) operator must be an lvalue¹ (modifiable²).

更改

a > b ? g = a : g = b ;  

至

a > b ? (g = a) : (g = b);    

或

g = a > b ? a : b;

---

<sub>1 。 C11-§6.5.16/ 2：赋值运算符的左操作数应具有可修改的左值。

2 。 §6.3.2.1/ 1： 左值是一个表达式（对象类型不是void）可能指定一个对象； ^64）如果左值在评估时未指定对象，则该行为未定义。当说一个对象具有特定类型时，该类型由用于指定该对象的左值指定。 可修改的左值是不具有数组类型，不完整的类型，不具有const限定类型的左值，并且如果是结构或联合，则不具有任何成员（包括递归） ，所有包含的集合或联合的任何成员或元素），其类型为const限定类型。。</sub>

---

<sub>1. C11-§6.5.16/2: An assignment operator shall have a modifiable lvalue as its left operand.
2. §6.3.2.1/1: An lvalue is an expression (with an object type other than void) that potentially designates an object;⁶⁴⁾ if an lvalue does not designate an object when it is evaluated, the behavior is undefined. When an object is said to have a particular type, the type is specified by the lvalue used to designate the object. A modifiable lvalue is an lvalue that does not have array type, does not have an incomplete type, does not have a const- qualified type, and if it is a structure or union, does not have any member (including, recursively, any member or element of all contained aggregates or unions) with a const- qualified type.</sub>

这篇关于三元表达式语句不起作用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！