将值分配给结构数组答案

【问题标题】：Assigning values to a struct array将值分配给结构数组
【发布时间】：2020-11-22 12:19:06
【问题描述】：

我正在尝试将值分配给结构数组。但是，我收到“预期表达式”错误。有什么我想念的吗？我正在使用 Xcode 以防万一。

#include <stdio.h>
#include <stdlib.h>

struct MEASUREMENT
{
    float relativeHumidity;
    float temperature;
    char timestamp[20];
};

int main()
{
    struct MEASUREMENT measurements[5];
    
    measurements[0] = {0.85, 23.5, "23.07.2019 08:00"}; //Expected expression error
    measurements[1] = {0.71, 19.0, "04.08.2019 10:21"}; //Expected expression error
    measurements[2] = {0.43, 10.2, "07.08.2019 02.00"}; //Expected expression error
    measurements[3] = {0.51, 14.3, "20.08.2019 14:45"}; //Expected expression error
    measurements[4] = {0.62, 10.9, "01.09.2019 01:00"}; //Expected expression error

谢谢！

【问题讨论】：

标签： c

【解决方案1】：

在大括号中列出值是声明中初始化的一种特殊语法。大括号中的列表本身并不构成可用于赋值的表达式。

你可以在定义数组时以这种形式提供初始值：

struct MEASUREMENT measurements[5] = {
        {0.85, 23.5, "23.07.2019 08:00"},
        {0.71, 19.0, "04.08.2019 10:21"},
        {0.43, 10.2, "07.08.2019 02.00"},
        {0.51, 14.3, "20.08.2019 14:45"},
        {0.62, 10.9, "01.09.2019 01:00"},
    };

在表达式中，您可以使用 复合文字 定义一个临时对象，然后将其值分配给另一个对象。复合文字由括号中的类型组成，后跟用大括号括起来的初始化器列表：

measurements[0] = (struct MEASUREMENT) {0.85, 23.5, "23.07.2019 08:00"};
measurements[1] = (struct MEASUREMENT) {0.71, 19.0, "04.08.2019 10:21"};
measurements[2] = (struct MEASUREMENT) {0.43, 10.2, "07.08.2019 02.00"};
measurements[3] = (struct MEASUREMENT) {0.51, 14.3, "20.08.2019 14:45"};
measurements[4] = (struct MEASUREMENT) {0.62, 10.9, "01.09.2019 01:00"};

【讨论】：